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Theorem 67.4 (in ZF). The following claims are equivalent:\n\n1. The Axiom of Well-Ordering\n\n2. Either \( A \preccurlyeq B \) or \( B \preccurlyeq A \), for any sets \( A \) and \( B \)
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Proof. (1) \( \Rightarrow \) (2). Fix \( A \) and \( B \) . Invoking (1), there are well-orderings \( \langle A, R\rangle \) and \( \langle B, S\rangle \) . Invoking Theorem 61.25, let \( f : \alpha \rightarrow \langle A, R\rangle \) and \( g : \beta \rightarrow \langle B, S\rangle \) be isomorphisms. By Trichotomy, either \( \alpha \in \beta \) or \( \alpha = \beta \) or \( \beta \in \alpha \) . In the first two cases \( \alpha \subseteq \beta \), so \( g \circ {f}^{-1} : A \rightarrow B \) is an injection, and hence \( A \preccurlyeq B \) . Similarly if \( \beta \in \alpha \) then \( B \preccurlyeq A \) .\n\n\( \left( 1\right) \Rightarrow \left( 2\right) \) . Fix \( A \) ; by Lemma 67.3 there is some ordinal \( \beta \) such that \( \beta \npreceq A \) . Invoking (2), we have \( A \preccurlyeq \beta \) . So there is some injection \( f : A \rightarrow \beta \), and we can use this injection to well-order the elements of \( A \), by defining an order \( \{ \langle a, b\rangle \in A \times A : f\left( a\right) \in f\left( b\right) \} .\)
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Yes
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Theorem 67.6 (in ZF). Well-Ordering and Choice are equivalent.
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Proof. Left-to-right. Let \( A \) be a set of sets. Then \( \bigcup A \) exists by the Axiom of Union, and so by Well-Ordering there is some \( < \) which well-orders \( \bigcup A \) . Now let \( f\left( x\right) = \) the \( < \) -least member of \( x \) . This is a choice function for \( A \) .\n\nRight-to-left. Fix \( A \) . By Choice, there is a choice function, \( f \), for \( \wp \left( A\right) \smallsetminus \{ \varnothing \} \) . Using Transfinite Recursion, define:\n\n\[ g\left( 0\right) = f\left( A\right) \]\n\n\[ g\left( \alpha \right) = \left\{ \begin{array}{ll} \text{ stop }! & \text{ if }A = g\left\lbrack \alpha \right\rbrack \\ f\left( {A \smallsetminus g\left\lbrack \alpha \right\rbrack }\right) & \text{ otherwise } \end{array}\right. \]\n\n(The indication to \
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Yes
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Lemma 67.7 (in \( {\mathbf{Z}}^{ - } \) ). Every finite set has a choice function.
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Proof. Let \( a = \left\{ {{b}_{1},\ldots ,{b}_{n}}\right\} \) . Suppose for simplicity that each \( {b}_{i} \neq \varnothing \) . So there are objects \( {c}_{1},\ldots ,{c}_{n} \) such that \( {c}_{1} \in {b}_{1},\ldots ,{c}_{n} \in {b}_{n} \) . Now by Proposition 60.5, the set \( \left\{ {\left\langle {{b}_{1},{c}_{1}}\right\rangle ,\ldots ,\left\langle {{b}_{n},{c}_{n}}\right\rangle }\right\} \) exists; and this is a choice function for \( a \) .
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Yes
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Theorem 67.8 (in \( {\mathbf{Z}}^{ - } \) + Countable Choice). For any \( A \), either \( \omega \preccurlyeq A \) or \( A \approx n \) for some \( n \in \omega \) .
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Proof. Suppose \( A ≉ n \) for all \( n \in \omega \) . Then in particular for each \( n < \omega \) there is subset \( {A}_{n} \subseteq A \) with exactly \( {2}^{n} \) elements. Using this sequence \( {A}_{0},{A}_{1},{A}_{2},\ldots \) , we define for each \( n \) :\n\n\[ \n{B}_{n} = {A}_{n} \smallsetminus \left( {{A}_{0} \cup {A}_{1} \cup \ldots \cup {A}_{n - 1}}\right) . \n\]\n\nNow note the following\n\n\[ \n\left| {{A}_{0} \cup {A}_{1} \cup \ldots \cup {A}_{n - 1}}\right| \leq \left| {A}_{0}\right| + \left| {A}_{1}\right| + \ldots + \left| {A}_{n - 1}\right| \n\]\n\n\[ \n= 1 + 2 + \ldots + {2}^{n - 1} \n\]\n\n\[ \n= {2}^{n} - 1 \n\]\n\n\[ \n< {2}^{n} = \left| {A}_{n}\right| \n\]\n\nHence each \( {B}_{n} \) has at least one member, \( {c}_{n} \) . Moreover, the \( {B}_{n}\mathrm{\;s} \) are pairwise disjoint; so if \( {c}_{n} = {c}_{m} \) then \( n = m \) . But every \( {c}_{n} \in A \) . So the function \( f\left( n\right) = {c}_{n} \) is an injection \( \omega \rightarrow A \) .
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Yes
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Theorem 67.9 (in \( {\mathbf{Z}}^{ - } \) + Countable Choice). If \( {A}_{n} \) is countable for each \( n \in \omega \) , then \( \mathop{\bigcup }\limits_{{n < \omega }}{A}_{n} \) is countable.
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Proof. Without loss of generality, suppose that each \( {A}_{n} \neq \varnothing \) . So for each \( n \in \omega \) there is a surjection \( {f}_{n} : \omega \rightarrow {A}_{n} \) . Define \( f : \omega \times \omega \rightarrow \mathop{\bigcup }\limits_{{n < \omega }}{A}_{n} \) by \( f\left( {m, n}\right) = \) \( {f}_{n}\left( m\right) \) . The result follows because \( \omega \times \omega \) is countable (Proposition 4.12) and \( f \) is a surjection.
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Yes
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Theorem 67.10 (in ZF). Choice is equivalent to the following principle. If the elements of \( A \) are disjoint and non-empty, then there is some \( C \) such that \( C \cap x \) is a singleton for every \( x \in A \) . (We call such a \( C \) a choice set for \( A \) .)
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The proof of this result is straightforward, and we leave it as an exercise for the reader.
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No
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Lemma 67.13. R forms an abelian group under composition of functions.
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Proof. Writing \( {0}_{R} \) for the rotation by 0 radians, this is an identity element for \( R \), since \( \rho \circ {0}_{R} = {0}_{R} \circ \rho = \rho \) for any \( \rho \in R \) .\n\nEvery element has an inverse. Where \( \rho \in R \) rotates by \( r \) radians, \( {\rho }^{-1} \in R \) rotates by \( {2\pi } - r \) radians, so that \( \rho \circ {\rho }^{-1} = {0}_{R} \) .\n\nComposition is associative: \( \left( {\tau \circ \sigma }\right) \circ \rho = \tau \circ \left( {\sigma \circ \rho }\right) \) for any \( \rho ,\sigma ,\tau \in R \)\n\nComposition is commutative: \( \sigma \circ \rho = \rho \circ \sigma \) for any \( \rho ,\sigma \in R \) .
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No
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Lemma 67.14. There is a partition of \( R \) into two disjoint sets, \( {R}_{1} \) and \( {R}_{2} \), both of which are a basis for \( R \) .
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Proof. Let \( {R}_{1} \) consist of the rotations by rational radian values in \( \lbrack 0,\pi ) \) ; let \( {R}_{2} = R \smallsetminus {R}_{1} \) . By elementary algebra, \( \left\{ {\rho \circ \rho : \rho \in {R}_{1}}\right\} = R \) . A similar result can be obtained for \( {R}_{2} \.
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No
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Lemma 67.15. \( \sim \) is an equivalence relation.
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Proof. Trivial, using Lemma 67.13.
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No
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Lemma 67.16. \( \mathbf{S} = \mathop{\bigcup }\limits_{{\rho \in R}}\rho \left\lbrack C\right\rbrack \) .
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Proof. Fix \( s \in \mathbf{S} \) ; there is some \( r \in C \) such that \( r \in {\left\lbrack s\right\rbrack }_{ \sim } \), i.e., \( r \sim s \), i.e., \( \rho \left( r\right) = s \) for some \( \rho \in R \) .
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Yes
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Lemma 67.17. If \( {\rho }_{1} \neq {\rho }_{2} \) then \( {\rho }_{1}\left\lbrack C\right\rbrack \cap {\rho }_{2}\left\lbrack C\right\rbrack = \varnothing \) .
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Proof. Suppose \( s \in {\rho }_{1}\left\lbrack C\right\rbrack \cap {\rho }_{2}\left\lbrack C\right\rbrack \) . So \( s = {\rho }_{1}\left( {r}_{1}\right) = {\rho }_{2}\left( {r}_{2}\right) \) for some \( {r}_{1},{r}_{2} \in C \) . Hence \( {\rho }_{2}^{-1}\left( {{\rho }_{1}\left( {r}_{1}\right) }\right) = {r}_{2} \), and \( {\rho }_{2}^{-1} \circ {\rho }_{1} \in R \), so \( {r}_{1} \sim {r}_{2} \) . So \( {r}_{1} = {r}_{2} \), as \( C \) selects exactly one member from each equivalence class under \( \sim \) . So \( s = {\rho }_{1}\left( {r}_{1}\right) = \) \( {\rho }_{2}\left( {r}_{1}\right) \), and hence \( {\rho }_{1} = {\rho }_{2} \) .
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Yes
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Lemma 67.18. There is a partition of \( \mathbf{S} \) into two disjoint sets, \( {D}_{1} \) and \( {D}_{2} \), such that \( {D}_{1} \) can be partitioned into countably many sets which can be rotated to form a copy of \( \mathbf{S} \) (and similarly for \( {D}_{2} \) ).
