Suppose $x$ has a distribution with a mean of 60 and a standard deviation of 21 . Random samples of size $n=36$ are drawn.
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(a) Describe the $\bar{x}$ distribution and compute the mean and standard deviation of the distribution.
\[
\bar{x} \text { has:- Select- } \quad \text { distribution with mean } \mu_{\bar{x}}=\square \text { and standard deviation } \sigma_{\bar{x}}=
\]
(b) Find the $z$ value corresponding to $\bar{x}=56.5$.
\[
z=
\]
(c) Find $P(\bar{x}< 56.5)$. (Round your answer to four decimal places.)
\[
P(\bar{x}< 56.5)=
\]
(d) Would it be unusual for a random sample of size 36 from the $x$ distribution to have a sample mean less than 56.5 ? Explain.
No, it would not be unusual because less than $5 \%$ of all such samples have means less than 56.5 .
Yes, it would be unusual because less than $5 \%$ of all such samples have means less than 56.5 .
No, it would not be unusual because more than $5 \%$ of all such samples have means less than 56.5 .
Yes, it would be unusual because more than $5 \%$ of all such samples have means less than 56.5 .
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Step 1 ：The distribution of \(\bar{x}\) is approximately normal due to the Central Limit Theorem, as the sample size is large (n=36).

Step 2 ：The mean of the distribution \(\mu_{\bar{x}}\) is equal to the mean of the population, which is 60.

Step 3 ：The standard deviation of the distribution \(\sigma_{\bar{x}}\) is equal to the standard deviation of the population divided by the square root of the sample size, which is 21/\(\sqrt{36}\) = 3.5.

Step 4 ：So, \(\bar{x}\) has a normal distribution with mean \(\mu_{\bar{x}}=60\) and standard deviation \(\sigma_{\bar{x}}=3.5\).

Step 5 ：The z value is calculated by subtracting the mean from the x value and then dividing by the standard deviation. So, for \(\bar{x}=56.5\), the z value is (56.5-60)/3.5 = -1.

Step 6 ：So, \(z=-1\).

Step 7 ：The probability \(P(\bar{x}<56.5)\) is equivalent to the probability \(P(z<-1)\). Looking up -1 in the z-table, we find that the probability is approximately 0.1587.

Step 8 ：So, \(P(\bar{x}<56.5)=0.1587\).

Step 9 ：It would be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 56.5 because less than 5% of all such samples have means less than 56.5. The probability we calculated in part (c) is 0.1587, which is greater than 0.05, so more than 5% of all such samples have means less than 56.5.

Step 10 ：Therefore, the correct answer is \(\boxed{\text{'No, it would not be unusual because more than 5% of all such samples have means less than 56.5.'}}\).