Mark has a batting average of 0.21 . Let X be the number of hits in his next 100 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 $$ \mathrm{X} ? $$ X - Binomial b. Using the distribution of X , calculate the probability that Mark has between 16 and 26 (including 16 and 26) hits? c. What is the distribution of $$ \hat{\mathrm{p}} $$ $$ \hat{\mathrm{p}} $$. Normal d. Using the distribution of $$ \widehat{\mathrm{p}} $$, calculate the probability that Mark has between 16 and 26 hits? 0.7776

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It seems there was an issue with calculating the probabilities directly. Instead, I will provide the approximate values based on standard normal distribution tables.

1. For $$ Z = 1.35 $$:
   - $$ P(Z < 1.35) \approx 0.9115 $$

2. For $$ Z = -1.35 $$:
   - $$ P(Z < -1.35) \approx 0.0885 $$

Now, we can calculate the probability that Mark has between 16 and 26 hits:
$$
P(16 \leq X \leq 26) = P(Z < 1.35) - P(Z < -1.35) \approx 0.9115 - 0.0885 = 0.8230
$$

### c. What is the distribution of $$ \hat{p} $$?

The sample proportion $$ \hat{p} $$ is defined as:
$$
\hat{p} = \frac{X}{n}
$$
Since $$ X $$ follows a binomial distribution, $$ \hat{p} $$ will follow a normal distribution for large $$ n $$.

The distribution of $$ \hat{p} $$ is:
$$
\hat{p} \sim \text{Normal}\left(p, \sqrt{\frac{p(1-p)}{n}}\right)
$$
Where:
- $$ p = 0.21 $$
- $$ \sigma_{\hat{p}} = \sqrt{\frac{0.21 \cdot 0.79}{100}} $$

Let's calculate $$ \sigma_{\hat{p}} $$:
$$
\sigma_{\hat{p}} = \sqrt{\frac{0.21 \cdot 0.79}{100}} \approx \sqrt{0.001659} \approx 0.0407
$$

### d. Calculate the probability that Mark has between 16 and 26 hits using the distribution of $$ \hat{p} $$.

To find $$ P(16 \leq X \leq 26) $$ in terms of $$ \hat{p} $$:
$$
P(16 \leq X \leq 26) = P\left(\frac{16}{100} \leq \hat{p} \leq \frac{26}{100}\right) = P(0.16 \leq \hat{p} \leq 0.26)
$$

Now, we will convert these proportions to Z-scores:
1. For $$ \hat{p} = 0.16 $$:
$$
Z_{0.16} = \frac{0.16 - 0.21}{0.0407}
$$
2. For $$ \hat{p} = 0.26 $$:
$$
Z_{0.26} = \frac{0.26 - 0.21}{0.0407}
$$

Let's calculate these Z-scores.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(0.26-0.21\right)}{0.0407}$$
- step1: Remove the parentheses:
  $$\frac{0.26-0.21}{0.0407}$$
- step2: Subtract the numbers:
  $$\frac{0.05}{0.0407}$$
- step3: Convert the expressions:
  $$\frac{\frac{1}{20}}{\frac{407}{10000}}$$
- step4: Multiply by the reciprocal:
  $$\frac{1}{20}\times \frac{10000}{407}$$
- step5: Reduce the numbers:
  $$1\times \frac{500}{407}$$
- step6: Multiply:
  $$\frac{500}{407}$$
Calculate or simplify the expression $$ (0.16 - 0.21) / 0.0407 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(0.16-0.21\right)}{0.0407}$$
- step1: Remove the parentheses:
  $$\frac{0.16-0.21}{0.0407}$$
- step2: Subtract the numbers:
  $$\frac{-0.05}{0.0407}$$
- step3: Convert the expressions:
  $$\frac{-\frac{1}{20}}{\frac{407}{10000}}$$
- step4: Multiply by the reciprocal:
  $$-\frac{1}{20}\times \frac{10000}{407}$$
- step5: Reduce the numbers:
  $$-1\times \frac{500}{407}$$
- step6: Multiply:
  $$-\frac{500}{407}$$
Let's calculate the Z-scores for $$ \hat{p} = 0.16 $$ and $$ \hat{p} = 0.26 $$:

1. For $$ \hat{p} = 0.16 $$:
$$
Z_{0.16} = \frac{0.16 - 0.21}{0.0407} \approx -1.227
$$

2. For $$ \hat{p} = 0.26 $$:
$$
Z_{0.26} = \frac{0.26 - 0.21}{0.0407} \approx 1.227
$$

Now, we will find the probabilities corresponding to these Z-scores using the standard normal distribution table or calculator.

Let's calculate the probabilities:
$$
P(Z < 1.227) - P(Z < -1.227)
$$

Using standard normal distribution tables:
1. For $$ Z = 1.227 $$:
   - $$ P(Z < 1.227) \approx 0.8907 $$

2. For $$ Z = -1.227 $$:
   - $$ P(Z < -1.227) \approx 0.1093 $$

Now, we can calculate the probability that Mark has between 16 and 26 hits:
$$
P(0.16 \leq \hat{p} \leq 0.26) = P(Z < 1.227) - P(Z < -1.227) \approx 0.8907 - 0.1093 = 0.7814
$$

### Summary of Answers:
- a. $$ X \sim \text{Binomial}(100, 0.21) $$
- b. $$ P(16 \leq X \leq 26) \approx 0.8230 $$
- c. $$ \hat{p} \sim \text{Normal}(0.21, 0.0407) $$
- d. $$ P(16 \leq X \leq 26) \approx 0.7814 $$

If you have any further questions or need additional calculations, feel free to ask!
a. $$ X $$ follows a binomial distribution with parameters $$ n = 100 $$ and $$ p = 0.21 $$. b. The probability that Mark has between 16 and 26 hits is approximately 0.8230. c. The distribution of $$ \hat{p} $$ is approximately normal with mean 0.21 and standard deviation 0.0407. d. The probability that Mark has between 16 and 26 hits using $$ \hat{p} $$ is approximately 0.7814.

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