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Proof. Using \( {R}_{1} \) and \( {R}_{2} \) from Lemma 67.14, let:\n\n\[ \n{D}_{1} = \mathop{\bigcup }\limits_{{\rho \in {R}_{1}}}\rho \left\lbrack C\right\rbrack \;{D}_{2} = \mathop{\bigcup }\limits_{{\rho \in {R}_{1}}}\rho \left\lbrack C\right\rbrack \n\] \n\nThis is a partition of \( \mathbf{S} \), by Lemma 67.16, and \( {D}_{1} \) and \( {D}_{2} \) are disjoint by Lemma 67.17. By construction, \( {D}_{1} \) can be partitioned into countably many sets, \( \rho \left\lbrack C\right\rbrack \) for each \( \rho \in {R}_{1} \) . And these can be rotated to form a copy of \( \mathbf{S} \), since \( \mathbf{S} = \mathop{\bigcup }\limits_{{\rho \in R}}\rho \left\lbrack C\right\rbrack = \) \( \mathop{\bigcup }\limits_{{\rho \in {R}_{1}}}\left( {\rho \circ \rho }\right) \left\lbrack C\right\rbrack \) by Lemma 67.14 and Lemma 67.16. The same reasoning applies to \( {D}_{2} \) .
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Yes
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Corollary 67.20 (Vitali). Let \( \mu \) be a measure such that \( \mu \left( \mathbf{S}\right) = 1 \), and such that \( \mu \left( X\right) = \mu \left( Y\right) \) if \( X \) and \( Y \) are congruent. Then \( \rho \left\lbrack C\right\rbrack \) is unmeasurable for all \( \rho \in R \) .
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Proof. For reductio, suppose otherwise. So let \( \mu \left( {\sigma \left\lbrack C\right\rbrack }\right) = r \) for some \( \sigma \in R \) and some \( r \in \mathbb{R} \) . For any \( \rho \in C,\rho \left\lbrack C\right\rbrack \) and \( \sigma \left\lbrack C\right\rbrack \) are congruent, and hence \( \mu \left( {\rho \left\lbrack C\right\rbrack }\right) = r \) for any \( \rho \in C \) . By Lemma 67.16 and Lemma 67.17, \( \mathbf{S} = \mathop{\bigcup }\limits_{{\rho \in R}}\rho \left\lbrack C\right\rbrack \) is a countable union of pairwise disjoint sets. So countable additivity dictates that \( \mu \left( \mathbf{S}\right) = 1 \) is the sum of the measures of each \( \rho \left\lbrack C\right\rbrack \), i.e.,\n\n\[1 = \mu \left( \mathbf{S}\right) = \mathop{\sum }\limits_{{\rho \in R}}\mu \left( {\rho \left\lbrack C\right\rbrack }\right) = \mathop{\sum }\limits_{{\rho \in R}}r\]\n\nBut if \( r = 0 \) then \( \mathop{\sum }\limits_{{\rho \in R}}r = 0 \), and if \( r > 0 \) then \( \mathop{\sum }\limits_{{\rho \in R}}r = \infty \) .
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Yes
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Proposition 68.1. For any sets \( A \) and \( B, A \cup B = B \cup A \) .
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In order to even start the proof, we need to know what it means for two sets to be identical; i.e., we need to know what the \
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No
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Proposition 68.4. For all sets \( A \) and \( B, A \subseteq A \cup B \) .
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Proof. Let \( A \) and \( B \) be arbitrary sets. We want to show that \( A \subseteq A \cup B \) . By definition of \( \subseteq \), this amounts to: for every \( x \), if \( x \in A \) then \( x \in A \cup B \) . So let \( x \in A \) be an arbitrary element of \( A \) . We have to show that \( x \in A \cup B \) . Since \( x \in A, x \in A \) or \( x \in B \) . Thus, \( x \in \{ x : x \in A \vee x \in B\} \) . But that, by definition of \( \cup \), means \( x \in A \cup B \) .
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Yes
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Proposition 68.5. Suppose \( B \subseteq D \) and \( C \subseteq E \) . Then \( B \cup C \subseteq D \cup E \) .
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Proof. Assume (a) that \( B \subseteq D \) and (b) \( C \subseteq E \) . By definition, any \( x \in B \) is also \( \in D \) (c) and any \( x \in C \) is also \( \in E \) (d). To show that \( B \cup C \subseteq D \cup E \), we have to show that if \( x \in B \cup C \) then \( x \in D \cup E \) (by definition of \( \subseteq \) ). \( x \in B \cup C \) iff \( x \in B \) or \( x \in C \) (by definition of \( \cup \) ). Similarly, \( x \in D \cup E \) iff \( x \in D \) or \( x \in E \) . So, we have to show: for any \( x \), if \( x \in B \) or \( x \in C \), then \( x \in D \) or \( x \in E \) .\n\nSo far we've only unpacked definitions! We've reformulated our proposition without \( \subseteq \) and \( \cup \) and are left with trying to prove a universal conditional claim. By what we've discussed above, this is done by assuming that \( x \) is something about which we assume the \
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No
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Proposition 68.6. Suppose that \( x \in B \) . Then there is an \( A \) such that \( A \subseteq B \) and \( A \neq \varnothing \) .
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Proof. Assume \( x \in B \) . Let \( A = \{ x\} \).\n\nHere we’ve defined the set \( A \) by enumerating its elements. Since we assume that \( x \) is an object, and we can always form a set by enumerating its elements, we don't have to show that we've succeeded in defining a set \( A \) here. However, we still have to show that \( A \) has the properties required by the proposition. The proof isn't complete without that!\n\nSince \( x \in A, A \neq \varnothing \) .\n\nThis relies on the definition of \( A \) as \( \{ x\} \) and the obvious facts that \( x \in \{ x\} \) and \( x \notin \varnothing \) .\n\nSince \( x \) is the only element of \( \{ x\} \), and \( x \in B \), every element of \( A \) is also an element of \( B \) . By definition of \( \subseteq, A \subseteq B \) .
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No
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Proposition 68.7. If \( A \neq \varnothing \), then \( A \cup B \neq \varnothing \) .
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Proof. Suppose \( A \neq \varnothing \) . So for some \( x, x \in A \) .\n\nHere we first just restated the hypothesis of the proposition. This hypothesis, i.e., \( A \neq \varnothing \), hides an existential claim, which you get to only by unpacking a few definitions. The definition of \( = \) tells us that \( A = \varnothing \) iff every \( x \in A \) is also \( \in \varnothing \) and every \( x \in \varnothing \) is also \( \in A \) . Negating both sides, we get: \( A \neq \varnothing \) iff either some \( x \in A \) is \( \notin \varnothing \) or some \( x \in \varnothing \) is \( \notin A \) . Since nothing is \( \in \varnothing \), the second disjunct can never be true, and \
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No
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Proposition 68.8. For any sets \( A, B \), and \( C, A \cup \left( {B \cap C}\right) = \left( {A \cup B}\right) \cap \left( {A \cup C}\right) \)
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Proof. We want to show that for any sets \( A, B \), and \( C, A \cup \left( {B \cap C}\right) = \left( {A \cup B}\right) \cap \left( {A \cup C}\right) \)\n\nFirst we unpack the definition of \
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Yes
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Proposition 68.10. If \( A \subseteq B \) and \( B = \varnothing \), then \( A \) has no elements.
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Proof. Suppose \( A \subseteq B \) and \( B = \varnothing \) . We want to show that \( A \) has no elements.\n\nSince this is a conditional claim, we assume the antecedent and want to prove the consequent. The consequent is: \( A \) has no elements. We can make that a bit more explicit: it's not the case that there is an \( x \in A \) .\n\n\( A \) has no elements iff it’s not the case that there is an \( x \) such that \( x \in A \) .\n\nSo we've determined that what we want to prove is really a negative claim \( \neg p \), namely: it’s not the case that there is an \( x \in A \) . To use proof by contradiction, we have to assume the corresponding positive claim \( p \), i.e., there is an \( x \in A \), and prove a contradiction from it. We indicate that we're doing a proof by contradiction by writing \
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Yes
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Proposition 68.11. \( A \subseteq A \cup B \) .
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Proof. We want to show that \( A \subseteq A \cup B \) . On the face of it, this is a positive claim: every \( x \in A \) is also in \( A \cup B \) . The negation of that is: some \( x \in A \) is \( \notin A \cup B \) . So we can prove the claim indirectly by assuming this negated claim, and showing that it leads to a contradiction. Suppose not, i.e., \( A \nsubseteq A \cup B \) . We have a definition of \( A \subseteq A \cup B \) : every \( x \in A \) is also \( \in A \cup B \) . To understand what \( A \nsubseteq A \cup B \) means, we have to use some elementary logical manipulation on the unpacked definition: it's false that every \( x \in A \) is also \( \in A \cup B \) iff there is some \( x \in A \) that is \( \notin C \) . (This is a place where you want to be very careful: many students' attempted proofs by contradiction fail because they analyze the negation of a claim like \
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No
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Proposition 68.12. If \( A \subseteq B \) and \( B \subseteq C \) then \( A \subseteq C \) .
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Proof. Suppose \( A \subseteq B \) and \( B \subseteq C \) . We want to show \( A \subseteq C \) .\n\nLet's proceed indirectly: we assume the negation of what we want to etablish.\n\nSuppose not, i.e., \( A \nsubseteq C \) .\n\nAs before, we reason that \( A \nsubseteq C \) iff not every \( x \in A \) is also \( \in C \) , i.e., some \( x \in A \) is \( \notin C \) . Don’t worry, with practice you won’t have to think hard anymore to unpack negations like this.\n\nIn other words, there is an \( x \) such that \( x \in A \) and \( x \notin C \) .\n\nNow we can use this to get to our contradiction. Of course, we'll have to use the other two assumptions to do it.\n\nSince \( A \subseteq B, x \in B \) . Since \( B \subseteq C, x \in C \) . But this contradicts \( x \notin C \) .
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Yes
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Proposition 68.13. If \( A \cup B = A \cap B \) then \( A = B \) .
