Question

Suppose that the number of gallons of milk sold per day at a local supermarket are normally distributed with mean and standard deviation of 486.9 and 24.01, respectively. What is the probability that on a given day the supermarket will sell between 477 and 525 gallons of milk

Answer 1

Answer:

0.6032 = 60.32% probability that on a given day the supermarket will sell between 477 and 525 gallons of milk

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Mean and standard deviation of 486.9 and 24.01, respectively.

This means that $\mu = 486.9, \sigma = 24.01$

What is the probability that on a given day the supermarket will sell between 477 and 525 gallons of milk?

This is the p-value of Z when X = 525 subtracted by the p-value of Z when X = 477.

X = 525

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{525 - 486.9}{24.01}$

$Z = 1.59$

$Z = 1.59$ has a p-value of 0.9441

X = 477

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{477 - 486.9}{24.01}$

$Z = -0.41$

$Z = -0.41$ has a p-value of 0.3409

0.9441 - 0.3409 = 0.6032

0.6032 = 60.32% probability that on a given day the supermarket will sell between 477 and 525 gallons of milk