Population has a mean of 180 and a standard deviation of 24. A sample of 64 observations will be taken. the probability that the mean from that sample will be between 183 and 186 is

Solution:

The given problem can be summarized in the figure given below:

Given n(sample size) = 64

Population mean(μ) = 180

Standard Deviation(σ) = 24

Standard error of the mean = σx-bar = σ/√n = 24/√64 = 24/8 = 3

Standardizing the sample mean we have

Z = (x-bar - μ)/σx-bar = (x-bar - μ)/σ/√n

x-bar = 180

Z(x-bar=186 at point C) = (186 - 180)/3 = 6/3 = 2

Z(x-bar=183 at point D) = (183 - 180)/3 = 6/3 = 1

The area ABCD is the probability that the sample mean will lie between 183 and 186

The shaded Area ABCD = (Area corresponding to Z = 2 or x-bar = 186) - (Area corresponding to Z = 1 or x-bar = 183)

Area corresponding to Z = 2 = 0.4772

Area corresponding to Z = 1 = 0.3413

The shaded Area ABCD = 0.4772 - 0.3413 = 0.1359

Therefore the probability that the sample mean will lie between 183 and 186 is 0.1359 or 13.59%.

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Population has a mean of 180 and a standard deviation of 24. A sample of 64 observations will be taken. the probability that the mean from that sample will be between 183 and 186 is

Summary:

Population has a mean of 180 and a standard deviation of 24. A sample of 64 observations will be taken. the probability that the mean from that sample will be between 183 and 186 is 0.1359 or 13.59%