Question 14 (Essay Worth 20 points) ( \( 04.05 \mathrm{HC} \mid 04.06 \mathrm{HC} \) ) While playing a board game, Isaiah noticed that the die landed on the number 5 more often than usual. Part A: Describe a simulation that could be run to test how many times out of 100 a fair die should land on the number 5 . State the representations and possible outcomes. Be sure to give enough detail that another person could replicate your simulation. ( 7 points) Part B: While running a simulation, the die landed on the number 5 a total of 29 times out of the 100 rolls. Construct and interpret a \( 95 \% \) confidence interval for the true proportion of rolls that will land on the number 5 . Show all work. ( 7 points) Part C: Does the confidence interval in part B support Isaiah's suspicions that the die is not fair? Explain your reasoning. (6 points) Paragraph \( T_{x} \) B \( \underline{u} \) \( x_{2} \)

**Part A** We want to simulate 100 rolls of a fair six‐sided die to record how many times a 5 appears. One way to do this is as follows: 1. Define a “success” as the outcome when the die shows a 5 and a “failure” as when it shows any other number (i.e., 1, 2, 3, 4, or 6). 2. Use a random number generator (or a physical random process) that gives an outcome equally likely for each number from 1 to 6. 3. Repeat the following process 100 times: - Roll the die (or simulate a roll) and record the outcome. - Note whether the outcome is a 5 (success) or not (failure). 4. The representation is a sequence of 100 independent trials, where each trial has two possible outcomes (“5” or “not 5”). Any person can replicate this by using a computer program that randomly generates integers between 1 and 6, a spinner divided into six equal regions numbered 1 to 6, or an actual die rolled on a flat surface. The simulation will give an estimate of the number of times, out of 100, that a fair die lands on 5. --- **Part B** In the simulation, the die landed on 5 a total of 29 times out of 100. We assume that the number of successes follows a binomial distribution. The sample proportion is \[ \hat{p} = \frac{29}{100} = 0.29. \] The standard error (SE) for the estimate is calculated by \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.29 \times 0.71}{100}}. \] Calculating the value: \[ 0.29 \times 0.71 \approx 0.2059,\quad \text{so} \quad SE \approx \sqrt{0.002059} \approx 0.0454. \] For a \(95\%\) confidence interval, we use the critical value \(z^{*} \approx 1.96\). The margin of error (ME) is \[ ME = 1.96 \times 0.0454 \approx 0.089. \] Thus, the \(95\%\) confidence interval for the true proportion \(p\) is \[ \hat{p} \pm ME = 0.29 \pm 0.089. \] In interval notation, this is \[ (0.29 - 0.089,\, 0.29 + 0.089) \approx (0.201,\, 0.379). \] --- **Part C** A fair die has a probability of landing on 5 equal to \(\frac{1}{6} \approx 0.167\). The \(95\%\) confidence interval for the true proportion from our simulation is approximately \((0.201,\, 0.379)\). Since \(\frac{1}{6} \approx 0.167\) is not contained within the interval, the evidence suggests that the true proportion of outcomes that are 5 is significantly higher than would be expected from a fair die. This supports Isaiah's suspicion that the die is not fair.