Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
