Question

for a sample of size 13, state the mean of the sample mean and the standard deviation of the sample mean. round to two decimal places if necessary.
\\( \mu _ { \overline { x } } = \\)
\\( \sigma _ { \overline { x } } = \\)
for a sample of size 13, find the probability that the sample mean is more than 242.
\\( p ( \overline { x } > 242 ) = \\)
if you take a sample of size 38, can you say what the shape of the distribution of the sample mean is? why?
if the sample size is 38, then you cant say anything about the sampling distribution of the sample mean, since the population of the random variable is not normally distributed and the sample size is greater than 30.
if the sample size is 38, then you can say the sampling distribution of the sample mean is normally distributed since the variable is normally distributed.
if the sample size is 38, then you can say the sampling distribution of the sample mean is not normally distributed since the sample size is greater than 30.
for a sample of size 38, statthe mean of the sample mean and the standard deviation of the sample mean. round to two decimal places if necessary. \\( \mu _ { \overline { x } } = \\)
\\( \sigma _ { \overline { x } } = \\)
for a sample of size 38, find the probability that the sample mean is more than 242. \\( p ( \overline { x } > 242 ) = \\)

Explanation:

Step1: Mean of the sample mean

The mean of the sample mean \(\mu_{\bar{x}}\) is equal to the population mean. Since for the sample of size \(13\), \(\mu_{\bar{x}} = 248\), for a sample of size \(n = 38\), \(\mu_{\bar{x}}=248\) (by the property \(\mu_{\bar{x}}=\mu\))

Step2: Standard deviation of the sample mean

The formula for the standard deviation of the sample mean (standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). For \(n = 13\), \(\sigma_{\bar{x}}=7.49=\frac{\sigma}{\sqrt{13}}\), so \(\sigma = 7.49\times\sqrt{13}\approx7.49\times3.606\approx27\). For \(n = 38\), \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{27}{\sqrt{38}}\approx\frac{27}{6.164}\approx4.38\)

Step3: Probability that the sample mean is more than \(242\)

First, we calculate the \(z -\)score: \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\). Here, \(\bar{x} = 242\), \(\mu_{\bar{x}}=248\), \(\sigma_{\bar{x}}\approx4.38\). So \(z=\frac{242 - 248}{4.38}=\frac{- 6}{4.38}\approx - 1.37\). Then \(P(\bar{X}>242)=1 - P(\bar{X}\leq242)\). Using the standard normal table, \(P(Z\leq - 1.37)=0.0853\). So \(P(\bar{X}>242)=1 - 0.0853 = 0.9147\approx0.91\)

Answer:

\(\mu_{\bar{x}} = 248\)

\(\sigma_{\bar{x}}\approx4.38\)

\(P(\bar{x}>242)\approx0.91\)