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

Answer:

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

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 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

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

$Z = \frac{24 - 24.5}{0.2}$

$Z = -2.5$

$Z = -2.5$ 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.