Answer:
Explanation:
We can use the formula for a confidence interval for the population mean:
CI = x¯¯¯ ± z*(σ/sqrt(n))
where:
x¯¯¯ is the sample mean, which is 81.
σ is the population standard deviation, which is 11.8 (assuming the sample is drawn from a normally distributed population).
n is the sample size, which is 175.
z is the z-score corresponding to the desired level of confidence, which is 2.33 for a 98% confidence level.
Substituting the known values, we get:
CI = 81 ± 2.33*(11.8/sqrt(175))
Calculating this expression, we get:
CI ≈ (79, 83)
Therefore, a 98% confidence interval for the population mean is approximately (79, 83).
Therefore, the answer is (B) (79, 83).