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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.)

Respuesta :

Answer:

0.71%

Step-by-step 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

[tex]z=\frac{x-102}{20}[/tex]

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) = [tex]P(Z<\frac{151-102}{20} \\=P(Z<2.45)\\=0.5-0.4929\\\\=0.0071[/tex]

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%