Suppose that the number of gallons of milk sold per day at a local supermarket are normally distributed with mean and standard deviation of 442.8 and 20.6, respectively. What is the probability that on a given day the supermarket will sell between 410 and 424 gallons of milk?

Respuesta :

Answer:

12.55% probability that on a given day the supermarket will sell between 410 and 424 gallons of milk.

Step-by-step explanation:

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 442.8, \sigma = 20.6[/tex]

What is the probability that on a given day the supermarket will sell between 410 and 424 gallons of milk?

This is the pvalue of Z when X = 424 subtracted by the pvalue of Z when X = 410.

X = 424

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{424 - 442.8}{20.6}[/tex]

[tex]Z = -0.91[/tex]

[tex]Z = -0.91[/tex] has a pvalue of 0.1814.

X = 410

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{410 - 442.8}{20.6}[/tex]

[tex]Z = -1.59[/tex]

[tex]Z = -1.59[/tex] has a pvalue of 0.0559.

So there is a 0.1814 - 0.0559 = 0.1255 = 12.55% probability that on a given day the supermarket will sell between 410 and 424 gallons of milk.