Answer :
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].
[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].
