Question

The manager of a healthy food store has determined that the weekly demand for a popular type of granola is a normally distributed random variable with mean 85 pounds and standard deviation 5 pounds. If the demand for a given week falls within the lowest 2.5% of all possible values for the weekly demand, the price of the granola will be reduced for the following week. Calculate the value in pounds (lbs) for the weekly demand below which the manager will have to reduce the price. (Use R and give code)

Answer 1

Solution :

Let X be a random variable which represents the weekly demand for a popular type of granola.

Given that, X ~ N(85, 5²)

Mean (μ) = 85 pounds

SD (σ) = 5 pounds

The manager will have to reduce the price if the demand for a given week falls within the lowest 2.5% of all possible values for the weekly demand.

Let the value that divide the lowest 2.5% of all possible weekly demand with the rest of the weekly demands is k.

Hence, P(X < k) = 0.025

We know that if X ~ N(μ, σ²) then,

..........................(1)

Using "qnorm" function of R we get, P(Z < -1.96) = 0.025.

Note : To obtain the z-score such that P(Z < z) = 0.025 in R simply type the function qnorm(0.025) in R and run, you will get the value -1.959964. On rounding this value to two decimal places we get 1.96.

On comparing P(Z < -1.96) = 0.025 and (1) we get,

Hence, the weekly demand below which the manager will have to reduce the price is 75.2 pounds.

The R code for the question is given below :

When we execute the above code in R, we shall get the following output :

Hence, the weekly demand below which the manager will have to reduce the price is 75.2 pounds.