Questions: Practice Homework for Exam Two Score: 0/0 Answered: 26/56 Question 39 Formulas: np ; sqrt(np(1-p)) ; sqrt(p(1-p)/n) ; σ/sqrt(n) Here is the quick link to the scientific calculator, and from there, you may navigate to any online statistics calculators using the "Content" tab on the upper-left-hand side. Mark has a batting average of 0.26. Let X be the number of hits in his next 80 at bats, and let p̂ be the sample proportion of Mark's hits. Please answer the following questions, and show your answers to 4 decimal places if necessary. a. What is the distribution of X? ( , ) b. Use the distribution of X, calculate the probability that Mark has between 16 and 21 (including 16 and 21) hits? c. What is the distribution of p̂? ( , ) d. Use the distribution of p̂, calculate the probability that Mark has between 16 and 21 hits? Submit Question

Practice Homework for Exam Two

Score: 0/0 Answered: 26/56

Question 39

Formulas: np ; sqrt(np(1-p)) ; sqrt(p(1-p)/n) ; σ/sqrt(n)

Here is the quick link to the scientific calculator, and from there, you may navigate to any online statistics calculators using the "Content" tab on the upper-left-hand side.

Mark has a batting average of 0.26. Let X be the number of hits in his next 80 at bats, and let p̂ be the sample proportion of Mark's hits. Please answer the following questions, and show your answers to 4 decimal places if necessary.

a. What is the distribution of X? ( , )

b. Use the distribution of X, calculate the probability that Mark has between 16 and 21 (including 16 and 21) hits?

c. What is the distribution of p̂? ( , )

d. Use the distribution of p̂, calculate the probability that Mark has between 16 and 21 hits?

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Transcript text: Practice Homework for Exam Two Score: 0/0 Answered: 26/56 Question 39 Formulas: $np ; \sqrt{np(1-p)} ; \sqrt{\frac{p(1-p)}{n}} ; \frac{\sigma}{\sqrt{n}}$ Here is the quick link to the scientific calculator, and from there, you may navigate to any online statistics calculators using the "Content" tab on the upper-left-hand side. Mark has a batting average of 0.26. Let X be the number of hits in his next 80 at bats, and let $\hat{p}$ be the sample proportion of Mark's hits. Please answer the following questions, and show your answers to 4 decimal places if necessary. a. What is the distribution of X? $\square$ ( $\square$ , $\square$ ) b. Use the distribution of X, calculate the probability that Mark has between 16 and 21 (including 16 and 21) hits? $\square$ c. What is the distribution of $\hat{p}$? $\square$ ( $\square$ , $\square$ ) d. Use the distribution of $\hat{p}$, calculate the probability that Mark has between 16 and 21 hits? $\square$ Submit Question

Solution

Solution Steps

Solution Approach

a. The distribution of $ X $ can be modeled using a binomial distribution with parameters $ n = 80 $ and $ p = 0.26 $.

b. To calculate the probability that Mark has between 16 and 21 hits, we can use the cumulative distribution function (CDF) of the binomial distribution.

c. The distribution of $ \hat{p} $ (the sample proportion) can be approximated using a normal distribution with mean $ p $ and standard deviation $ \sqrt{\frac{p(1-p)}{n}} $.

Step 1: Determine the Distribution of $ X $

The number of hits $ X $ in 80 at bats can be modeled using a binomial distribution with parameters $ n = 80 $ and $ p = 0.26 $. Therefore, the distribution of $ X $ is: \[ X \sim \text{Binomial}(80, 0.26) \]

Step 2: Calculate the Probability of 16 to 21 Hits

To find the probability that Mark has between 16 and 21 hits (inclusive), we use the cumulative distribution function (CDF) of the binomial distribution: \[ P(16 \leq X \leq 21) = P(X \leq 21) - P(X \leq 15) \] Using the CDF values: \[ P(X \leq 21) \approx 0.5194 \] \[ P(X \leq 15) \approx 0.0260 \] Thus: \[ P(16 \leq X \leq 21) \approx 0.5194 - 0.0260 = 0.4934 \]

Step 3: Determine the Distribution of $ \hat{p} $

The sample proportion $ \hat{p} $ can be approximated using a normal distribution with mean $ p $ and standard deviation $ \sqrt{\frac{p(1-p)}{n}} $. Therefore, the distribution of $ \hat{p} $ is: \[ \hat{p} \sim \mathcal{N}\left(0.26, \sqrt{\frac{0.26 \cdot 0.74}{80}}\right) \] Calculating the standard deviation: \[ \sigma_{\hat{p}} = \sqrt{\frac{0.26 \cdot 0.74}{80}} \approx 0.0490 \] Thus, the distribution of $ \hat{p} $ is: \[ \hat{p} \sim \mathcal{N}(0.26, 0.0490) \]