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Proof. Suppose \( A \cup B = A \cap B \) . We want to show that \( A = B \) .\n\nThe beginning is now routine:\n\nAssume, by way of contradiction, that \( A \neq B \) .\n\nOur assumption for the proof by contradiction is that \( A \neq B \) . Since \( A = B \) iff \( A \subseteq B \) an \( B \subseteq A \), we get that \( A \neq B \) iff \( A \nsubseteq B \) or \( B \nsubseteq A \) . (Note how important it is to be careful when manipulating negations!) To prove a contradiction from this disjunction, we use a proof by cases and show that in each case, a contradiction follows.\n\n\( A \neq B \) iff \( A \nsubseteq B \) or \( B \nsubseteq A \) . We distinguish cases.\n\nIn the first case, we assume \( A \nsubseteq B \), i.e., for some \( x, x \in A \) but \( \notin B \) . \( A \cap B \) is defined as those elements that \( A \) and \( B \) have in common, so if something isn't in one of them, it's not in the intersection. \( A \cup B \) is \( A \) together with \( B \), so anything in either is also in the union. This tells us that \( x \in A \cup B \) but \( x \notin A \cap B \), and hence that \( A \cap B \neq B \cap A \) .\n\nCase 1: \( A \nsubseteq B \) . Then for some \( x, x \in A \) but \( x \notin B \) . Since \( x \notin B \), then \( x \notin A \cap B \) . Since \( x \in A, x \in A \cup B \) . So, \( A \cap B \neq B \cap A \), contradicting the assumption that \( A \cap B = A \cup B \) .\n\nCase 2: \( B \nsubseteq A \) . Then for some \( y, y \in B \) but \( y \notin A \) . As before, we have \( y \in A \cup B \) but \( y \notin A \cap B \), and so \( A \cap B \neq A \cup B \), again contradicting \( A \cap B = \) \( A \cup B \) .
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Yes
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Proposition 68.14 (Absorption). For all sets \( A, B \) , \[ A \cap \left( {A \cup B}\right) = A \]
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Proof. If \( z \in A \cap \left( {A \cup B}\right) \), then \( z \in A \), so \( A \cap \left( {A \cup B}\right) \subseteq A \) . Now suppose \( z \in A \) . Then also \( z \in A \cup B \), and therefore also \( z \in A \cap \left( {A \cup B}\right) \) .
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Yes
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Problem 68.3. Expand the following proof of \( A \cup \left( {A \cap B}\right) = A \), where you mention all the inference patterns used, why each step follows from assumptions or claims established before it, and where we have to appeal to which definitions.
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Proof. If \( z \in A \cup \left( {A \cap B}\right) \) then \( z \in A \) or \( z \in A \cap B \) . If \( z \in A \cap B, z \in A \) . Any \( z \in A \) is also \( \in A \cup \left( {A \cap B}\right) \) .
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No
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Theorem 69.1. With \( n \) dice one can throw all \( {5n} + 1 \) possible values between \( n \) and \( {6n} \) .
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Proof. Let \( P\left( n\right) \) be the claim: \
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No
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Proposition 69.2. \( {s}_{n} = n\left( {n + 1}\right) /2 \) .
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Proof. We have to prove (1) that \( {s}_{0} = 0 \cdot \left( {0 + 1}\right) /2 \) and (2) if \( {s}_{k} = k\left( {k + 1}\right) /2 \) then \( {s}_{k + 1} = \left( {k + 1}\right) \left( {k + 2}\right) /2 \) . (1) is obvious. To prove (2), we assume the inductive hypothesis: \( {s}_{k} = k\left( {k + 1}\right) /2 \) . Using it, we have to show that \( {s}_{k + 1} = \) \( \left( {k + 1}\right) \left( {k + 2}\right) /2 \) .\n\nWhat is \( {s}_{k + 1} \) ? By the definition, \( {s}_{k + 1} = {s}_{k} + \left( {k + 1}\right) \) . By inductive hypothesis, \( {s}_{k} = k\left( {k + 1}\right) /2 \) . We can substitute this into the previous equation, and then just need a bit of arithmetic of fractions:\n\n\[ \n{s}_{k + 1} = \frac{k\left( {k + 1}\right) }{2} + \left( {k + 1}\right) = \n\]\n\n\[ \n= \frac{k\left( {k + 1}\right) }{2} + \frac{2\left( {k + 1}\right) }{2} = \n\]\n\n\[ \n= \frac{k\left( {k + 1}\right) + 2\left( {k + 1}\right) }{2} = \n\]\n\n\[ \n= \frac{\left( {k + 2}\right) \left( {k + 1}\right) }{2}\text{.} \n\]
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Yes
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Proposition 69.4. For any \( n \), the number of \( \lbrack \) in a nice term of length \( n \) is \( < n/2 \) .
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Proof. To prove this result by (strong) induction, we have to show that the following conditional claim is true:\n\nIf for every \( l < k \), any nice term of length \( l \) has \( l/2 \) ’s, then any nice term of length \( k \) has \( k/2 \) ’s.\n\nTo show this conditional, assume that its antecedent is true, i.e., assume that for any \( l < k \), nice terms of length \( l \) contain \( < l/2 \) [’s. We call this assumption the inductive hypothesis. We want to show the same is true for nice terms of length \( k \) .\n\nSo suppose \( t \) is a nice term of length \( k \) . Because nice terms are inductively defined, we have two cases: (1) \( t \) is a letter by itself, or (2) \( t \) is \( \left\lbrack {{s}_{1} \circ {s}_{2}}\right\rbrack \) for some nice terms \( {s}_{1} \) and \( {s}_{2} \) .\n\n1. \( t \) is a letter. Then \( k = 1 \), and the number of \( \lbrack \) in \( t \) is 0 . Since \( 0 < 1/2 \), the claim holds.\n\n2. \( t \) is \( \left\lbrack {{s}_{1} \circ {s}_{2}}\right\rbrack \) for some nice terms \( {s}_{1} \) and \( {s}_{2} \) . Let’s let \( {l}_{1} \) be the length of \( {s}_{1} \) and \( {l}_{2} \) be the length of \( {s}_{2} \) . Then the length \( k \) of \( t \) is \( {l}_{1} + {l}_{2} + 3 \) (the lengths of \( {s}_{1} \) and \( {s}_{2} \) plus three symbols \( \left\lbrack {,\circ ,}\right\rbrack ) \) . Since \( {l}_{1} + {l}_{2} + 3 \) is always greater than \( {l}_{1},{l}_{1} < k \) . Similarly, \( {l}_{2} < n \) . That means that the induction hypothesis applies to the terms \( {s}_{1} \) and \( {s}_{2} \) : the number \( {m}_{1} \) of \( \left\lbrack {\text{in}{s}_{1}}\right. \) is \( \left. { < {l}_{1}/2}\right\rbrack \), and the number \( {m}_{2} \) of [in \( {s}_{2} \) is \( < {l}_{2}/2 \) .\n\nThe number of [ in \( t \) is the number of [ in \( {s}_{1} \), plus the number of [ in \( {s}_{2} \) , plus 1, i.e., it is \( {m}_{1} + {m}_{2} + 1 \) . Since \( {m}_{1} < {l}_{1}/2 \) and \( {m}_{2} < {l}_{2}/2 \) we have:\n\n\[ \n{m}_{1} + {m}_{2} + 1 < \frac{{l}_{1}}{2} + \frac{{l}_{2}}{2} + 1 = \frac{{l}_{1} + {l}_{2} + 2}{2} < \frac{{l}_{1} + l - 2 + 3}{2} = k/2. \n\]\n\nIn each case, we’ve shown that the number of [in \( t \) is \( < k/2 \) (on the basis of the inductive hypothesis). By strong induction, the proposition follows.
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Yes
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Proposition 69.5. The number of [equals the number of] in any nice term \( t \) .
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Proof. We use structural induction. Nice terms are inductively defined, with letters as initial objects and the operations \( o \) for constructing new nice terms out of old ones.\n\n1. The claim is true for every letter, since the number of [ in a letter by itself is 0 and the number of \( \rbrack \) in it is also 0 .\n\n2. Suppose the number of [ in \( {s}_{1} \) equals the number of ], and the same is true for \( {s}_{2} \) . The number of \( \left\lbrack {\text{in}o\left( {{s}_{1},{s}_{2}}\right) }\right. \), i.e., in \( \left\lbrack {{s}_{1} \circ {s}_{2}}\right\rbrack \), is the sum of the number of [in \( {s}_{1} \) and \( {s}_{2} \) . The number of ] in \( o\left( {{s}_{1},{s}_{2}}\right) \) is the sum of the number of \( \rbrack \) in \( {s}_{1} \) and \( {s}_{2} \) . Thus, the number of \( \left\lbrack \right. \) in \( o\left( {{s}_{1},{s}_{2}}\right) \) equals the number of \( \rbrack \) in \( o\left( {{s}_{1},{s}_{2}}\right) \) .
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Yes
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Proposition 69.6. Every proper initial segment of a nice term \( t \) has more \( \left\lbrack {{}^{\prime }s\text{than}}\right\rbrack \) ’s.
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Proof. By induction on \( t \) :\n\n1. \( t \) is a letter by itself: Then \( t \) has no proper initial segments.\n\n2. \( t = \left\lbrack {{s}_{1} \circ {s}_{2}}\right\rbrack \) for some nice terms \( {s}_{1} \) and \( {s}_{2} \) . If \( r \) is a proper initial segment of \( t \), there are a number of possibilities:\n\na) \( r \) is just \( \lbrack \) : Then \( r \) has one more \( \left\lbrack \text{than it does}\right\rbrack \) .\n\nb) \( r \) is \( \left\lbrack {r}_{1}\right. \) where \( {r}_{1} \) is a proper initial segment of \( {s}_{1} \) : Since \( {s}_{1} \) is a nice term, by induction hypothesis, \( {r}_{1} \) has more [than ] and the same is true for \( \left\lbrack {r}_{1}\right. \) .\n\nc) \( r \) is \( \left\lbrack {{s}_{1}\text{or}\left\lbrack {{s}_{1} \circ : }\right. }\right. \) By the previous result, the number of \( \left\lbrack \text{and}\right\rbrack \) in \( {s}_{1} \) are equal; so the number of [ in [ \( {s}_{1} \) or [ \( {s}_{1} \circ \) is one more than the number of \( \rbrack \) .\n\nd) \( r \) is \( \left\lbrack {{s}_{1} \circ {r}_{2}}\right. \) where \( {r}_{2} \) is a proper initial segment of \( {s}_{2} \) : By induction hypothesis, \( {r}_{2} \) contains more \( \left\lbrack \text{than}\right\rbrack \) . By the previous result, the number of [and of] in \( {s}_{1} \) are equal. So the number of [in \( \left\lbrack {{s}_{1} \circ {r}_{2}}\right. \) is greater than the number of \( \rbrack \) .\n\ne) \( r \) is \( \left\lbrack {{s}_{1} \circ {s}_{2}}\right. \) : By the previous result, the number of \( \left\lbrack \text{and}\right\rbrack \) in \( {s}_{1} \) are equal, and the same for \( {s}_{2} \) . So there is one more \( \left\lbrack {\text{in}\left\lbrack {{s}_{1} \circ {s}_{2}}\right. }\right. \) than there are \( \rbrack \) .
