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------------------------------------------------ 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, [tex]\mu[/tex], of all top executives?

A. 63.5 < [tex]\mu[/tex] < 66.1
B. 65.3 < [tex]\mu[/tex] < 68.5
C. 62.8 < [tex]\mu[/tex] < 66.8
D. 66.1 < [tex]\mu[/tex] < 67.7

Answer :

Answer: Option 'd' is correct.

Step-by-step explanation:

Since we have given that

n = 250

Average = 66.9 inches

standard deviation = 6.2 inches

We need to find the 95% confidence interval for the mean.

So, z = 1.96

Interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=66.9\pm 1.96\times \dfrac{6.2}{\sqrt{250}}\\\\=66.9\pm 0.77\\\\=(66.9-0.77,66.9+0.77)\\\\=(66.13,67.67)[/tex]

Hence, option 'd' is correct.

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