Question

Suppose a normally distributed set of data has a mean of 102 and a standard deviation of 20. Use the 68-95-99.7 Rule to determine to determine the percent of scores in the data set expected to be below a score of 151. Give your answer as a percent and includeas many decimal places as the 68-95-99.7 rule dictates. (For example, enter 99.7 instead of 0.997.)

Answer 1

0.71%

Explanation:

Given that a normally distributed set of data has a mean of 102 and a standard deviation of 20.

Let X be the random variable

Then X is N(102, 20)

We can convert this into standard Z score by

`z=(x-102)/(20)`

We are to find the probability and after wards percentage of scores in the data set expected to be below a score of 151.

First let us find out probability using std normal table

P(X<151) =
`P(Z<(151-102)/(20) \\=P(Z<2.45)\\=0.5-0.4929\\\\=0.0071`

We can convert this into percent as muliplying by 100

percent of scores in the data set expected to be below a score of 151.

=0.71%