andy/Fall-2026-Math-1041
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Update to PreTeXt project source.

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andy
2026-04-09 19:54:48 -04:00
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source/exams/exam-02.ptx
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@@ -187,7 +187,7 @@
<solution> <solution>
<p> <p>
<m>S</m> is binomially distributed with <m>E(S) = (180)(0.7) = 126</m> and <m>\Var(S) = (180)(0.7)(0.3) = 37.8</m>. Therefore: <m>S</m> is binomially distributed with <m>\E(S) = (180)(0.7) = 126</m> and <m>\Var(S) = (180)(0.7)(0.3) = 37.8</m>. Therefore:
<md> <md>
<mrow> \Pr(120 \leq S \leq 130) \amp \approx \Pr\left( \frac{119.5 - 126}{\sqrt{37.8}} \leq Z \leq \frac{130.5 - 126}{\sqrt{37.8}}\right) </mrow> <mrow> \Pr(120 \leq S \leq 130) \amp \approx \Pr\left( \frac{119.5 - 126}{\sqrt{37.8}} \leq Z \leq \frac{130.5 - 126}{\sqrt{37.8}}\right) </mrow>
<mrow> \amp \approx \Pr(-1.06 \leq Z \leq 0.73) </mrow> <mrow> \amp \approx \Pr(-1.06 \leq Z \leq 0.73) </mrow>
@@ -204,7 +204,7 @@
<statement> <statement>
<p> <p>
The weights of five apples are measured and recorded below. The weights of five apples are measured and recorded below.
Find the sample mean, sample variance, and a 95\% confidence interval around the sample mean for the weights of the apples. Find the sample mean, sample variance, and a 95% confidence interval around the sample mean for the weights of the apples.
(Pretend that 5 measurements is enough for the CLT to apply. (Pretend that 5 measurements is enough for the CLT to apply.
Do not use the <m>t</m>-distribution.) Do not use the <m>t</m>-distribution.)
</p> </p>
@@ -329,11 +329,11 @@
If the coin is fair, then <m>\E(S) = (160)(0.5) = 80</m> and <m>\Var(S) = (160)(0.5)(0.5) = 40</m>. If the coin is fair, then <m>\E(S) = (160)(0.5) = 80</m> and <m>\Var(S) = (160)(0.5)(0.5) = 40</m>.
The probability of rejecting <m>H_0</m> is then: The probability of rejecting <m>H_0</m> is then:
<md> <md>
<mrow> \Pr(S \geq 75) \amp \approx \Pr\left(Z \geq \frac{75 - 80}{\sqrt{40}}\right) </mrow> <mrow> \Pr(S \geq 75) \amp \approx \Pr\left(Z \geq \frac{74.5 - 80}{\sqrt{40}}\right) </mrow>
<mrow> \amp \approx \Pr(Z \geq -0.79) </mrow> <mrow> \amp \approx \Pr(Z \geq -0.87) </mrow>
<mrow> \amp \approx 1 - \Phi(-0.79) </mrow> <mrow> \amp \approx 1 - \Phi(-0.87) </mrow>
<mrow> \amp \approx 1 - 0.2148 </mrow> <mrow> \amp \approx 1 - 0.1922 </mrow>
<mrow> \amp = 0.7852. </mrow> <mrow> \amp = 0.8078. </mrow>
</md> </md>
</p> </p>
</solution> </solution>