Answer :
We are given the sample mean [tex]$\bar{x} = 102.1$[/tex], population standard deviation [tex]$\sigma = 11.4$[/tex], and sample size [tex]$n = 62$[/tex]. For a 95% confidence interval when the population standard deviation is known, we use the [tex]$z$[/tex]-score corresponding to the 95% level, which is [tex]$z = 1.96$[/tex].
The standard error of the mean is calculated as:
[tex]$$
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{11.4}{\sqrt{62}}.
$$[/tex]
Evaluating the square root,
[tex]$$
\sqrt{62} \approx 7.874,
$$[/tex]
so
[tex]$$
\text{SE} \approx \frac{11.4}{7.874} \approx 1.4478.
$$[/tex]
Next, the margin of error is given by:
[tex]$$
\text{ME} = z \times \text{SE} = 1.96 \times 1.4478 \approx 2.8377.
$$[/tex]
The confidence interval is then computed by subtracting and adding the margin of error to the sample mean:
[tex]$$
\text{Lower bound} = \bar{x} - \text{ME} \approx 102.1 - 2.8377 \approx 99.2623,
$$[/tex]
[tex]$$
\text{Upper bound} = \bar{x} + \text{ME} \approx 102.1 + 2.8377 \approx 104.9377.
$$[/tex]
Thus, the 95% confidence interval for the population mean [tex]$\mu$[/tex] is approximately:
[tex]$$
99.3 < \mu < 104.9.
$$[/tex]
Among the options, this corresponds to:
[tex]$$
99.3 < \mu < 104.9.
$$[/tex]
The standard error of the mean is calculated as:
[tex]$$
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{11.4}{\sqrt{62}}.
$$[/tex]
Evaluating the square root,
[tex]$$
\sqrt{62} \approx 7.874,
$$[/tex]
so
[tex]$$
\text{SE} \approx \frac{11.4}{7.874} \approx 1.4478.
$$[/tex]
Next, the margin of error is given by:
[tex]$$
\text{ME} = z \times \text{SE} = 1.96 \times 1.4478 \approx 2.8377.
$$[/tex]
The confidence interval is then computed by subtracting and adding the margin of error to the sample mean:
[tex]$$
\text{Lower bound} = \bar{x} - \text{ME} \approx 102.1 - 2.8377 \approx 99.2623,
$$[/tex]
[tex]$$
\text{Upper bound} = \bar{x} + \text{ME} \approx 102.1 + 2.8377 \approx 104.9377.
$$[/tex]
Thus, the 95% confidence interval for the population mean [tex]$\mu$[/tex] is approximately:
[tex]$$
99.3 < \mu < 104.9.
$$[/tex]
Among the options, this corresponds to:
[tex]$$
99.3 < \mu < 104.9.
$$[/tex]
