Question

The weight of corn chips dispensed into a 24-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 24.5 ounces and a standard deviation of 0.2 ounce. What proportion of the 24-ounce bags contain less than the advertised 24 ounces of chips

Answer 1

0.0062 = 0.62% of the 24-ounce bags contain less than the advertised 24 ounces of chips.

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 = (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 of 24.5 ounces and a standard deviation of 0.2 ounce.

This means that
`\mu = 24.5, \sigma = 0.2`

What proportion of the 24-ounce bags contain less than the advertised 24 ounces of chips?

This is the p-value of Z when X = 24. So

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

`Z = (24 - 24.5)/(0.2)`

`Z = -2.5`

`Z = -2.5` has a p-value of 0.0062.

0.0062 = 0.62% of the 24-ounce bags contain less than the advertised 24 ounces of chips.