A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98% confidence interval for the true average age of all students in the university is

24.4185<x<25.5815

Step-by-step explanation:

Given the following:

n = 64

mean x = 25

s = 2

z is the z score at 98% CI = 2.326

Get the Confidence Interval:

CI = x±z*s/√n

CI = 25±2.326*2/√64

CI = 25±2.326*2/8

CI = 25±0.5815

CI = (25-0.5815, 25+0.5815)

CI = (24.4185, 25.5815)

CI = 24.4185<x<25.5815

Hence the 98% confidence interval for the true average age of all students in the university is 24.4185<x<25.5815

the anwser is d) 75°

step-by-step explanation:

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step-by-step explanation: