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A stationery store wants to estimate the mean retail value of greeting cards that it has...

A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 121 greeting cards indicates a mean value of $ 3.86 and a standard deviation of $ 0.57. a. Assuming a normal​ distribution, construct a 90 % confidence interval estimate of the mean value of all greeting cards in the​ store's inventory. b. Suppose there were 1,000 greeting cards in the​ store's inventory. How are the results in part​ (a) useful in assisting the store owner to estimate the total value of her​ inventory?(Round to two decimal places as​ needed.)

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Answer #1

Given,

n= 121. 3.86. s=0.57

Since, sample size is large (n>30), so z-interval is use

90% confidence interval for mu is

ar{x} pm z_{alpha/2} rac{s}{sqrt{n}}

Confidence interval =90% , a1-0.90 0.10 , ~o/2-20. 10/2 1.645

0.57 3.86 +: 1.645* i121

3.86 pm 0.08524091

(3.774759,3.945241

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