Question

The weights of items produced by a company are normally distributed with a mean of 5 ounces and a standard deviation of 0.2 ounces. What is the minimum weight of the heaviest 9.85% of all items produced?

Answer 1

Answer:

The minimum weight of the heaviest 9.85% of all items produced is 5.26 ounces.

Step-by-step 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:

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

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 question, we have that:

$\mu = 5, \sigma = 0.2$

What is the minimum weight of the heaviest 9.85% of all items produced?

This is the 100 - 9.85 = 90.15th percentile, which is X when Z has a pvalue of 0.9015. So X when Z = 1.29.

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

$1.29 = \frac{X - 5}{0.2}$

$X - 5 = 1.29*0.2$

$X = 5.26$

The minimum weight of the heaviest 9.85% of all items produced is 5.26 ounces.