A population has a mean of 300 and a standard deviation of 90. Suppose a sample

Question

A population has a mean of 300 and a standard deviation of 90. Suppose a sample of size 100 is selected and (x-bar) is used to estimate (mu). What is the probability that the sample mean will be within +/- 8 of the population mean (to 4 decimals)?

A market research firm conducts telephone surveys with a 38% historical response rate. What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions? In other words, what is the probability that the sample proportion will be at least 150/400 = .375?

Use excel to figure out the answers

Explanation / Answer

1) here std error of mean =std deviation/(n)1/2 =90/(100)1/2 =9

therefore probability that the sample mean will be within +/- 8 of the population mean

=P(300-8<X<300+8)=P(-8/9<Z<8/9)=P(-0.8889<Z<0.8889)=0.8130-0.1871 =0.6259

(You may use excel function ==NORM.DIST(308,300,9,TRUE)-NORM.DIST(292,300,9,TRUE) to calulcate above from excel)

2)

here std error of proportion =(p(1-p)/n)1/2 =(0.38*(1-0.38)/400)1/2 =0.02427

therefore probability that the sample proportion will be at least 0.375=P(X>0.375)=1-P(X<0.375)

=1-P(Z<(0.375-0.38)/0.02427)=1-P(Z<0.2060)=1-0.5816=0.4184

(you may use excel function =1-NORM.DIST(0.375,0.38,0.02427,TRUE) to calculate above)

plese revert for any help

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