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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

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Answer:

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.

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