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the average price of a gallon of unleaded regular gasoline was reported to be $2

ID: 3335751 • Letter: T

Question

the average price of a gallon of unleaded regular gasoline was reported to be $2.34 in northern kentucky (the cincinnati enquire, january 21, 2006). Use this price as the population mean, and assume the popularion standard deviation is $0.20.

A) what is the probbability that the mean price for a sample of 30 service station is within $0.03 of the population mean?

B) what is the probability that the mean price for a sample of 50 service stations is within $0.03 of the population mean?

C) what is the probability that the mean price for a sample of 100 service stations is within $0.03 of the population mean?

D) which, if any, of the sample sizes in parts a,b, and c, would you recommend to have at least a .95 probability that the sample mean is within $0.03 of the population mean?

***please show how to find answers in microsoft excel***

Explanation / Answer

here you can use normsdist function to calculate probability from z score: revert if you need further help

here std error of mean =std deviaiton/(n)1/2

a) std error of mean =std deviaiton/(n)1/2 =0.20/(30)1//2 =0.0365

hence probbability that the mean price for a sample of 30 service station is within $0.03 of the population mean

=P(-0.03/0.0365<Z<0.03/0.0365)=P(-0.8216<Z<0.8216)=normsdist(0.8216)-normsdist(-0.8216) =0.5887

b) std error of mean =std deviaiton/(n)1/2 =0.20/(50)1//2 =0.0283

probability that the mean price for a sample of 50 service stations is within $0.03 of the population mean

=P(-0.03/0.0283<Z<0.03/0.0283)=P(-1.0607<Z<1.0607)=normsdist(1.0607)-normsdist(-1.0607) =0.7112

c)

std error of mean =std deviaiton/(n)1/2 =0.20/(100)1//2 =0.02

probability that the mean price for a sample of 100 service stations is within $0.03 of the population mean

=P(-0.03/0.02<Z<0.03/0.02)=P(-1.5<Z<1.5)=0.9332-0.0668=0.8664

D)

we will recommend part C for having  probability near to 0.95