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A population has a mean of 180 and a standard deviation of 36. A sample of 84 observations will be taken. The probability that the sample mean will be between 181 and 185 is
Given n(sample size) = 84
Population mean(μ) = 180
Standard Deviation(σ) = 36
Standard error of the mean = σx-bar = σ/√n = 36/√84 = 36/9.165 = 3.927
Standardizing the sample mean we have
Z = (x-bar - μ)/σx-bar = (x-bar - μ)/σ/√n
x-bar = 180
Z(x-bar=185 at point C) = (185 - 180)/3.927 = 5/3.927 = 1.273
Z(x-bar=181 at point D) = (181 - 180)/3.927 = 1/3.927 = 0.254
The area ABCD is the probability that the sample mean will lie between 181 and 185.
The shaded Area ABCD = (Area corresponding to Z = 2 or x-bar = 185) - (Area corresponding to Z = 1 or x-bar = 181)
Area corresponding to Z = 1.273 = 0.898
Area corresponding to Z = 0.254 = 0.598
The shaded Area ABCD = 0.898-0.598 = 0.300
Therefore the probability that the sample mean will lie between 181 and 185 is 0.300.
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