The mean amount of money that U.S. adults spend on food each week is $151 and th

Question

The mean amount of money that U.S. adults spend on food each week is $151 and the standard deviation is $49. Random samples of size 40 are drawn from this population and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of the sampling means. What is the probability that the mean amount spent on food in a week for a certain sample is more than $160? What is the probability that the mean amount spent on food in a week for a certain sample is between $135 and $150?

Explanation / Answer

Here mean=151 and sd=49 and also n=40>30

a. As per central limit theorem as n>30, sampling mean is normally distribution with mean=151 and sd=sd/sqrt(40)=7.748

b. P(xbar>160)=P(z>160-151/7.748)=P(z>1.162)=0.5-P(0<=z<=1.162)=0.5-0.3774=0.1226

c. P(135<x<150)=P(-2.07<z<-0.13)=0.4291

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