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Yes
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Proposition 69.9. Suppose \( t \) is a nice term. Then either \( t \) is a letter by itself, or there are uniquely determined nice terms \( {s}_{1},{s}_{2} \) such that \( t = \left\lbrack {{s}_{1} \circ {s}_{2}}\right\rbrack \) .
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Proof. If \( t \) is a letter by itself, the condition is satisfied. So assume \( t \) isn’t a letter by itself. We can tell from the inductive definition that then \( t \) must be of the form \( \left\lbrack {{s}_{1} \circ {s}_{2}}\right\rbrack \) for some nice terms \( {s}_{1} \) and \( {s}_{2} \) . It remains to show that these are uniquely determined, i.e., if \( t = \left\lbrack {{r}_{1} \circ {r}_{2}}\right\rbrack \), then \( {s}_{1} = {r}_{1} \) and \( {s}_{2} = {r}_{2} \) . So suppose \( t = \left\lbrack {{s}_{1} \circ {s}_{2}}\right\rbrack \) and also \( t = \left\lbrack {{r}_{1} \circ {r}_{2}}\right\rbrack \) for nice terms \( {s}_{1},{s}_{2},{r}_{1},{r}_{2} \) . We have to show that \( {s}_{1} = {r}_{1} \) and \( {s}_{2} = {r}_{2} \) . First, \( {s}_{1} \) and \( {r}_{1} \) must be identical, for otherwise one is a proper initial segment of the other. But by Proposition 69.6, that is impossible if \( {s}_{1} \) and \( {r}_{1} \) are both nice terms. But if \( {s}_{1} = {r}_{1} \), then clearly also \( {s}_{2} = {r}_{2} \) .
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Yes
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Theorem 71.1. \( L \approx S \)
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Proof: first part.. Fix \( a, b \in \mathrm{L} \) . Write them in binary notation, so that we have infinite sequences of 0 s and \( 1\mathrm{\;s},{a}_{1},{a}_{2},\ldots \), and \( {b}_{1},{b}_{2},\ldots \), such that:\n\n\[ a = 0.{a}_{1}{a}_{2}{a}_{3}{a}_{4}\ldots \]\n\n\[ b = 0.{b}_{1}{b}_{2}{b}_{3}{b}_{4}\ldots \]\n\nNow consider the function \( f : \mathrm{S} \rightarrow \mathrm{L} \) given by\n\n\[ f\left( {a, b}\right) = 0.{a}_{1}{b}_{1}{a}_{2}{b}_{2}{a}_{3}{b}_{3}{a}_{4}{b}_{4}\ldots \]\n\nNow \( f \) is an injection, since if \( f\left( {a, b}\right) = f\left( {c, d}\right) \), then \( {a}_{n} = {c}_{n} \) and \( {b}_{n} = {d}_{n} \) for all \( n \in \mathbb{N} \), so that \( a = c \) and \( b = d \) .\n\nUnfortunately, as Dedekind pointed out to Cantor, this does not answer the original question. Consider \( {0.10} = {0.1010101010}\ldots \) We need that \( f\left( {a, b}\right) = \) \( {0.1}\dot{0} \), where:\n\n\[ a = 0.\dot{1}\dot{1} = {0.111111}\ldots \]\n\n\[ b = 0 \]\n\nBut \( a = {0.11} = 1 \) . So, when we say \
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No
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Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience). 1. A soccer coach selects 6 players from a group of boys aged 8 to 10,7 players from a group of boys aged 11 to 12, and 3 players from a group of boys aged 13 to 14 to form a recreational soccer team.
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1. stratified
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Yes
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Do you think that either of these samples is representative of (or is characteristic of) the entire 10,000 part-time student population?
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No. The first sample probably consists of science-oriented students. Besides the chemistry course, some of them are taking first-term calculus. Books for these classes tend to be expensive. Most of these students are, more than likely, paying more than the average part-time student for their books. The second sample is a group of senior citizens who are, more than likely, taking courses for health and interest. The amount of money they spend on books is probably much less than the average part-time student. Both samples are biased. Also, in both cases, not all students have a chance to be in either sample.
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Yes
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Is the sample biased?
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The sample is unbiased, but a larger sample would be recommended to increase the likelihood that the sample will be close to representative of the population. However, for a biased sampling technique, even a large sample runs the risk of not being representative of the population.
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No
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For Susan Dean's spring pre-calculus class, scores for the first exam were as follows (smallest to largest):\n\n\\( {33};{42};{49};{49};{53};{55};{55};{61};{63};{67};{68};{68};{69};{69};{72};{73};{74};{78};{80};{83};{88};{88};{88};{90};{92};{94};{94};{94};{94}; \)\n\n96; 100
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Stem-and-Leaf Diagram\n\n<table><thead><tr><th>Stem</th><th>Leaf</th></tr></thead><tr><td>3</td><td>3</td></tr><tr><td>4</td><td>299</td></tr><tr><td>5</td><td>355</td></tr><tr><td>6</td><td>1378899</td></tr><tr><td>7</td><td>2348</td></tr><tr><td>8</td><td>03888</td></tr><tr><td>9</td><td>0244446</td></tr><tr><td>10</td><td>0</td></tr></table>\n\nTable 2.1\n\nThe stem plot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately \\( {26}\\% \\) of the scores were in the \\( {90}^{\\prime }\\mathrm{s} \\) or 100, a fairly high number of As.\n\nThe stem plot is a quick way to graph and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers. In the example above, there were no outliers.
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Yes
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Create a stem plot using the data:\n\n\( {1.1};{1.5};{2.3};{2.5};{2.7};{3.2};{3.3};{3.3};{3.5};{3.8};{4.0};{4.2};{4.5};{4.5};{4.7};{4.8};{5.5};{5.6};{6.5};{6.7};{12.3} \)\n\nThe data are the distance (in kilometers) from a home to the nearest supermarket.
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Problem (Solution on p. 114.)
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No
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The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The heights are continuous data since height is measured.\n\n$ {60};{60.5};{61};{61};{61.5} $\n\n$ {63.5};{63.5};{63.5} $\n\n$ {64};{64};{64};{64};{64};{64};{64.5};{64.5};{64.5};{64.5};{64.5};{64.5};{64.5};{64.5};{64.5} $\n\n$ {66};{66};{66};{66};{66};{66};{66};{66};{66};{66};{66.5};{66.5};{66.5};{66.5};{66.5};{66.5};{66.5};{66.5};{66.5};{66.5};{66.5};{66.5};{66.5};{67};{67};{67}; $\n\n67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5\n\n68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5; 69.5\n\n70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71\n\n72; 72; 72; 72.5; 72.5; 73; 73.5\n\n74
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The smallest data value is 60 . Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5 , $ {0.05},{0.005} $, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.\n\n$ {60} - {0.05} = {59.95} $ which is more precise than, say,61.5 by one decimal place. The starting point is, then, 59.95.\n\nThe largest value is $ {74.74} + {0.05} = {74.05} $ is the ending value.\n\nNext, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose 8 bars.\n\n$ \\frac{{74.05} - {59.95}}{8} = {1.76} $\n\n(2.0)\n\nNOTE: We will round up to 2 and make each bar or class interval 2 units wide. Rounding up to 2 is one way to prevent a value from falling on a boundary. Rounding to the next number is necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work.\n\nThe boundaries are:\n\n- 59.95\n\n- $ {59.95} + 2 = {61.95} $\n\n- $ {61.95} + 2 = {63.95} $\n\n- $ {63.95} + 2 = {65.95} $\n\n- $ {65.95} + 2 = {67.95} $\n\n- $ {67.95} + 2 = {69.95} $\n\n- $ {69.95} + 2 = {71.95} $\n\n- $ {71.95} + 2 = {73.95} $\n\n- $ {73.95} + 2 = {75.95} $\n\nThe heights 60 through 61.5 inches are in the interval 59.95 - 61.95 . The heights that are 63.5 are in the interval 61.95 - 63.95. The heights that are 64 through 64.5 are in the interval 63.95 - 65.95 . The heights 66 through 67.5 are in the interval 65.95 - 67.95 . The heights 68 through 69.5 are in the interval 67.95 - 69.95. The heights 70 through 71 are in the interval 69.95 - 71.95. The heights 72 through 73.5 are in the interval 71.95 - 73.95. The height 74 is in the interval 73.95 - 75.95.\n\nThe following histogram displays the heights on the x-axis and relative frequency on the y-axis.\n\nRelative Frequency 
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Yes
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Construct a box plot:
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Using the TI-83, 83+, 84, 84+ Calculator\n\n- Enter data into the list editor (Press STAT 1:EDIT). If you need to clear the list, arrow up to the name L1, press CLEAR, arrow down.\n\n- Put the data values in list L1.\n\n- Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1.\n\n- Press ENTER\n\n- Use the down and up arrow keys to scroll.\n\n- Smallest value \( = {59} \)\n\n- Largest value \( = {77} \)\n\n- Q1: First quartile \( = {64.5} \)\n\n- Q2: Second quartile or median= 66\n\n- Q3: Third quartile \( = {70} \)\n\nUsing the TI-83, 83+, 84, 84+ to Construct the Box Plot\n\nGo to 14:Appendix for Notes for the TI-83, 83+, 84, 84+ Calculator. To create the box plot:\n\n- Press Y=. If there are any equations, press CLEAR to clear them.\n\n- Press 2nd Y=.\n\n- Press 4:Plotsoff. Press ENTER\n\n---\n\n\n\n- Press 2nd Y=\n\n- Press 1:Plot1. Press ENTER.\n\n- Arrow down and then use the right arrow key to go to the 5th picture which is the box plot. Press ENTER.\n\n- Arrow down to Xlist: Press 2nd 1 for L1\n\n- Arrow down to Freq: Press ALPHA. Press 1.\n\n- Press ZOOM. Press 9:ZoomStat.\n\n- Press TRACE and use the arrow keys to examine the box plot.
