Question

The amount of cereal in a box is normal with mean 16.5 ounces. If the packager is required to fill at least 90% of the cereal boxes with 16 or more ounces of cereal, what is the largest standard deviation for the amount of cereal in a box?

Answer 1

Answer:

The largest standard deviation for the amount of cereal in a box is 0.3906.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

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

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 16$

The z-score of X = 16 has a pvalue of 0.1. So it is $Z = -1.28$. Now we have to find the value of $\sigma$.

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

$-1.28 = \frac{16 - 16.5}{\sigma}$

$-1.28\sigma = -0.5$

$\sigma = 0.3906$

The largest standard deviation for the amount of cereal in a box is 0.3906.