The mean amount purchased by a typical customer at Churchil\'s Grocery Store is

Question

The mean amount purchased by a typical customer at Churchil's Grocery Store is $21.00 with a standard devlation of $700. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 42 customers, answer the following questions a. What is the lkellhood the sample mean is at least $2250? (Round your z value to 2 decimal pleces and final answer to 4 decimal places.) b. What is the likelhood the sample mean is greater than $20.00 but less than $22.50? (Round your z value to 2 decimal pleces and finel enswer to 4 decimal pleces.) c. Within what limits will 99 percent of the sample means occur? (Round your answers to 2 decimal places

Explanation / Answer

Ans:

Given that

mean=21

standard deviation=7

n=42

a)

z(22.5)=(22.5-21)/(7/sqrt(42))

z(22.5)=1.39

P(z>=1.39)=0.0823

b)

z(20)=(20-21)/(7/sqrt(42))

z(20)=-0.93

P(-0.93<z<1.39)=P(z<1.39)-P(z<-0.93)=0.9177-0.1761=0.7416

c)

z cutt off values for middle 99% are +/-2.576

lower limit=21-2.58*(7/sqrt(42))=18.22

upper limit=21+2.58*(7/sqrt(42))=23.78

*(if z is taken 2.58,then limits will be 18.21 and 23.79)

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