Question

A baker knows that the daily demand for strawberry pies is a random variable that follows the normal distribution with a mean of 31.8 pies and a standard deviation of 4.5 pies. Find the demand that has an 8% probability of being exceeded.

Answer 1

A demand of 38.1225 pies has an 8% probability of being excedeed.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 31.8, \sigma = 4.5`

Find the demand that has an 8% probability of being exceeded:

This is the value of X when Z has a pvalue of 0.92. So it is X when Z = 1.405.

`Z = (X - \mu)/(\sigma)`

`1.405 = (X - 31.8)/(4.5)`

`X - 31.8 = 1.405*4.5`

`X = 38.1225`

A demand of 38.1225 pies has an 8% probability of being excedeed.

Answer 2

The demand for strawberry pies that is exceeded with a probability of 0.08 is 38.1230 strawberry pies.

Explanation:

A normal random variable with mean Mu = 31.8 and standard deviation sd = 4.5, is standardized with the transformation:

Z = (X - Mu) / sd = (X - 31.8) / 4.5

For a probability of 0.08, P (Z > k) = P ([(X - 31.8) / 4.5] > k) = 0.08.

P (Z > 1.4051) = 0.08, thus, k = 1.4051

Now, if k = [(X - 31.8) / 4.5], then X = 4.5k + 31.8 = 4.5 (1.4051) + 31.8 = 38.1230.

The demand for strawberry pies that is exceeded with a probability of 0.08 is 38.1230 strawberry pies.