The height (in inches) of adult men in the United States is believed to be Normally distributed with mean μ . The average height of a random sample of 25 American adult men is found to be ¯ x = 69.72 inches, and the standard deviation of the 25 heights is found to be s = 4.15 . A 90% confidence interval for μ is: a. 69.72 ± 1.37 b. 69.72 ± 1.42 c. 69.72 ± 1.09

a. 69.72 ± 1.37

Step-by-step explanation:

We want to determine a 90% confidence interval for the mean height (in inches) of adult men in the United States.

Number of sample, n = 25

Mean, u = 69.72 inches

Standard deviation, s = 4.15

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

69.72 ± 1.645 × 4.15/√25

= 69.72 ± 1.645 × 0.83

= 69.72 ± 1.37

The lower end of the confidence interval is 69.72 - 1.37 = 68.35

The upper end of the confidence interval is 69.72 + 1.37 = 71.09

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step-by-step explanation: