Find a $$ 95 \% $$ confidence interval for the population mean body temperature if for a simple random sample of 47 individuals, the mean temperature was 98.5 . Sigma is 0.67 . $$ (98.308,98.692) $$ $$ (98.336,98.664) $$ $$ (98.339,98.661) $$ $$ (98.303,98.697) $$

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We are given a simple random sample of $$ n = 47 $$ individuals with a sample mean of  
$$
\bar{x} = 98.5,
$$  
and the population standard deviation is  
$$
\sigma = 0.67.
$$

Because $$\sigma$$ is known, we use the $$z$$-interval formula for the confidence interval for the population mean:
$$
\bar{x} \pm z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}}
$$

For a $$95\%$$ confidence interval, the critical value is 
$$
z_{0.025} = 1.96.
$$

1. Calculate the standard error (SE):
$$
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{0.67}{\sqrt{47}}.
$$
Since $$\sqrt{47} \approx 6.85565$$, we have:
$$
\text{SE} \approx \frac{0.67}{6.85565} \approx 0.0977.
$$

2. Compute the margin of error (ME):
$$
\text{ME} = 1.96 \times 0.0977 \approx 0.1915.
$$

3. Determine the confidence interval:
- Lower bound:
$$
\bar{x} - \text{ME} = 98.5 - 0.1915 \approx 98.3085,
$$
- Upper bound:
$$
\bar{x} + \text{ME} = 98.5 + 0.1915 \approx 98.6915.
$$

Thus, the $$95\%$$ confidence interval for the population mean is approximately:
$$
(98.308, 98.692).
$$
The $$95\%$$ confidence interval for the population mean body temperature is approximately $$ (98.308, 98.692) $$.

Find a confidence interval for the population mean body temperature if for a
simple random sample of 47 individuals, the mean temperature was 98.5 . Sigma is
0.67 .

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Find a confidence interval for the population mean body temperature if for a simple random sample of 47 individuals, the mean temperature was 98.5 . Sigma is 0.67 .