Question

Scores on a test have a mean of 78.3 and 7 percent of the scores are above 89. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth, if necessary.

Answer 1

Mean, = 78.3

Standard deviation =

P(X < A) = P(Z < (A - )/)

P(X > 89) = 0.07

P(X < 89) = 1 - 0.07

P(Z < (89 - 78.3)/) = 0.93

Take the value of Z corresponding to 0.93 from standard normal distribution table.

(89 - 78.3)/ = 1.48

= 7.2