Question

A department store will place a sale item in a special display for a one-day sale. Previous experience suggests that 43 percent of all customers who pass such a special display will purchase the item. If 1,462 customers will pass the display on the day of the sale, and if a one-item-per-customer limit is placed on the sale item, how many units of the sale item should the store stock in order to have at most a 1 percent chance of running short of the item on the day of the sale? Assume here that customers make independent purchase decisions. (Round your answer to nearest whole number.)
Number of units?

Answer 1

Answer:

672.7436 units of the sale item should the store stock in order to have at most a 1 percent chance of running short of the item on the day of the sale

Step-by-step explanation:

No. of customers pass the display on the day of sale = 1462

Proportion of people will purchase if pass such a special display = 0.43

Mean = E(X)=np

E(X)=$1462 \times 0.43$

E(X)=628.66

Standard deviation = $\sqrt{npq}$

Standard deviation =$\sqrt{1462 \times 0.43 \times (1-0.43)}=18.92$

Now we are supposed to find how many units of the sale item should the store stock in order to have at most a 1 percent chance of running short of the item on the day of the sale

P(X<x)=0.01

Using z table

$P(\frac{x-\mu}{\sigma})= 2.33 \\\frac{x-628.66}{18.92}=2.33\\x=(2.33 \times 18.92)+628.66\\x=672.7436$

Hence 672.7436 units of the sale item should the store stock in order to have at most a 1 percent chance of running short of the item on the day of the sale