Answer :
Let's solve the given problem step-by-step.
### Part (a) Describe the sampling distribution of [tex]\(\bar{x}\)[/tex].
Given:
- Sample size [tex]\( n = 49 \)[/tex]
- Population mean [tex]\( \mu = 79 \)[/tex]
- Population standard deviation [tex]\( \sigma = 28 \)[/tex]
According to the Central Limit Theorem, for a sufficiently large sample size, the sampling distribution of the sample mean [tex]\(\bar{x}\)[/tex] will be approximately normally distributed, regardless of the shape of the population distribution. Since our sample size is [tex]\( n = 49 \)[/tex], which is generally considered large enough, the sampling distribution of [tex]\(\bar{x}\)[/tex] can be said to be approximately normal.
Choice for part (a): [tex]\(\boxed{\text{B. The distribution is approximately normal.}}\)[/tex]
### Mean and Standard Deviation of the Sampling Distribution of [tex]\(\bar{x}\)[/tex]
The mean of the sampling distribution of [tex]\(\bar{x}\)[/tex] ([tex]\(\mu_{\bar{x}}\)[/tex]) is the same as the mean of the population ([tex]\(\mu\)[/tex]). Thus,
[tex]\[ \mu_{\bar{x}} = 79 \][/tex]
The standard deviation of the sampling distribution of [tex]\(\bar{x}\)[/tex] ([tex]\(\sigma_{\bar{x}}\)[/tex]) is given by:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
Substituting the given values, we have:
[tex]\[ \sigma_{\bar{x}} = \frac{28}{\sqrt{49}} = \frac{28}{7} = 4 \][/tex]
### Part (b) Calculate [tex]\( P(76.4 < \bar{x} < 88.8) \)[/tex]
To find the probability [tex]\( P(76.4 < \bar{x} < 88.8) \)[/tex], we first need to convert the sample means to z-scores using the sampling distribution's mean and standard deviation.
1. Calculate the z-score for [tex]\( \bar{x} = 76.4 \)[/tex]:
[tex]\[ z_{76.4} = \frac{76.4 - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{76.4 - 79}{4} = \frac{-2.6}{4} = -0.65 \][/tex]
2. Calculate the z-score for [tex]\( \bar{x} = 88.8 \)[/tex]:
[tex]\[ z_{88.8} = \frac{88.8 - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{88.8 - 79}{4} = \frac{9.8}{4} = 2.45 \][/tex]
Now, we use the standard normal distribution to find the cumulative probabilities for these z-scores.
3. Find the cumulative probability for [tex]\( z = -0.65 \)[/tex]:
[tex]\[ P(Z \le -0.65) \approx 0.2578 \][/tex]
4. Find the cumulative probability for [tex]\( z = 2.45 \)[/tex]:
[tex]\[ P(Z \le 2.45) \approx 0.9928 \][/tex]
5. Subtract the cumulative probabilities to find [tex]\( P(-0.65 < Z < 2.45) \)[/tex]:
[tex]\[ P(76.4 < \bar{x} < 88.8) = P(-0.65 < Z < 2.45) = 0.9928 - 0.2578 = 0.735 \][/tex]
Thus, [tex]\( P(76.4 < \bar{x} < 88.8) = 0.7350 \)[/tex] (rounded to four decimal places).
Summary of results:
1. Shape of the distribution of [tex]\(\bar{x}\)[/tex]: [tex]\(\boxed{\text{B. The distribution is approximately normal.}}\)[/tex]
2. Mean and standard deviation of the sampling distribution:
[tex]\[ \mu_{\bar{x}} = 79, \quad \sigma_{\bar{x}} = 4 \][/tex]
3. Probability [tex]\( P(76.4 < \bar{x} < 88.8) = 0.7350 \)[/tex]
### Part (a) Describe the sampling distribution of [tex]\(\bar{x}\)[/tex].
Given:
- Sample size [tex]\( n = 49 \)[/tex]
- Population mean [tex]\( \mu = 79 \)[/tex]
- Population standard deviation [tex]\( \sigma = 28 \)[/tex]
According to the Central Limit Theorem, for a sufficiently large sample size, the sampling distribution of the sample mean [tex]\(\bar{x}\)[/tex] will be approximately normally distributed, regardless of the shape of the population distribution. Since our sample size is [tex]\( n = 49 \)[/tex], which is generally considered large enough, the sampling distribution of [tex]\(\bar{x}\)[/tex] can be said to be approximately normal.
Choice for part (a): [tex]\(\boxed{\text{B. The distribution is approximately normal.}}\)[/tex]
### Mean and Standard Deviation of the Sampling Distribution of [tex]\(\bar{x}\)[/tex]
The mean of the sampling distribution of [tex]\(\bar{x}\)[/tex] ([tex]\(\mu_{\bar{x}}\)[/tex]) is the same as the mean of the population ([tex]\(\mu\)[/tex]). Thus,
[tex]\[ \mu_{\bar{x}} = 79 \][/tex]
The standard deviation of the sampling distribution of [tex]\(\bar{x}\)[/tex] ([tex]\(\sigma_{\bar{x}}\)[/tex]) is given by:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
Substituting the given values, we have:
[tex]\[ \sigma_{\bar{x}} = \frac{28}{\sqrt{49}} = \frac{28}{7} = 4 \][/tex]
### Part (b) Calculate [tex]\( P(76.4 < \bar{x} < 88.8) \)[/tex]
To find the probability [tex]\( P(76.4 < \bar{x} < 88.8) \)[/tex], we first need to convert the sample means to z-scores using the sampling distribution's mean and standard deviation.
1. Calculate the z-score for [tex]\( \bar{x} = 76.4 \)[/tex]:
[tex]\[ z_{76.4} = \frac{76.4 - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{76.4 - 79}{4} = \frac{-2.6}{4} = -0.65 \][/tex]
2. Calculate the z-score for [tex]\( \bar{x} = 88.8 \)[/tex]:
[tex]\[ z_{88.8} = \frac{88.8 - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{88.8 - 79}{4} = \frac{9.8}{4} = 2.45 \][/tex]
Now, we use the standard normal distribution to find the cumulative probabilities for these z-scores.
3. Find the cumulative probability for [tex]\( z = -0.65 \)[/tex]:
[tex]\[ P(Z \le -0.65) \approx 0.2578 \][/tex]
4. Find the cumulative probability for [tex]\( z = 2.45 \)[/tex]:
[tex]\[ P(Z \le 2.45) \approx 0.9928 \][/tex]
5. Subtract the cumulative probabilities to find [tex]\( P(-0.65 < Z < 2.45) \)[/tex]:
[tex]\[ P(76.4 < \bar{x} < 88.8) = P(-0.65 < Z < 2.45) = 0.9928 - 0.2578 = 0.735 \][/tex]
Thus, [tex]\( P(76.4 < \bar{x} < 88.8) = 0.7350 \)[/tex] (rounded to four decimal places).
Summary of results:
1. Shape of the distribution of [tex]\(\bar{x}\)[/tex]: [tex]\(\boxed{\text{B. The distribution is approximately normal.}}\)[/tex]
2. Mean and standard deviation of the sampling distribution:
[tex]\[ \mu_{\bar{x}} = 79, \quad \sigma_{\bar{x}} = 4 \][/tex]
3. Probability [tex]\( P(76.4 < \bar{x} < 88.8) = 0.7350 \)[/tex]
