Question

Scores on a test have a mean of 71.4 and 6 percent of the scores are above 90. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth, if necessary.

Answer 1

Given:Mean, =71.4 and P(X > 90) =6% =0.06 (where P =Probability)

P(X > x) =P(Z > (X - )/)

P(X > 90) =P(Z > (90 - 71.4)/) =0.06

P(Z > 18.6/) =0.06

We know that P(Z > 1.555) =0.06

Therefore, 18.6/ =1.555 =18.6/1.555 =11.9614. Rounding to the nearest tenth would be 12.

Hence, the standard deviation, =12

[NOTE: How to know that P(Z > 1.555) =0.06? From "Cumulative Z-Table", the corresponding Z-value at the area(probability) of 0.94 is 1.555 and this 0.94 is P(Z 1.555) and thus, P(Z > 1.555) =1 - 0.94 =0.06].