Asurvey 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, μ, of all top executives? a. 63.5 < μ < 66.1 b. 65.3 < μ < 68.5 c. 62.8 < μ < 66.8 d. 66.1 < μ < 67.7

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

Hence, option 'd' is correct.

idk

step-by-step explanation:

(23*2)-12=x     x=34

step-by-step explanation: