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. The Z-score corresponding to 81% in the standard normal distribution table is approximately 0.88. We then use the formula $X = \mu + Z\sigma$, where $X$ is the score we're looking for, $\mu$ is the mean (63.2), $Z$ is the Z-score (0.88), and $\sigma$ is the standard deviation (11.7). Plugging in the values, we get $X = 63.2 + 0.88 \times 11.7 = 73.5$.

Scores on a Test Are Normally Distributed with a Mean P81P _ { 81 }P81​

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