Question

Items produced by a manufacturing process are supposed to weigh 90 grams. The manufacturing procon

Answer 1

Using the normal distribution, it is found that 0.26% of the items will either  weigh less than 87 grams or more than  93 grams.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem:

• The mean is of 90 grams, hence $\mu = 90$.

• The standard deviation is of 1 gram, hence $\sigma = 1$.

We want to find the probability of an item differing more than 3 grams from the mean, hence:

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

$Z = \frac{3}{1}$

$Z = 3$

The probability is P(|Z| > 3), which is 2 multiplied by the p-value of Z = -3.

• Looking at the z-table, Z = -3 has a p-value of 0.0013.

2 x 0.0013 = 0.0026

0.0026 x 100% = 0.26%

0.26% of the items will either  weigh less than 87 grams or more than  93 grams.

For more on the normal distribution, you can check brainly.com/question/24663213