Question

Suppose a normal distribution has a mean of 38 and a standard deviation of

2. What is the probability that a data value is between 36 and 41? Round your
answer to the nearest tenth of a percent.​

Answer 1

77.4% probability that a data value is between 36 and 41

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

`\mu = 38, \sigma = 2`

What is the probability that a data value is between 36 and 41?

This is the pvalue of Z when X = 41 subtracted by the pvalue of Z when X = 36.

X = 41

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

`Z = (41 - 38)/(2)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.933

X = 36

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

`Z = (36 - 38)/(2)`

`Z = -1`

`Z = -1` has a pvalue of 0.159

0.933 - 0.159 = 0.774

77.4% probability that a data value is between 36 and 41