Question

Suppose a brewery has a filling machine that fills 12 ounce bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean of 12.4 ounces and a standard deviation of 0.04 ounce. Find the probability that the bottle contains between 12.3 and 12.36 ounces.

Answer 1

15.25% probability that the bottle contains between 12.3 and 12.36 ounces.

Explanation:

Problems of normally distributed samples are 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 = 12.4, \sigma = 0.04`

Find the probability that the bottle contains between 12.3 and 12.36 ounces.

This is the pvalue of Z when X = 12.36 subtracted by the pvalue of Z when X = 12.3

X = 12.36

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

`Z = (12.36 - 12.4)/(0.04)`

`Z = -1`

`Z = -1` has a pvalue of 0.1587

X = 12.3

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

`Z = (12.3 - 12.4)/(0.04)`

`Z = -2.5`

`Z = -2.5` has a pvalue of 0.0062

0.1587 - 0.0062 = 0.1525

15.25% probability that the bottle contains between 12.3 and 12.36 ounces.