Question

This Question: 1pl Astery store wants to estimate the mean real value of greetrg cards that it has inis inventory A random sample of 121 greeting cards indicates a mean value of 5371 and a standard deviaton of 50 Assuming a normal disbution construct a 90% confidence interval estimate of the mean value of all greeting cards in the store's inventory Suppose there were 2.500 greeting cards in the store's inventory How are the results in part (a) used in assisting the store owner to estmate the total value of her inventory? Dick here hav e the table cercal values for the totbuton Click here to view page 2 of the table scrical values for the distributo Roundt two decimal places as needed) b. Choose the correct awer below OA Tomate the totale multiply the confidence interval by 2,500 OB. The stof para) is the confidence interval of the total value O The valid for the sample Estimates for the total value of the vertory cannot be made from this contence interval OD. To estimate the total value, add 2.500 to the confidence interval

Answer 1

Solution. Let it be the four mean netail value of greeting card that it has in life inventory and it be the corresponding vemple mien & R= 69.74 Dad21 S = 0.49 (0)= 90% confidence Internal for sell ig given by, nt tn, 5/8 g Hen noudt which is very large 5o, value of En 1,5% & Z54 90, Xt Z5+X5 = 3.71 2 2.64 x 0.49 (3.64, 3.78) $ 13,64) EML9[3.78] V121 * To Estimate the total valu, multiply the Confidence Interval by 2,500.