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.

Jannat Arora asked Jul 28, 2023

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Jannat Arora asked Jul 28, 2023

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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:

Answer

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

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