A department store will place a sale item in a special display for a one-day sal

Question

A department store will place a sale item in a special display for a one-day sale. Previous experience suggests that 44 percent of all customers who pass such a special display will purchase the item. If 1, 147 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 __________

Explanation / Answer

Number of customers N = 1147 customers

proportion of people will purchase if pass such a special display = 0.44

so mean number of customers who will purchase the item E(x)= 1147 * 0.44 = 504.68

Standard deviation of the number of customers who will purchase the item = sqrt [p (1-p) * N] = sqrt [ 0.44 * 0.56 * 1147] = 16.8113

so here we have to find that number of units of the sale item the store stock in order to have at most a 1 percent chance of running short of the item

so that means mathematically that Pr(X >x) = 0.01

where x is the sale of the item

so the Z - value for p- value = 0.01 is

Z = + 2.33

so Z = [X - E(x)] /

2.33 = (X - 504.68)/ 16.81

X = 543.85

so, Number of units = 544 units must be there to

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