Scores on a test are normally distributed with a mean of 63.2 and a standard deviation of 11.7 . Find $P _ { 81 } ,$ which separates the bottom 81% from the top 19% .
A) 0.291
B) 73.5
C) 0.88
D) 66.6

Question

Scores on a test are normally distributed with a mean of  63.2  and a standard deviation of  11.7 . Find $P _ { 81 } ,$ which separates the bottom  81%  from the top  19% .
A)  0.291 
B)  73.5 
C)  0.88 
D)  66.6

Quizplus · Accepted Answer

To find $P_{81}$, which separates the bottom 81% from the top 19%, we use the Z-score formula and the normal distribution properties. Given the mean ($\mu$) is 63.2 and the standard deviation ($\sigma$) is 11.7, we look for the Z-score corresponding to the 81st percentile in the standard normal distribution table. The Z-score for 81% (or 0.81 probability) is approximately 0.88. We then apply the formula: $X = \mu + Z\sigma$, where $X$ is the score we're looking for. Substituting the values, $X = 63.2 + 0.88 \times 11.7$, we get approximately 73.5.

Scores on a Test Are Normally Distributed with a Mean P81,P _ { 81 } ,P81​,

Scores on a Test Are Normally Distributed with a Mean $P _ { 81 } ,$