Question

interpret the standard deviation.
in srss of size n = 500, the sample proportion of people who received a jury summons in the previous 12 months typically varies from the true proportion by about 0.234.
(b) justify that the sampling distribution of \\(\hat{p}\\) is approximately normal.
\\(np = \square\\) and \\(n(1 - p) = \square\\).
the sampling distribution of \\(\hat{p}\\) is approximately normal because \\(np\\) and \\(n(1 - p)\\) are \\(\square\\).
(c) calculate the probability that at most 13% of the residents in
the sample received a jury summons in the previous 12 months.
(round to 4 decimal places. leave your answer in decimal form.)

Explanation:

To solve this problem, we'll address each part step by step. However, we need to assume the value of the population proportion \( p \) (usually, if not given, maybe from a previous part or a common scenario, but let's assume \( p = 0.15 \) for illustration, or maybe it's from a context we missed. Wait, actually, maybe the first part (a) had \( p \), but since it's not here, let's proceed with the general method.

Part (b): Justify Sampling Distribution of \( \hat{p} \) is Normal

For the sampling distribution of the sample proportion \( \hat{p} \) to be approximately normal, we need \( np \geq 10 \) and \( n(1 - p) \geq 10 \) (or sometimes 5, but 10 is standard).

Let’s assume \( p = 0.15 \) (common in jury summons, but adjust if needed). Given \( n = 500 \):

• \( np = 500 \times p \)
• \( n(1 - p) = 500 \times (1 - p) \)

If \( p = 0.15 \):
\( np = 500 \times 0.15 = 75 \)
\( n(1 - p) = 500 \times 0.85 = 425 \)

Both \( 75 \) and \( 425 \) are greater than \( 10 \) (or 5), so the sampling distribution of \( \hat{p} \) is approximately normal because \( np \) and \( n(1 - p) \) are both at least 10 (or 5, depending on the rule).

Part (c): Probability that at most 13% (0.13) of the sample received a jury summons

To calculate \( P(\hat{p} \leq 0.13) \), we use the normal approximation to the sampling distribution of \( \hat{p} \).

Step 1: Find the mean and standard deviation of \( \hat{p} \)

• Mean of \( \hat{p} \): \( \mu_{\hat{p}} = p \)
• Standard deviation of \( \hat{p} \): \( \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \)

Assume \( p = 0.15 \) (from part (b) assumption):
\( \mu_{\hat{p}} = 0.15 \)
\( \sigma_{\hat{p}} = \sqrt{\frac{0.15 \times 0.85}{500}} = \sqrt{\frac{0.1275}{500}} = \sqrt{0.000255} \approx 0.01597 \)

Step 2: Calculate the z-score

The z-score for \( \hat{p} = 0.13 \) is:
\( z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.13 - 0.15}{0.01597} \approx \frac{-0.02}{0.01597} \approx -1.25 \)

Step 3: Find the probability from the z-table

\( P(Z \leq -1.25) \) is the area to the left of \( z = -1.25 \) in the standard normal distribution. From the z-table, \( P(Z \leq -1.25) = 0.1056 \)

Final Answers (Assuming \( p = 0.15 \))

• (b) \( np = 75 \), \( n(1 - p) = 425 \); both are at least 10 (or 5).
• (c) Probability ≈ \( 0.1056 \)

(Note: If the actual \( p \) is different, recalculate with the correct \( p \). For example, if \( p = 0.16 \), \( np = 80 \), \( n(1 - p) = 420 \), z-score would be \( \frac{0.13 - 0.16}{\sqrt{\frac{0.16 \times 0.84}{500}}} \approx \frac{-0.03}{0.0164} \approx -1.83 \), and \( P(Z \leq -1.83) \approx 0.0336 \). Adjust based on the correct \( p \) from the problem context.)

Answer:

To solve this problem, we'll address each part step by step. However, we need to assume the value of the population proportion \( p \) (usually, if not given, maybe from a previous part or a common scenario, but let's assume \( p = 0.15 \) for illustration, or maybe it's from a context we missed. Wait, actually, maybe the first part (a) had \( p \), but since it's not here, let's proceed with the general method.

Part (b): Justify Sampling Distribution of \( \hat{p} \) is Normal

For the sampling distribution of the sample proportion \( \hat{p} \) to be approximately normal, we need \( np \geq 10 \) and \( n(1 - p) \geq 10 \) (or sometimes 5, but 10 is standard).

Let’s assume \( p = 0.15 \) (common in jury summons, but adjust if needed). Given \( n = 500 \):

• \( np = 500 \times p \)
• \( n(1 - p) = 500 \times (1 - p) \)

If \( p = 0.15 \):
\( np = 500 \times 0.15 = 75 \)
\( n(1 - p) = 500 \times 0.85 = 425 \)

Both \( 75 \) and \( 425 \) are greater than \( 10 \) (or 5), so the sampling distribution of \( \hat{p} \) is approximately normal because \( np \) and \( n(1 - p) \) are both at least 10 (or 5, depending on the rule).

Part (c): Probability that at most 13% (0.13) of the sample received a jury summons

To calculate \( P(\hat{p} \leq 0.13) \), we use the normal approximation to the sampling distribution of \( \hat{p} \).

Step 1: Find the mean and standard deviation of \( \hat{p} \)

• Mean of \( \hat{p} \): \( \mu_{\hat{p}} = p \)
• Standard deviation of \( \hat{p} \): \( \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \)

Assume \( p = 0.15 \) (from part (b) assumption):
\( \mu_{\hat{p}} = 0.15 \)
\( \sigma_{\hat{p}} = \sqrt{\frac{0.15 \times 0.85}{500}} = \sqrt{\frac{0.1275}{500}} = \sqrt{0.000255} \approx 0.01597 \)

Step 2: Calculate the z-score

The z-score for \( \hat{p} = 0.13 \) is:
\( z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.13 - 0.15}{0.01597} \approx \frac{-0.02}{0.01597} \approx -1.25 \)

Step 3: Find the probability from the z-table

\( P(Z \leq -1.25) \) is the area to the left of \( z = -1.25 \) in the standard normal distribution. From the z-table, \( P(Z \leq -1.25) = 0.1056 \)

Final Answers (Assuming \( p = 0.15 \))

• (b) \( np = 75 \), \( n(1 - p) = 425 \); both are at least 10 (or 5).
• (c) Probability ≈ \( 0.1056 \)

(Note: If the actual \( p \) is different, recalculate with the correct \( p \). For example, if \( p = 0.16 \), \( np = 80 \), \( n(1 - p) = 420 \), z-score would be \( \frac{0.13 - 0.16}{\sqrt{\frac{0.16 \times 0.84}{500}}} \approx \frac{-0.03}{0.0164} \approx -1.83 \), and \( P(Z \leq -1.83) \approx 0.0336 \). Adjust based on the correct \( p \) from the problem context.)