Question

Jackson is interested in the number of memories people can recall. He knows that the number of memories recalled is normally distributed in the general population, with a population mean of 30 and a population standard deviation of 5. What is the probability that Jackson randomly selects a person from this population who recalls between 26 and 29 memories

Answer 1

0.2088 = 20.88% probability that Jackson randomly selects a person from this population who recalls between 26 and 29 memories.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

`Z = (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 p-value, we get the probability that the value of the measure is greater than X.

Population mean of 30 and a population standard deviation of 5.

This means that
`\mu = 30, \sigma = 5`

What is the probability that Jackson randomly selects a person from this population who recalls between 26 and 29 memories?

This is the pvalue of Z when X = 29 subtracted by the pvalue of Z when X = 26. So

X = 29

`Z = (X - \mu)/(\sigma)`

`Z = (29 - 30)/(5)`

`Z = -0.2`

`Z = -0.2` has a pvalue of 0.4207

X = 26

`Z = (X - \mu)/(\sigma)`

`Z = (26 - 30)/(5)`

`Z = -0.8`

`Z = -0.8` has a pvalue of 0.2119

0.4207 - 0.2119 = 0.2088

0.2088 = 20.88% probability that Jackson randomly selects a person from this population who recalls between 26 and 29 memories.