Question

A popular soft drink is sold in 2-liter (2,000-milliliter) bottles. Because of variation in the filling process, bottles have a mean of 2,000 milliliters and a standard deviation of 20, normally distributed. a. If the process overfills the bottle by more than 50 milliliters, the overflow will cause a machine malfunction. What is the probability of this occurring

Answer 1

0.0062 = 0.62% probability of this occurring

Explanation:

When the distribution is normal, we use 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 question, we have that:

`\mu = 2000, \sigma = 20`

a. If the process overfills the bottle by more than 50 milliliters, the overflow will cause a machine malfunction. What is the probability of this occurring

Probability of filling the bottle with more than 2000 + 50 = 2050 milliliters, which is 1 subtracted by the pvalue of Z when X = 2050. So

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

`Z = (2050 - 2000)/(20)`

`Z = 2.5`

`Z = 2.5` has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.0062 = 0.62% probability of this occurring