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Introduction to Probability and Statistics for the Life Sciences Spring 2026 MATH 1044

Andy Eisenberg

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Worksheet Quiz 3

The following work should be completed individually. Use of notes or textbooks is not allowed. You may use a scientific calculator, not a graphing calculator or phone app.
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Show all work unless instructed otherwise.
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1.

Write either True or False for each of the following statements. No justification is required.
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(a)

Suppose a parameter \(\theta \in (-\infty, \infty)\) has likelihood function \(\mathcal{L}(\theta)\) with \(\L'(\theta) = (1 - \theta)e^{\theta}\text{.}\) Then the maximum likelihood estimation is \(\widehat{\theta} = 1\text{.}\)
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Solution.
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(b)

Suppose \(X_1, \dotsc, X_n\) are random variables, and \(S = X_1 + \dotsb + X_n\text{.}\) Then, for sufficiently large \(n\text{,}\) \(S \approx \operatorname{N}(0, 1)\text{.}\)
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Solution.
False.
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(c)

Suppose we collect data to estimate the value of a parameter \(\theta\text{.}\) Based on the collected data, we find a 95% confidence interval \([a, b]\) and a 90% confidence interval \([c, d]\text{.}\) Then \(a \leq c \leq d \leq b\text{.}\)
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Solution.
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2.

The heights of five plants are measured and recorded below. Find the sample mean, sample variance, and a 95% confidence interval around the sample mean for the heights of the plants. (Pretend that 5 measurements is enough for the CLT to apply.)
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Plant \(i\) \(H_i\) (in)
\(1\) \(13\)
\(2\) \(14\)
\(3\) \(10\)
\(4\) \(15\)
\(5\) \(17\)
Solution.
Let \(A = \frac{H_1 + \dotsb + H_5}{5}\) be the average height of the sample. Then:
\begin{align*} A \amp = 13.8 \amp s^2 \amp = \frac{\sum H_i^2 - n(A^2)}{n - 1} = \frac{979 - 5(13.8^2)}{4} = 6.7 \\ \sum H_i^2 \amp = 979 \amp \sqrt{\frac{s^2}{n}} \amp = \sqrt{\frac{6.7}{5}} \approx 1.16 \end{align*}
So the 95% confidence limits are:
\begin{align*} \mu_{\ell} \amp = 13.8 - 1.96(1.16) \approx \boxed{11.53} \\ \mu_{h} \amp = 13.8 + 1.96(1.16) \approx \boxed{16.07} \end{align*}
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3.

Suppose a coin comes up heads with probability 0.6. We flip the coin 80 times, and let \(S\) be the number of heads. Estimate \(\Pr(42 \leq S \leq 50)\text{.}\) (Use a continuity correction if appropriate.)
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Solution.
\(S \sim \Bin(80, 0.6)\text{,}\) so \(\E(S) = (80)(0.6) = 48\) and \(\Var(S) = (80)(0.6)(0.4) = 19.2\text{.}\) By the CLT, \(S \approx \Norm(48, 19.2)\text{,}\) so:
\begin{align*} \Pr(42 \leq S \leq 50) \amp \approx \Pr\left( \frac{41.5 - 48}{\sqrt{19.2}} \leq Z \leq \frac{50.5 - 48}{\sqrt{19.2}}\right) \\ \amp \approx \Pr(-1.48 \leq Z \leq 0.57) \\ \amp = \Phi(0.57) - \Phi(-1.48) \\ \amp \approx 0.7157 - 0.0694 \\ \amp = \boxed{0.6463} \end{align*}
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Table 126. \(\Phi(z)\) Values
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\(z\) 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
−1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455
−1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559
−1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681
−1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823
−1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549
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