Question

The mean score on an exam was 77.7 and the standard deviation was 12.1Find the probability that a person scored above 90 on this exam.

Answer 1

Step by step explanation

Lets first find the z-score for 90 in this normal distribution:

`z=(90-77.7)/(12.1)\approx1.0165`

Reviewing the table, for this value we have:

`P(x>90)=0.1547`

Answer

The probability that a person scored above 90 is 0.1547, that is 15.47%