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Yes
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For the following 13 real estate prices, calculate the IQR and determine if any prices are outliers. Prices are in dollars. (Source: San Jose Mercury News)\n\n389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000
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Order the data from smallest to largest.\n\n114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000\n\n\[ M = {488},{800} \]\n\n\[ {Q}_{1} = \frac{{230500} + {387000}}{2} = {308750} \]\n\n\[ {Q}_{3} = \frac{{639000} + {659000}}{2} = {649000} \]\n\n\[ {IQR} = {649000} - {308750} = {340250} \]\n\n\[ \left( {1.5}\right) \left( {IQR}\right) = \left( {1.5}\right) \left( {340250}\right) = {510375} \]\n\n\[ {Q}_{1} - \left( {1.5}\right) \left( {IQR}\right) = {308750} - {510375} = - {201625} \]\n\n\[ {Q}_{3} + \left( {1.5}\right) \left( {IQR}\right) = {649000} + {510375} = {1159375} \]\n\nNo house price is less than -201625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier.
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Yes
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On a timed math test, the first quartile for times for finishing the exam was 35 minutes. Interpret the first quartile in the context of this situation.
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- 25% of students finished the exam in 35 minutes or less.\n- 75% of students finished the exam in 35 minutes or more.\n- A low percentile could be considered good, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.)
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Yes
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Verify the mean and standard deviation calculated above on your calculator or computer.
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Using the TI-83,83+,84+ Calculators\n\n- Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.\n\n- Put the data values \( \\left( {9,{9.5},{10},{10.5},{11},{11.5}}\\right) \) into list L1 and the frequencies \( \\left( {1,2,4,4,6,3}\\right) \) into list L2. Use the arrow keys to move around.\n\n- Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.\n\n- \( \\bar{x} = {10.525} \)\n\n- Use Sx because this is sample data (not a population): \( {Sx} = {0.715891} \)\n\n- For the following problems, recall that value \( = \) mean \( + \\left( {\\# \\text{ofSTDEVs}}\\right) \) (standard deviation)\n\n- For a sample: \( x = \\bar{x} + \\left( {\\# \\text{ofSTDEVs}}\\right) \\left( \\mathrm{s}\\right) \)\n\n- For a population: \( x = \\mu + \\left( {\\# \\text{ofSTDEVs}}\\right) \\left( \\sigma \\right) \)\n\n- For this example, use \( x = \\bar{x} + \\left( {\\# \\text{ofSTDEVs}}\\right) \\left( \\mathrm{s}\\right) \) because the data is from a sample
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Yes
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Find the value that is 1 standard deviation above the mean. Find \( \left( {\bar{x} + {1s}}\right) \) .
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\( \left( {\bar{x} + {1s}}\right) = {10.53} + \left( 1\right) \left( {0.72}\right) = {11.25} \)
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Yes
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Find the value that is two standard deviations below the mean. Find \( \left( {\bar{x} - {2s}}\right) \) .
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\[ \left( {\bar{x} - {2s}}\right) = {10.53} - \left( 2\right) \left( {0.72}\right) = {9.09} \]
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Yes
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Find the values that are 1.5 standard deviations from (below and above) the mean.
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\[ \text{-}\left( {\bar{x} - {1.5s}}\right) = {10.53} - \left( {1.5}\right) \left( {0.72}\right) = {9.45} \] \[ \text{-}\left( {\bar{x} + {1.5s}}\right) = {10.53} + \left( {1.5}\right) \left( {0.72}\right) = {11.61} \]
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Yes
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a. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
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<table><thead><tr><th>Data</th><th>Frequency</th><th>Relative Frequency</th><th>Cumulative Relative Frequency</th></tr></thead><tr><td>33</td><td>1</td><td>0.032</td><td>0.032</td></tr><tr><td>42</td><td>1</td><td>0.032</td><td>0.064</td></tr><tr><td>49</td><td>2</td><td>0.065</td><td>0.129</td></tr><tr><td>53</td><td>1</td><td>0.032</td><td>0.161</td></tr><tr><td>55</td><td>2</td><td>0.065</td><td>0.226</td></tr><tr><td>61</td><td>1</td><td>0.032</td><td>0.258</td></tr><tr><td>63</td><td>1</td><td>0.032</td><td>0.29</td></tr><tr><td>67</td><td>1</td><td>0.032</td><td>0.322</td></tr><tr><td>68</td><td>2</td><td>0.065</td><td>0.387</td></tr><tr><td>69</td><td>2</td><td>0.065</td><td>0.452</td></tr><tr><td>72</td><td>1</td><td>0.032</td><td>0.484</td></tr><tr><td>73</td><td>1</td><td>0.032</td><td>0.516</td></tr><tr><td>74</td><td>1</td><td>0.032</td><td>0.548</td></tr><tr><td>78</td><td>1</td><td>0.032</td><td>0.580</td></tr><tr><td>80</td><td>1</td><td>0.032</td><td>0.612</td></tr><tr><td>83</td><td>1</td><td>0.032</td><td>0.644</td></tr><tr><td>88</td><td>3</td><td>0.097</td><td>0.741</td></tr><tr><td>90</td><td>1</td><td>0.032</td><td>0.773</td></tr><tr><td>92</td><td>1</td><td>0.032</td><td>0.805</td></tr><tr><td>94</td><td>4</td><td>0.129</td><td>0.934</td></tr><tr><td>96</td><td>1</td><td>0.032</td><td>0.966</td></tr><tr><td>100</td><td>1</td><td>0.032</td><td>0.998 (Why isn’t this value 1?)</td></tr></table>
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Yes
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Flip two fair coins. (This is an experiment.)
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The sample space is \( \{ {HH},{HT},{TH},{TT}\} \) where \( T = \) tails and \( H = \) heads. The outcomes are \( {HH} \) , \( {HT},{TH} \), and \( {TT} \) . The outcomes \( {HT} \) and \( {TH} \) are different. The \( {HT} \) means that the first coin showed heads and the second coin showed tails. The \( {TH} \) means that the first coin showed tails and the second coin showed heads.\n\n- Let \( A = \) the event of getting at most one tail. (At most one tail means 0 or 1 tail.) Then \( A \) can be written as \( \{ {HH},{HT},{TH}\} \) . The outcome \( {HH} \) shows 0 tails. \( {HT} \) and \( {TH} \) each show 1 tail.\n\n- Let \( B = \) the event of getting all tails. \( B \) can be written as \( \{ {TT}\} \) . \( B \) is the complement of \( A \) . So, \( B = {A}^{\prime } \) . Also, \( P\left( A\right) + P\left( B\right) = P\left( A\right) + P\left( {A}^{\prime }\right) = 1 \).\n\n- The probabilities for \( A \) and for \( B \) are \( P\left( A\right) = \frac{3}{4} \) and \( P\left( B\right) = \frac{1}{4} \) .\n\n- Let \( C = \) the event of getting all heads. \( C = \{ {HH}\} \) . Since \( B = \{ {TT}\}, P\left( {BANDC}\right) = 0 \) . \( B \) and \( C \) are mutually exclusive. ( \( B \) and \( C \) have no members in common because you cannot have all tails and all heads at the same time.)\n\n- Let \( D = \) event of getting more than one tail. \( D = \{ {TT}\} .P\left( D\right) = \frac{1}{4} \).\n\n- Let \( E = \) event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) \( E = \{ {HT},{HH}\} .P\left( E\right) = \frac{2}{4} \).\n\n- Find the probability of getting at least one (1 or 2) tail in two flips. Let \( F = \) event of getting at least one tail in two flips. \( F = \{ {HT},{TH},{TT}\} .\mathrm{P}\left( \mathrm{F}\right) = \frac{3}{4} \
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Yes
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Show that \( \mathrm{P}\left( {\mathrm{G} \mid \mathrm{H}}\right) = \mathrm{P}\left( \mathrm{G}\right) \) .
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\[ \mathrm{P}\left( {\mathrm{G} \mid \mathrm{H}}\right) = \frac{\mathrm{P}\left( {\mathrm{G}\mathrm{{AND}}\mathrm{H}}\right) }{\mathrm{P}\left( \mathrm{H}\right) } = \frac{0.3}{0.5} = {0.6} = \mathrm{P}\left( \mathrm{G}\right) \]
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Yes
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Show \( \mathrm{P}\left( {\mathrm{G}\mathrm{{ANDH}}}\right) = \mathrm{P}\left( \mathrm{G}\right) \cdot \mathrm{P}\left( \mathrm{H}\right) \) .
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\( P\left( G\right) \cdot P\left( H\right) = {0.6} \cdot {0.5} = {0.3} = \mathrm{P}\left( \mathrm{{GANDH}}\right) \)
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Yes
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What is the probability that the member is a novice swimmer?
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\\( \\frac{28}{150} \\)
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Yes
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Are being a novice swimmer and practicing 4 times a week independent events? Why or why not?
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No, these are not independent events.\n\n\[ \text{P(novice AND practices 4 times per week)} = {0.0667} \]\n\n(3.0)\n\nP(novice) \( \cdot \) P(practices 4 times per week) \( = {0.0996} \)\n\n(3.0)\n\n\[ {0.0667} \neq {0.0996} \]\n\n(3.0)
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Yes
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What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
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\( \mathrm{P}\left( \mathrm{B}\right) = {0.143};\mathrm{P}\left( \mathrm{N}\right) = {0.85} \)
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Yes
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What is the probability that the woman has breast cancer AND tests negative?
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\( \mathrm{P}\left( {\mathrm{B}\;\mathrm{{AND}}\;\mathrm{N}}\right) = \mathrm{P}\left( \mathrm{B}\right) \cdot \mathrm{P}\left( {\mathrm{N}\;\mathrm{I}\;\mathrm{B}}\right) = \left( {0.143}\right) \; \cdot \;\left( {0.02}\right) = \;{0.0029} \)
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Yes
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What is the probability that the woman has breast cancer or tests negative?
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\( \begin{aligned} P\left( {BORN}\right) & = P\left( B\right) + P\left( N\right) - P\left( {BANDN}\right) \\ & = {0.143} + {0.85} - {0.0029} = {0.9901} \end{aligned} \)
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Yes
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Are having breast cancer and testing negative independent events?
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No. \( \mathrm{P}\left( \mathrm{N}\right) = {0.85};\mathrm{P}\left( {\mathrm{N} \mid \mathrm{B}}\right) = {0.02} \) . So, \( \mathrm{P}\left( {\mathrm{N} \mid \mathrm{B}}\right) \) does not equal \( \mathrm{P}\left( \mathrm{N}\right) \)
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Yes
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Are having breast cancer and testing negative mutually exclusive?
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No. P(B AND N) \( = {0.0029} \) . For \( B \) and \( N \) to be mutually exclusive, P(B AND N) must be 0 .
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Yes
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Problem 1\n\n\( \mathrm{P} \) (person is a car phone user) \( = \)
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Solution\n\n\( \frac{\text{ number of car phone users }}{\text{ total number in study }} = \frac{305}{755} \)
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Yes
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Problem 2\n\n\( \mathrm{P} \) (person had no violation in the last year) \( = \)
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Solution\n\n\( \frac{\text{ number that had no violation }}{\text{ total number in study }} = \frac{685}{755} \)
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Yes
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P(person had no violation in the last year AND was a car phone user) \( = \)
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\( \frac{280}{755} \)
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Yes
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What is the probability that Alissa does not catch Muddy?
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\( \frac{41}{60} \)
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Yes
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Suppose an experiment has the outcomes \( 1,2,3,\ldots ,{12} \) where each outcome has an equal chance of occurring. Let event \( A = \{ 1,2,3,4,5,6\} \) and event \( B = \{ 6,7,8,9\} \) . Then \( {AANDB} = \{ 6\} \) and \( {AORB} = \{ 1,2,3,4,5,6,7,8,9\} \) .
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The Venn diagram is as follows:\n\nS 
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Yes
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Flip 2 fair coins. Let \( A = \) tails on the first coin. Let \( B = \) tails on the second coin. Then \( A = \{ {TT},{TH}\} \) and \( B = \{ {TT},{HT}\} \) . Therefore, A AND B \( = \{ {TT}\} \) . A OR B \( = \{ {TH},{TT},{HT}\} \) .
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The sample space when you flip two fair coins is \( S = \{ {HH},{HT},{TH},{TT}\} \) . The outcome \( {HH} \) is in neither \( A \) nor \( B \) . The Venn diagram is as follows:\n\n---
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Yes
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List the \( {24BR} \) outcomes: \( {B1R1},{B1R2},{B1R3},\ldots \)
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Solution on p. 161.
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No
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Using the tree diagram, calculate P(RR).
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\\( P\\left( {RR}\\right) \\; = \\frac{3}{11}\\; \\cdot \\frac{3}{11}\\; = \\frac{9}{121} \\)
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Yes
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Using the tree diagram, calculate P(RB OR BR).
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\\( P\\left( \\text{RB OR BR}\\right) \\; = \\frac{3}{11}\\; \\cdot \\frac{8}{11}\\; + \\frac{8}{11}\\; \\cdot \\frac{3}{11}\\; = \\frac{48}{121} \\)
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Yes
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\( \mathrm{P}\left( {\mathrm{B}\text{ on }2\mathrm{{nd}} \mid \mathrm{R}\text{ on }1\mathrm{{st}}}\right) = \)
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There are 6 + 24 outcomes that have \( R \) on the first draw (6 \( {RR} \) and 24 \( {RB} \) ). The 6 and the 24 are frequencies. They are also the numerators of the fractions \( \frac{6}{110} \) and \( \frac{24}{110} \) . The sample space is no longer \( {110} \) but \( 6 + {24} = {30} \) . Twenty-four of the 30 outcomes have \( B \) on the second draw. The probability is then \( \frac{24}{30} \) . Did you get this answer?
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Yes
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A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let \( X = \) the number of times a newborn wakes its mother after midnight. For this example, \( x = 0,1,2 \) , \( 3,4,5 \) .
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X takes on the values \( 0,1,2,3,4,5 \) . This is a discrete \( {PDF} \) because\n1. Each \( \mathrm{P}\left( \mathrm{x}\right) \) is between 0 and 1, inclusive.\n2. The sum of the probabilities is 1 , that is,\n\n\[ \frac{2}{50} + \frac{11}{50} + \frac{23}{50} + \frac{9}{50} + \frac{4}{50} + \frac{1}{50} = 1 \]
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Yes
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Find the expected value for the example about the number of times a newborn baby's crying wakes its mother after midnight. The expected value is the expected number of times a newborn wakes its mother after midnight.
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<table><thead><tr><th>\( x \)</th><th>P(X)</th><th>\( x\mathbf{P}\left( \mathbf{X}\right) \)</th></tr></thead><tr><td>0</td><td>\( P\left( {x = 0}\right) = \frac{2}{50} \)</td><td>(0) \( \left( \frac{2}{50}\right) = 0 \)</td></tr><tr><td>1</td><td>\( P\left( {x = 1}\right) = \frac{11}{50} \)</td><td>\( \left( 1\right) \left( \frac{11}{50}\right) = \frac{11}{50} \)</td></tr><tr><td>2</td><td>\( P\left( {x = 2}\right) = \frac{23}{50} \)</td><td>(2) \( \left( \frac{23}{50}\right) = \frac{46}{50} \)</td></tr><tr><td>3</td><td>\( P\left( {x = 3}\right) = \frac{9}{50} \)</td><td>(3) \( \left( \frac{9}{50}\right) = \frac{27}{50} \)</td></tr><tr><td>4</td><td>\( P\left( {x = 4}\right) = \frac{4}{50} \)</td><td>\( \left( 4\right) \left( \frac{4}{50}\right) = \frac{16}{50} \)</td></tr><tr><td>5</td><td>\( P\left( {x = 5}\right) = \frac{1}{50} \)</td><td>(5) \( \left( \frac{1}{50}\right) = \frac{5}{50} \)</td></tr></table>\n\nTable 4.5: You expect a newborn to wake its mother after midnight 2.1 times, on the average.\n\nAdd the last column to find the expected value. \( \mu = \) Expected Value \( = \frac{105}{50} = {2.1} \)
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Yes
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What is the expected value, \( \mu \) ? Do you come out ahead?
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Like data, probability distributions have standard deviations. To calculate the standard deviation \( \left( \sigma \right) \) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root . To understand how to do the calculation, look at the table for the number of days per week a men's soccer team plays soccer. To find the standard deviation, add the entries in the column labeled \( {\left( x - \mu \right) }^{2} \cdot P\left( x\right) \) and take the square root.\n\n<table><thead><tr><th>\( x \)</th><th>P(x)</th><th>\( x\mathrm{P}\left( \mathrm{x}\right) \)</th><th>\( {\left( x - \mu \right) }^{2}P\left( x\right) \)</th></tr></thead><tr><td>0</td><td>0.2</td><td>\( \left( 0\right) \left( {0.2}\right) = 0 \)</td><td>\( {\left( 0 - {1.1}\right) }^{2}\left( {.2}\right) = {0.242} \)</td></tr><tr><td>1</td><td>0.5</td><td>\( \left( 1\right) \left( {0.5}\right) = {0.5} \)</td><td>\( {\left( 1 - {1.1}\right) }^{2}\left( {.5}\right) = {0.005} \)</td></tr><tr><td>2</td><td>0.3</td><td>(2)(0.3) \( = {0.6} \)</td><td>\( {\left( 2 - {1.1}\right) }^{2}\left( {.3}\right) = {0.243} \)</td></tr></table>\n\n## Table 4.8\n\nAdd the last column in the table. \( {0.242} + {0.005} + {0.243} = {0.490} \) . The standard deviation is the square root of 0.49 . \( \sigma = \sqrt{0.49} = {0.7} \
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No
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Approximately \( {70}\% \) of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly.
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This is a binomial problem because there is only a success or a ___, there are a definite number of trials, and the probability of a success is 0.70 for each trial.
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No
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It has been stated that about \( {41}\% \) of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?
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Let \( X = \) the number of workers who have a high school diploma but do not pursue any further education.\n\n\( X \) takes on the values \( 0,1,2,\ldots ,{20} \) where \( n = {20} \) and \( p = {0.41}.q = 1 - {0.41} = {0.59}.X \sim B\left( {{20},{0.41}}\right) \)\n\nFind \( P\left( {x \leq {12}}\right) .P\left( {x \leq {12}}\right) = {0.9738} \) . (calculator or computer)\n\nUsing the TI-83+ or the TI-84 calculators, the calculations are as follows. Go into 2nd DISTR. The syntax for the instructions are\n\nTo calculate \( \left( {x = \text{value}}\right) \) : binompdf \( \left( {n, p\text{, number}}\right) \) If \
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Yes
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A safety engineer feels that \( {35}\% \) of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in thepile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. On the average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least 3 reports until she finds a report showing an accident caused by employee failure to follow instructions?
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Let \( X = \) the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions. \( X \) takes on the values \( 1,2,3,\ldots \) . The first question asks you to find the expected value or the mean. The second question asks you to find \( P\left( {x \geq 3}\right) \) . (\
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No
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Suppose that you are looking for a student at your college who lives within five miles of you. You know that \( 55\% \) of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he/she lives within five miles of you. What is the probability that you need to contact four people?
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This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he/she lives within five miles of you. There is no definite number of trials (number of times you ask a student).
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No
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Let \( X = \) the number of students you must ask until one says yes.
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Let \( X = \) the number of students you must ask until one says yes.
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No
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Assume that the probability of a defective computer component is 0.02 . Components are randomly selected. Find the probability that the first defect is caused by the 7th component tested. How many components do you expect to test until one is found to be defective?
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Let \( X = \) the number of computer components tested until the first defect is found.\n\n\( X \) takes on the values \( 1,2,3,\ldots \) where \( p = {0.02}.X \sim \mathrm{G}\left( {0.02}\right) \)\n\nFind \( P\left( {x = 7}\right) .P\left( {x = 7}\right) = {0.0177} \) . (calculator or computer)\n\nTI-83+ and TI-84: For a general discussion, see this example (binomial). The syntax is similar. The geometric parameter list is (p, number) If \
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No
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A school site committee is to be chosen randomly from 6 men and 5 women. If the committee consists of 4 members chosen randomly, what is the probability that 2 of them are men? How many men do you expect to be on the committee?
