Question

A set of sunglasses prices are normally distributed with a mean of 76 dollars and a standard deviation of 10 dollars. Find the probability that a randomly selected sunglasses will have a price between 63 dollars and 90 dollars.

Answer 1

Using the normal distribution, it is found that there is a 0.8224 = 82.24% probability that a randomly selected sunglasses will have a price between 63 dollars and 90 dollars.

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:

• Mean of 76, thus, $\mu = 76$.
• Standard deviation of 10, thus, $\sigma = 10$.

The probability of a price between 63 dollars and 90 dollars is the p-value of Z when X = 90 subtracted by the p-value of Z when X = 63, hence:

X = 90:

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

$Z = \frac{90 - 76}{10}$

$Z = 1.4$

$Z = 1.4$ has a p-value of 0.9192.

X = 63:

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

$Z = \frac{63 - 76}{10}$

$Z = -1.3$

$Z = -1.3$ has a p-value of 0.0968.

0.9192 - 0.0968 = 0.8224.

0.8224 = 82.24% probability that a randomly selected sunglasses will have a price between 63 dollars and 90 dollars.

A similar problem is given at https://brainly.com/question/24663213