Question

the population mean and standard deviation are given below. find the indicated probability and determine whether the given sample mean would be considered unusual.
for a sample of ( n = 37 ), find the probability of a sample mean being less than 12,749 or greater than 12,752 when ( mu = 12,749 ) and ( sigma = 2.1 ).
for the given sample, the probability of a sample mean being less than 12,749 or greater than 12,752 is
(round to four decimal places as needed.)
would the given sample mean be considered unusual?
a. the sample mean would be considered unusual because there is a probability less than 0.05 of the sample mean being within this range.
b. the sample mean would not be considered unusual because there is a probability greater than 0.05 of the sample mean being within this range.
c. the sample mean would be considered unusual because there is a probability greater than 0.05 of the sample mean being within this range.
d. the sample mean would not be considered unusual because there is a probability less than 0.05 of the sample mean being within this range.

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 2.1$ and $n = 37$, then $\sigma_{\bar{x}}=\frac{2.1}{\sqrt{37}}\approx\frac{2.1}{6.0828}\approx0.3453$.

Step2: Calculate the z - scores

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$.
For $\bar{x}=12749$, $z_1=\frac{12749 - 12749}{0.3453}=0$.
For $\bar{x}=12752$, $z_2=\frac{12752 - 12749}{0.3453}=\frac{3}{0.3453}\approx8.69$.

Step3: Calculate the probability

We want to find $P(\bar{X}<12749\cup\bar{X}>12752)$.
Since $P(\bar{X}<12749) = P(Z < 0)=0.5$ (from the standard normal distribution table, where the cumulative distribution function $\varPhi(0)=0.5$) and $P(\bar{X}>12752)=P(Z>8.69)\approx0$ (because for a standard normal distribution $Z$, $P(Z > 3.5)\approx0$).
So $P(\bar{X}<12749\cup\bar{X}>12752)=P(\bar{X}<12749)+P(\bar{X}>12752)=0.5 + 0=0.5$.

Answer:

The probability is $0.5000$.
A. The sample mean would be considered unusual because there is a probability less than $0.05$ of the sample mean being within this range.