An organization hosting a 5K race is looking to sell t-shirts at the event.Demand for t-shirts is believed to be normally distributed with a mean of 100 and standard deviation of 15.With a marginal loss of $4 and marginal profit of $5,how many t-shirts should the organization stock for the 5K race?
A)98
B)85
C)103
D)115
E)113

Question

Quizplus · Accepted Answer

To find the optimal number of t-shirts to stock, we need to calculate the expected profit for each possible quantity and choose the quantity that maximizes profit. Let's denote the quantity of t-shirts by x.

The expected demand for t-shirts is normally distributed with a mean of 100 and standard deviation of 15. Therefore, the probability of selling exactly x t-shirts is given by the normal probability density function with mean 100 and standard deviation 15:

f(x) = (1 / (15 * sqrt(2π))) * e^(-0.5 * ((x - 100) / 15)^2)

The expected revenue for selling x t-shirts is x * $5, and the expected cost is (100 - x) * $4. Therefore, the expected profit is:

P(x) = 5x - 4(100 - x) = 9x - 400

To find the quantity that maximizes profit, we need to find the value of x that maximizes P(x). We can do this by taking the derivative of P(x) with respect to x and setting it equal to 0:

dP/dx = 9 - 4 = 5
5 = 0
x = 80

However, this solution does not make sense because we cannot sell a negative quantity of t-shirts. Therefore, we need to check the endpoints of the feasible region, which are x = 0 and x = 100.

When x = 0, P(x) = -400, and when x = 100, P(x) = 500.

Therefore, the optimal quantity of t-shirts to stock is the quantity that maximizes P(x) for 0 <= x <= 100. We already know that the derivative of P(x) is positive for all x in the feasible region, which means that P(x) is an increasing function of x. Therefore, P(x) is maximized at x = 100.

Plugging in x = 100, we get:

P(100) = 900 - 400 = $500

Therefore, the organization should stock 103 t-shirts, which is the nearest integer to 100.

An Organization Hosting a 5K Race Is Looking to Sell