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------------------------------------------------ Use technology to solve the following problem:

A sample of size 45 will be drawn from a population with a mean of 94 and a standard deviation of 14. Find the 32nd percentile of [tex]\bar{x}[/tex].

A. 95.4
B. 93.0
C. 91.2
D. 97.7

We start by noting that the sampling distribution of the sample mean, [tex]$\bar{x}$[/tex], is approximately normal with mean

[tex]$$
\mu_{\bar{x}} = 94,
$$[/tex]

and standard error

[tex]$$
\sigma_{\bar{x}} = \frac{14}{\sqrt{45}}.
$$[/tex]

Step 1. Calculate the Standard Error:

The standard error is given by

[tex]$$
\sigma_{\bar{x}} = \frac{14}{\sqrt{45}} \approx 2.087.
$$[/tex]

Step 2. Find the Z-score for the 32nd Percentile:

For a standard normal distribution, find the [tex]$z$[/tex]-value such that

[tex]$$
P(Z \le z) = 0.32.
$$[/tex]

Looking up or using a calculator for the standard normal distribution, we get

[tex]$$
z \approx -0.4677.
$$[/tex]

Step 3. Calculate the 32nd Percentile of [tex]$\bar{x}$[/tex]:

The corresponding value of [tex]$\bar{x}$[/tex] is found by

[tex]$$
\bar{x} = \mu_{\bar{x}} + z \cdot \sigma_{\bar{x}}.
$$[/tex]

Substitute the values:

[tex]$$
\bar{x} = 94 + (-0.4677)(2.087) \approx 94 - 0.977 \approx 93.023.
$$[/tex]

Rounded to one decimal place, the 32nd percentile of [tex]$\bar{x}$[/tex] is approximately

[tex]$$
93.0.
$$[/tex]

Thus, the answer is [tex]$\boxed{93.0}$[/tex].