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Let \( X = \) the number of men on the committee of 4 . The men are the group of interest (first group).\n\n\( X \) takes on the values \( 0,1,2,3,4 \), where \( r = 6, b = 5 \), and \( n = 4.X \sim H\left( {6,5,4}\right) \)\n\nFind \( P\left( {x = 2}\right) .P\left( {x = 2}\right) = {0.4545} \) (calculator or computer)\n\nNOTE: Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There are a number of computer packages, including Microsoft Excel, that do.\n\nThe probability that there are 2 men on the committee is about 0.45 .\n\nThe graph of \( X \sim H\left( {6,5,4}\right) \) is:\n\n\n\nThe \( y \) -axis contains the probability of \( X \), where \( X = \) the number of men on the committee.\n\nYou would expect \( m = {2.18} \) (about 2) men on the committee.\n\nThe formula for the mean is \( \mu = \frac{n \cdot r}{r + b} = \frac{4 \cdot 6}{6 + 5} = {2.18} \)
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Yes
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Leah's answering machine receives about 6 telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than 1 call in the next 15 minutes?
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Let \( X = \) the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or \( \frac{1}{4} \) hour.)\n\n\[ x = 0,1,2,3,\ldots \]\n\nIf Leah receives, on the average, 6 telephone calls in 2 hours, and there are eight 15 minutes intervals in 2 hours, then Leah receives\n\n\[ \frac{1}{8} \cdot 6 = {0.75} \]\n\ncalls in 15 minutes, on the average. So, \( \mu = {0.75} \) for this problem.\n\n\[ X \sim \mathrm{P}\left( {0.75}\right) \]\n\nFind \( P\left( {x > 1}\right) .P\left( {x > 1}\right) = {0.1734} \) (calculator or computer)\n\nTI-83+ and TI-84: For a general discussion, see this example (Binomial). The syntax is similar. The Poisson parameter list is ( \( \mu \) for the interval of interest, number). For this problem:\n\nPress 1- and then press 2nd DISTR. Arrow down to C:poissoncdf. Press ENTER. Enter .75,1). The result is \( P\left( {x > 1}\right) = {0.1734} \) . NOTE: The TI calculators use \( \lambda \) (lambda) for the mean.\n\nThe probability that Leah receives more than 1 telephone call in the next fifteen minutes is about 0.1734 .
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Yes
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Consider the function \( f\left( x\right) = \frac{1}{20} \) for \( 0 \leq x \leq {20}.x = \) a real number. The graph of \( f\left( x\right) = \frac{1}{20} \) is a horizontal line. However, since \( 0 \leq x \leq {20}, f\left( x\right) \) is restricted to the portion between \( x = 0 \) and \( x = {20} \), inclusive.
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The graph of \( f\left( x\right) = \frac{1}{20} \) is a horizontal line segment when \( 0 \leq x \leq {20} \). The area between \( f\left( x\right) = \frac{1}{20} \) where \( 0 \leq x \leq {20} \) and the \( \mathrm{x} \) -axis is the area of a rectangle with base \( = {20} \) and height \( = \frac{1}{20} \). AREA \( = {20} \cdot \frac{1}{20} = 1 \)
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Yes
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Illustrate the uniform distribution. The data that follows are 55 smiling times, in seconds, of an eight-week old baby.
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We will assume that the smiling times, in seconds, follow a uniform distribution between 0 and 23 seconds, inclusive. This means that any smiling time from 0 to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.\n\nLet \( X = \) length, in seconds, of an eight-week old baby’s smile.\n\nThe notation for the uniform distribution is\n\n\( X \sim U\left( {a, b}\right) \) where \( a = \) the lowest value of \( x \) and \( b = \) the highest value of \( x \) .\n\nThe probability density function is \( f\left( x\right) = \frac{1}{b - a} \) for \( a \leq x \leq b \) .\n\nFor this example, \( x \sim U\left( {0,{23}}\right) \) and \( f\left( x\right) = \frac{1}{{23} - 0} \) for \( 0 \leq x \leq {23} \) .\n\nFormulas for the theoretical mean and standard deviation are\n\n\[ \mu = \frac{a + b}{2}\text{and}\sigma = \sqrt{\frac{{\left( b - a\right) }^{2}}{12}} \]\n\nFor this problem, the theoretical mean and standard deviation are\n\n\[ \mu = \frac{0 + {23}}{2} = {11.50}\text{ seconds and }\sigma = \sqrt{\frac{{\left( {23} - 0\right) }^{2}}{12}} = {6.64}\text{ seconds } \]\n\nNotice that the theoretical mean and standard deviation are close to the sample mean and standard deviation.
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Yes
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What is the probability that a randomly chosen eight-week old baby smiles between 2 and 18 seconds?
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Find \( P\left( {2 < x < {18}}\right) \).\n\n\[ P\left( {2 < x < {18}}\right) = \text{(base) (height)} = \left( {{18} - 2}\right) \cdot \frac{1}{23} = \frac{16}{23}\text{.} \]
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Yes
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What is the probability that a person waits fewer than 12.5 minutes?
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Let \( X = \) the number of minutes a person must wait for a bus. \( a = 0 \) and \( b = {15}.x \sim U\left( {0,{15}}\right) \) . Write the probability density function. \( f\left( x\right) = \frac{1}{{15} - 0} = \frac{1}{15} \) for \( 0 \leq x \leq {15} \) .\n\nFind \( P\left( {x < {12.5}}\right) \) . Draw a graph.\n\n\[ P\left( {x < k}\right) = \text{(base) (height)} = \left( {{12.5} - 0}\right) \cdot \frac{1}{15} = {0.8333} \]\n\nThe probability a person waits less than 12.5 minutes is 0.8333 .
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Yes
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On the average, how long must a person wait? Find the mean, \( \mu \), and the standard deviation, \( \sigma \) .
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\( \mu = \frac{a + b}{2} = \frac{{15} + 0}{2} = {7.5} \) . On the average, a person must wait 7.5 minutes. \( \sigma = \sqrt{\frac{{\left( b - a\right) }^{2}}{12}} = \sqrt{\frac{{\left( {15} - 0\right) }^{2}}{12}} = {4.3} \) . The Standard deviation is 4.3 minutes.
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Yes
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Find the probability that a different nine-year old child eats a donut in more than 2 minutes given that the child has already been eating the donut for more than 1.5 minutes.
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First way: Since you already know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at \( a = {0.5} \) minutes. Your starting point is 1.5 minutes.\n\nWrite a new \( f\\left( x\\right) \) :\n\n\[ f\\left( x\\right) = \\frac{1}{4 - {1.5}} = \\frac{2}{5}\\;\\text{ for }{1.5} \\leq x \\leq 4. \]\n\nFind \( P\\left( {x > 2 \\mid x > {1.5}}\\right) \) . Draw a graph.\n\n\( \\mathrm{f}\\left( \\mathrm{x}\\right) \) \n\n\( P\\left( {x > 2 \\mid x > {1.5}}\\right) = \) (base) (new height) \( = \\left( {4 - 2}\\right) \\left( {2/5}\\right) = \) ?\n\nThe probability that a nine-year old child eats a donut in more than 2 minutes given that the child has already been eating the donut for more than 1.5 minutes is \( \\frac{4}{5} \).\n\nSecond way: Draw the original graph for \( x \\sim U\\left( {{0.5},4}\\right) \) . Use the conditional formula\n\n\[ P\\left( {x > 2|x > {1.5}}\\right) = \\frac{P\\left( {x > 2\\text{ AND }x > {1.5}}\\right) }{P\\left( {x > {1.5}}\\right) } = \\frac{P\\left( {x > 2}\\right) }{P\\left( {x > {1.5}}\\right) } = \\frac{\\frac{2}{3.5}}{\\frac{2.5}{3.5}} = {0.8} = \\frac{4}{5} \]
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Yes
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Find the probability that a randomly selected furnace repair requires longer than 2 hours.
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To find \( f\left( x\right) : f\left( x\right) = \frac{1}{4 - {1.5}} = \frac{1}{2.5} \) so \( f\left( x\right) = {0.4} \)\n\n\( \mathrm{P}\left( {\mathrm{x} > 2}\right) = \) (base)(height) \( = \left( {4 - 2}\right) \left( {0.4}\right) = {0.8} \) Example 4 Figure 1\n\n\n\nFigure 5.4: Uniform Distribution between 1.5 and 4 with shaded area between 2 and 4 representing the probability that the repair time \( x \) is greater than 2
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Yes
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The longest \( {25}\% \) of furnace repair times take at least how long? (Find the minimum time for the longest \( {25}\% \) of repairs.)
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\n\nFigure 5.7: Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest \( {25}\% \) of repair times.\n\n\[ P\left( {x > k}\right) = {0.25} \]\n\n\[ P\left( {x > k}\right) = \text{(base) (height)} = \left( {4 - k}\right) \cdot \left( {0.4}\right) \]\n\n\[ {0.25} = \left( {4 - \mathrm{k}}\right) \left( {0.4}\right) \text{; Solve for}\mathrm{k}\text{:} \]\n\n\( {0.625} = 4 - \mathrm{k} \), obtained by dividing both sides by 0.4\n\n\( - {3.375} = - \mathrm{k} \), obtained by subtracting 4 from both sides\n\n\( \mathrm{k} = {3.375} \)\n\nThe longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer).\n\nNote: Since \( {25}\% \) of repair times are 3.375 hours or longer, that means that \( {75}\% \) of repair times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times.
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Yes
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Illustrates the exponential distribution: Let \( X = \) amount of time (in minutes) a postal clerk spends with his/her customer. The time is known to have an exponential distribution with the average amount of time equal to 4 minutes.
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X is a continuous random variable since time is measured. It is given that \( \mu = 4 \) minutes. To do any calculations, you must know \( m \), the decay parameter.\n\n\( m = \frac{1}{\mu } \) . Therefore, \( m = \frac{1}{4} = {0.25} \)\n\nThe standard deviation, \( \sigma \), is the same as the mean. \( \mu = \sigma \)\n\nThe distribution notation is \( X \sim \operatorname{Exp}\left( m\right) \) . Therefore, \( X \sim \operatorname{Exp}\left( {0.25}\right) \).\n\nThe probability density function is \( f\left( x\right) = m \cdot {e}^{-m \cdot x} \) The number \( e = {2.71828182846}\ldots \) It is a number that is used often in mathematics. Scientific calculators have the key \
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Yes
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Find the probability that a clerk spends four to five minutes with a randomly selected customer.
