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Answer :
We are given the following information:
- Sample size: [tex]$n = 625$[/tex]
- Population standard deviation: [tex]$\sigma = 8.8$[/tex]
- Sample mean: [tex]$\bar{x} = 97.6$[/tex]
- Confidence level: [tex]$90\%$[/tex]
Since the population standard deviation is known and the sample size is large, we use the [tex]$z$[/tex]-score for the confidence interval. For a [tex]$90\%$[/tex] confidence level, the critical value is [tex]$z^* = 1.645$[/tex].
Step 1. Calculate the standard error (SE):
The standard error of the mean is given by
[tex]$$
\text{SE} = \frac{\sigma}{\sqrt{n}}.
$$[/tex]
Substitute the given values:
[tex]$$
\text{SE} = \frac{8.8}{\sqrt{625}} = \frac{8.8}{25} = 0.352.
$$[/tex]
Step 2. Calculate the margin of error (ME):
The margin of error is
[tex]$$
\text{ME} = z^* \times \text{SE}.
$$[/tex]
Thus,
[tex]$$
\text{ME} = 1.645 \times 0.352 \approx 0.57904.
$$[/tex]
Step 3. Determine the confidence interval:
The confidence interval for the population mean [tex]$\mu$[/tex] is calculated as
[tex]$$
\bar{x} \pm \text{ME}.
$$[/tex]
So the lower bound is
[tex]$$
\bar{x} - \text{ME} = 97.6 - 0.57904 \approx 97.02096,
$$[/tex]
and the upper bound is
[tex]$$
\bar{x} + \text{ME} = 97.6 + 0.57904 \approx 98.17904.
$$[/tex]
Step 4. Round to the nearest tenth:
Rounding the lower bound, we obtain [tex]$97.0$[/tex], and rounding the upper bound, we obtain [tex]$98.2$[/tex].
Thus, the [tex]$90\%$[/tex] confidence interval for the population mean [tex]$\mu$[/tex] is
[tex]$$
97.0 < \mu < 98.2.
$$[/tex]
- Sample size: [tex]$n = 625$[/tex]
- Population standard deviation: [tex]$\sigma = 8.8$[/tex]
- Sample mean: [tex]$\bar{x} = 97.6$[/tex]
- Confidence level: [tex]$90\%$[/tex]
Since the population standard deviation is known and the sample size is large, we use the [tex]$z$[/tex]-score for the confidence interval. For a [tex]$90\%$[/tex] confidence level, the critical value is [tex]$z^* = 1.645$[/tex].
Step 1. Calculate the standard error (SE):
The standard error of the mean is given by
[tex]$$
\text{SE} = \frac{\sigma}{\sqrt{n}}.
$$[/tex]
Substitute the given values:
[tex]$$
\text{SE} = \frac{8.8}{\sqrt{625}} = \frac{8.8}{25} = 0.352.
$$[/tex]
Step 2. Calculate the margin of error (ME):
The margin of error is
[tex]$$
\text{ME} = z^* \times \text{SE}.
$$[/tex]
Thus,
[tex]$$
\text{ME} = 1.645 \times 0.352 \approx 0.57904.
$$[/tex]
Step 3. Determine the confidence interval:
The confidence interval for the population mean [tex]$\mu$[/tex] is calculated as
[tex]$$
\bar{x} \pm \text{ME}.
$$[/tex]
So the lower bound is
[tex]$$
\bar{x} - \text{ME} = 97.6 - 0.57904 \approx 97.02096,
$$[/tex]
and the upper bound is
[tex]$$
\bar{x} + \text{ME} = 97.6 + 0.57904 \approx 98.17904.
$$[/tex]
Step 4. Round to the nearest tenth:
Rounding the lower bound, we obtain [tex]$97.0$[/tex], and rounding the upper bound, we obtain [tex]$98.2$[/tex].
Thus, the [tex]$90\%$[/tex] confidence interval for the population mean [tex]$\mu$[/tex] is
[tex]$$
97.0 < \mu < 98.2.
$$[/tex]
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