Suppose that you have a sample of 100 values from a populati

Suppose that you have a sample of 100 values from a population with mean mu = 500 and with standard deviation sigma = 80. a) What is the probability that the sample mean will be in the interval (490, 510)? Give an interval that covers the middle 95% of the distribution of the sample mean.

Solution

Sample size = n = 100

mu = 500

sigma = 80

Sample follows a normal distirbution with mu = 500 and std error = 80/10 = 8

Hence we convert x to z score and use std normal variable probabilities

x = 490 means z score = -10/8 = -1.25

x = 510, z = 1.25

Hence P(490<x<510) = P(-1.25<z<1.25) = 0.3944+0.3944 = 0.7888

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b) for 95% os sample mean z alpha/2 = 1.96

Hence x score for -1.96 z = 500-1.96(8) = 484.32

x score for 1.96z = 500+1.96(8) = 515.68

Hence interval that covers 95% is (484.32, 515.68)