Answer :
Sure, let's go through each part of the question step-by-step:
1) What is the mean of the sampling distribution of the means?
The mean of the sampling distribution of the means ([tex]\(\mu_{\bar{X}}\)[/tex]) is equal to the population mean ([tex]\(\mu\)[/tex]). Therefore:
[tex]\[
\mu_{\bar{X}} = 172 \text{ pounds}
\][/tex]
2) What is the standard deviation of the sampling distribution of the means?
The standard deviation of the sampling distribution of the means ([tex]\(\sigma_{\bar{X}}\)[/tex]) is calculated by dividing the population standard deviation ([tex]\(\sigma\)[/tex]) by the square root of the sample size ([tex]\(n\)[/tex]).
Given:
- [tex]\(\sigma = 29\)[/tex] pounds
- [tex]\(n = 36\)[/tex]
We calculate [tex]\(\sigma_{\bar{X}}\)[/tex] as follows:
[tex]\[
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{29}{\sqrt{36}} = \frac{29}{6} = 4.833333333333333 \text{ pounds}
\][/tex]
3) What is the probability that [tex]\(n=36\)[/tex] randomly selected individuals will have a mean weight of at least 180 pounds?
To find this probability, we first calculate the z-score for the sample mean ([tex]\(\bar{X}\)[/tex]) using the formula:
[tex]\[
z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}}
\][/tex]
where:
[tex]\[
\bar{X} = 180 \text{ pounds}, \quad \mu = 172 \text{ pounds}, \quad \sigma_{\bar{X}} = 4.833333333333333 \text{ pounds}
\][/tex]
Plugging in the values, we get:
[tex]\[
z = \frac{180 - 172}{4.833333333333333} = 1.6551724137931036
\][/tex]
Next, we look up this z-score in the standard normal distribution table or use a standard normal distribution calculator. A z-score of approximately 1.655172 suggests that the cumulative probability to the left of this z-score is about 0.9510552353517007.
The probability that the sample mean is at least 180 pounds is the complement of this cumulative probability:
[tex]\[
P(\bar{X} \geq 180) = 1 - P(\bar{X} < 180) = 1 - 0.9510552353517007 = 0.04894476464829933
\][/tex]
So, the probability that the mean weight of the sample is at least 180 pounds is:
[tex]\[
0.04894476464829933
\][/tex]
Summarizing the answers:
1. The mean of the sampling distribution of the means is 172 pounds.
2. The standard deviation of the sampling distribution of the means is 4.833333333333333 pounds.
3. The probability that [tex]\(n=36\)[/tex] randomly selected individuals will have a mean weight of at least 180 pounds is 0.04894476464829933.
1) What is the mean of the sampling distribution of the means?
The mean of the sampling distribution of the means ([tex]\(\mu_{\bar{X}}\)[/tex]) is equal to the population mean ([tex]\(\mu\)[/tex]). Therefore:
[tex]\[
\mu_{\bar{X}} = 172 \text{ pounds}
\][/tex]
2) What is the standard deviation of the sampling distribution of the means?
The standard deviation of the sampling distribution of the means ([tex]\(\sigma_{\bar{X}}\)[/tex]) is calculated by dividing the population standard deviation ([tex]\(\sigma\)[/tex]) by the square root of the sample size ([tex]\(n\)[/tex]).
Given:
- [tex]\(\sigma = 29\)[/tex] pounds
- [tex]\(n = 36\)[/tex]
We calculate [tex]\(\sigma_{\bar{X}}\)[/tex] as follows:
[tex]\[
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{29}{\sqrt{36}} = \frac{29}{6} = 4.833333333333333 \text{ pounds}
\][/tex]
3) What is the probability that [tex]\(n=36\)[/tex] randomly selected individuals will have a mean weight of at least 180 pounds?
To find this probability, we first calculate the z-score for the sample mean ([tex]\(\bar{X}\)[/tex]) using the formula:
[tex]\[
z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}}
\][/tex]
where:
[tex]\[
\bar{X} = 180 \text{ pounds}, \quad \mu = 172 \text{ pounds}, \quad \sigma_{\bar{X}} = 4.833333333333333 \text{ pounds}
\][/tex]
Plugging in the values, we get:
[tex]\[
z = \frac{180 - 172}{4.833333333333333} = 1.6551724137931036
\][/tex]
Next, we look up this z-score in the standard normal distribution table or use a standard normal distribution calculator. A z-score of approximately 1.655172 suggests that the cumulative probability to the left of this z-score is about 0.9510552353517007.
The probability that the sample mean is at least 180 pounds is the complement of this cumulative probability:
[tex]\[
P(\bar{X} \geq 180) = 1 - P(\bar{X} < 180) = 1 - 0.9510552353517007 = 0.04894476464829933
\][/tex]
So, the probability that the mean weight of the sample is at least 180 pounds is:
[tex]\[
0.04894476464829933
\][/tex]
Summarizing the answers:
1. The mean of the sampling distribution of the means is 172 pounds.
2. The standard deviation of the sampling distribution of the means is 4.833333333333333 pounds.
3. The probability that [tex]\(n=36\)[/tex] randomly selected individuals will have a mean weight of at least 180 pounds is 0.04894476464829933.
