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 16 percent of all customers who pass such a special display will purchase the item. If 2,749 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.)

A department store will place a sale item in a special display for a one-day sale. Previous experience suggests that 16 percent of all customers who pass such a special display will purchase the item. If 2,749 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.)

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 2749 * 0.01 = 27.49
standard deviation ( npq )= 2749*0.01*0.99 = 5.2168
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
P ( Z < x ) = 0.01
Value of z to the cumulative probability of 0.01 from normal table is -2.3263
P( x-u/s.d < x - 27.49/5.2168 ) = 0.01
That is, ( x - 27.49/5.2168 ) = -2.3263
--> x = -2.3263 * 5.2168 + 27.49 = 15.3539 ~ 16 units

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