The mean age at which men marry in the United States is 24.8 years, with a stand
ID: 3322422 • Letter: T
Question
The mean age at which men marry in the United States is 24.8 years, with a standard deviation of 2.5 year. A simple random sample of 64 men is taken. a. What theory allows us to use a normal distribution to approximate the sampling b. Compute the standard error of the mean for this sample of 64 men. What is the probability that the sample mean with be within 0.25 years of the population mean? What is the probability that the sample mean with be within 0.5 years of the population mean? Researchers would like to achieve a 0.95 probability that their sample mean will be within 0.5 years of the population mean. What will need to happen to their sample size? c. d. e.Explanation / Answer
X: Age at which men marry.
X ~ N ( mu=24.8 , sigma2=6.25)
mu=24.8, sigma=2.5 and n=64
a) The Sampling distribution of sample mean is
Xbar ~ N( mu, sigma2/n) i.e. Xbar ~ N(24.8, 0.09765)
b) standard error of mean of 64 samples
= 2.5/sqrt(64)
= 0.3125
c) Required probability
= P ( -0.25<(xbar-mu)<0.25) since z = (xbar-mu)/(sigma/sqrt(n))~N(0,1)
= P(-0.25/(2.5/8) < z < 0.25/(2.5/8))
= P(-0.80< z < 0.80)
From normal probabilty table
P(-0.80< z < 0.80)=0.5763
d) Required probability
= P ( -0.50<(xbar-mu)<0.50) since z = (xbar-mu)/(sigma/sqrt(n))~N(0,1)
= P(-0.50/(2.5/8) < z < 0.50/(2.5/8))
= P(-1.6< z < 1.6)
From normal probabilty table
P(-1.6< z < 1.6)=0.8904
e) margin of error =E=0.5 and level of significance =alpha= 0.05 Z alpha/2 = 1.96
minimum sample size
n= (Zalpha/2*sigma/E)^2
= ( 1.96*2.5/0.5)^2
=96.04
n=97
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