Question

Items produced by a manufacturing process are supposed to weigh 90 grams. The manufacturing procon
such, however, that there is variability in the items produced and they do not all weigh exactly 90
distribution of weights can be approximated by a Normal distribution with mean 90 grams and a standard
deviation of 1 gram. About what percentage of the items will either
weigh less than 87 grams or more than
93 grams?

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 https://brainly.com/question/24663213