A survey randomly selected 250 top executives. The average height of these executives was 66.9 inches with a standard deviation of 6.2 inches. What is a 95\% confidence interval for the mean height, $\mu$, of all top executives?
A) $66.1

Question

A survey randomly selected 250 top executives. The average height of these executives was 66.9 inches with a standard deviation of 6.2 inches. What is a 95\% confidence interval for the mean height, $\mu$, of all top executives?
A) $66.1<\mu<67.7$
B) $65.3<\mu<68.5$
C) $62.8<\mu<66.8$
D) $63.5<\mu<66.1$

Quizplus · Accepted Answer

To calculate a 95% confidence interval for the mean height, we use the formula: $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level (for 95%, $z = 1.96$), $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the given values: $66.9 \pm 1.96\frac{6.2}{\sqrt{250}}$, we get the interval $66.1<\mu<67.7$.

A Survey Randomly Selected 250 Top Executives μ\muμ , of All Top Executives? A)

A Survey Randomly Selected 250 Top Executives $\mu$ , of All Top Executives?
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