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

Question

1 year ago

9

exam.

1 answer:

Darya [45] · Answer 1 · 2024-06-01T04:44:54+03:00

8 0

Step by step explanation

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

$z=\frac{90-77.7}{12.1}\approx1.0165$

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%