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Find \( P\left( {4 < x < 5}\right) \) .\n\nThe cumulative distribution function (CDF) gives the area to the left.\n\n\[ P\left( {x < x}\right) = 1 - {e}^{-m \cdot x} \]\n\n\[ P\left( {x < 5}\right) = 1 - {e}^{-{0.25} \cdot 5} = {0.7135}\text{and}P\left( {x < 4}\right) = 1 - {e}^{-{0.25} \cdot 4} = {0.6321} \]\n\nThe probability that a postal clerk spends four to five minutes with a randomly selected customer is\n\n\[ \begin{matrix} P\left( {4 < x < 5}\right) = P\left( {x < 5}\right) - P\left( {x < 4}\right) = {0.7135} - {0.6321} = {0.0814} \end{matrix} \]\n\nNOTE: TI-83+ and TI-84: On the home screen, enter (1-e^(-.25*5))-(1-e^(-.25*4)) or enter e^(-.25*4)- \( {\mathrm{e}}^{ \land }\left( {-{.25} * 5}\right) \) .
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Yes
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What is the probability that a computer part lasts more than 7 years?
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Let \( x = \) the amount of time (in years) a computer part lasts.\n\n\[ \mu = {10}\text{so}m = \frac{1}{\mu } = \frac{1}{10} = {0.1} \]\n\nFind \( P\left( {x > 7}\right) \) . Draw a graph.\n\n\[ P\left( {x > 7}\right) = 1 - P\left( {x < 7}\right) . \]\n\nSince \( P\left( {X < x}\right) = 1 - {e}^{-{mx}} \) then \( P\left( {X > x}\right) = 1 - \left( {1 - {e}^{-m \cdot x}}\right) = {e}^{-m \cdot x} \)\n\n\( P\left( {x > 7}\right) = {e}^{-{0.1} \cdot 7} = {0.4966} \) . The probability that a computer part lasts more than 7 years is 0.4966 .
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Yes
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Suppose \( X \sim N\left( {5,6}\right) \). This says that \( X \) is a normally distributed random variable with mean \( \mu = 5 \) and standard deviation \( \sigma = 6 \). Suppose \( x = {17} \). Then:
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\[ z = \frac{x - \mu }{\sigma } = \frac{{17} - 5}{6} = 2 \] This means that \( x = {17} \) is 2 standard deviations \( \left( {2\sigma }\right) \) above or to the right of the mean \( \mu = 5 \). The standard deviation is \( \sigma = 6 \). Notice that: \[ 5 + 2 \cdot 6 = {17}\;\text{ (The pattern is }\mu + {z\sigma } = x.\text{) } \]
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Yes
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Suppose the random variables \( X \) and \( Y \) have the following normal distributions: \( X \sim N\left( {5,6}\right) \) and \( Y \sim N\left( {2,1}\right) \) . If \( x = {17} \), then \( z = 2 \) . (This was previously shown.) If \( y = 4 \), what is \( z \) ?
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\[ z = \frac{y - \mu }{\sigma } = \frac{4 - 2}{1} = 2\;\text{ where }\mu = 2\text{ and }\sigma = 1. \]
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Yes
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Find the probability that a household personal computer is used between 1.8 and 2.75 hours per day.
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Let \( X = \) the amount of time (in hours) a household personal computer is used for entertainment. \( x \sim N\left( {2,{0.5}}\right) \) where \( \mu = 2 \) and \( \sigma = {0.5} \) .\n\nFind \( P\left( {{1.8} < x < {2.75}}\right) \) .\n\nThe probability for which you are looking is the area between \( x = {1.8} \) and \( x = \) 2.75. \( P\left( {{1.8} < x < {2.75}}\right) = {0.5886} \)\n\n\n\nnormalcdf(1.8,2.75,2,0.5) \( = {0.5886} \)\n\nThe probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886 .
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Yes
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Find the probability that the sample mean is between 85 and 92.
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Let \( X = \) one value from the original unknown population. The probability question asks you to find a probability for the sample mean.\n\nLet \( \bar{X} = \) the mean of a sample of size 25 . Since \( {\mu }_{X} = {90},{\sigma }_{X} = {15} \), and \( n = {25} \) ;\n\nthen \( \bar{X} \sim N\left( {{90},\frac{15}{\sqrt{25}}}\right) \)\n\nFind \( P\left( {{85} < \bar{x} < {92}}\right) \; \) Draw a graph.\n\n\( P\left( {{85} < \bar{x} < {92}}\right) = {0.6997} \)\n\nThe probability that the sample mean is between 85 and 92 is 0.6997 .\n\n\n\nTI-83 or 84: normal cdf(lower value, upper value, mean, standard error of the mean)\n\nThe parameter list is abbreviated (lower value, upper value, \( \mu ,\frac{\sigma }{\sqrt{n}} \) )\n\n\( \operatorname{normal}\operatorname{cdf}\left( {{85},{92},{90},\frac{15}{\sqrt{25}}}\right) = {0.6997} \)
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Yes
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Find the value that is 2 standard deviations above the expected value (it is 90) of the sample mean.
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To find the value that is 2 standard deviations above the expected value 90 , use the formula\n\n\\( \\mathrm{{value}} = {\\mu }_{X} + \\left( {\\# \\mathrm{{of}}\\mathrm{{STDEVs}}}\\right) \\left( \\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right) \\)\n\nvalue \\( = {90} + 2 \\cdot \\frac{15}{\\sqrt{25}} = {96} \\)\n\nSo, the value that is 2 standard deviations above the expected value is 96 .
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Yes
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a. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7500.
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Let \( X \) = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.\n\n\( {\sum X} = \) the sum or total of 80 values. Since \( {\mu }_{X} = {90},{\sigma }_{X} = {15} \), and \( n = {80} \), then\n\n\( {\sum X} \sim N\left( {{80} \cdot {90},\sqrt{80} \cdot {15}}\right) \)\n\n-. mean of the sums \( = n \cdot {\mu }_{X} = \left( {80}\right) \left( {90}\right) = {7200} \)\n\n-. standard deviation of the sums \( = \sqrt{n} \cdot {\sigma }_{X} = \sqrt{80} \cdot {15} \)\n\n-. sum of 80 values \( = {\sum x} = {7500} \)\n\na: Find \( P\left( {{\sum x} > {7500}}\right) \)\n\n---\n\n\( P\left( {{\sum x} > {7500}}\right) = {0.0127} \)\n\n\n\nnormal cdf(lower value, upper value, mean of sums, stdev of sums)\n\nThe parameter list is abbreviated (lower, upper, \( n \cdot {\mu }_{X},\sqrt{n} \cdot {\sigma }_{X} \) )\n\nnormal cdf \( ({7500},1\mathrm{E}{99},{80} \cdot {90},\sqrt{80} \cdot {15} = {0.0127} \)\n\nReminder: \( {1E99} = {10}^{99} \). Press the EE key for E.
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Yes
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The probability that the mean stress score for the 75 students is less than 2.
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Since the individual stress scores follow a uniform distribution, \( X \sim U\left( {1,5}\right) \) where \( a = 1 \) and \( b = 5 \) (See Continuous Random Variables (Section 5.1) for the uniform).\n\n\[{\mu }_{X} = \frac{a + b}{2} = \frac{1 + 5}{2} = 3\]\n\n\[{\sigma }_{X} = \sqrt{\frac{{\left( b - a\right) }^{2}}{12}} = \sqrt{\frac{{\left( 5 - 1\right) }^{2}}{12}} = {1.15}\]\n\nFor problems 1. and 2., let \( \bar{X} = \) the mean stress score for the 75 students. Then,\n\n\[\bar{X} \sim N\left( {3,\frac{1.15}{\sqrt{75}}}\right) \;\text{ where }n = {75}.\]
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No
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Find \( P\left( {\bar{x} < 2}\right) \) . Draw the graph.
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\[ P\left( {\bar{x} < 2}\right) = 0 \]\n\nThe probability that the mean stress score is less than 2 is about 0 .\n\n\[ \mathrm{P}\left( {\overline{\mathrm{x}} < 2}\right) \]\n\n\n\nnormalcdf \( \left( {1,2,3,\frac{1.15}{\sqrt{75}}}\right) = 0 \)\n\nREMINDER: The smallest stress score is 1 . Therefore, the smallest mean for 75 stress scores is 1 .
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Yes
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Find \( P\left( {{\sum x} < {200}}\right) \) . Draw the graph.
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The mean of the sum of 75 stress scores is \( {75} \cdot 3 = {225} \)\n\nThe standard deviation of the sum of 75 stress scores is \( \sqrt{75} \cdot {1.15} = {9.96} \)\n\n\[ P\left( {{\sum x} < {200}}\right) = 0 \]\n\n\n\nThe probability that the total of 75 scores is less than 200 is about 0 .\n\nnormal cdf \( \left( {{75},{200},{75} \cdot 3,\sqrt{75} \cdot {1.15}}\right) = 0. \)\n\nREMINDER: The smallest total of 75 stress scores is 75 since the smallest single score is 1 .
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Yes
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a. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find \( P\\left( {\\bar{x} > {20}}\\right) \)
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\[ P\\left( {\\bar{x} > {20}}\\right) = {0.7919}\\text{using normalcdf}\\left( {{20},{1E99},{22},\\frac{22}{\\sqrt{80}}}\\right) \]\n\nThe probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.
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Yes
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Find the probability that at least 150 favor a charter school.
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For Problem 1., you include \( \mathbf{{150}} \) so \( P\left( {x \geq {150}}\right) \) has normal approximation \( P\left( {Y \geq {149.5}}\right) = {0.8641}. \n\nnormal cdf \( \left( {{149.5},{10}^{ \land }{99},{159},{8.6447}}\right) = {0.8641} \).
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Yes
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Suppose we have collected data from a sample. We know the sample mean but we do not know the mean for the entire population. The sample mean is 7 and the error bound for the mean is 2.5.
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The confidence interval is \( \left( {7 - {2.5},7 + {2.5}}\right) \) ; calculating the values gives \( \left( {{4.5},{9.5}}\right) \). If the confidence level (CL) is \( {95}\% \), then we say that \
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No
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