Question

The mean score on a math quiz is 11.9 and 45.2​% of the students scored between 10.1 and 13.7. Complete parts a. and b.

A. determine the standard deviation of the quiz scores.

B. determine the percentage of test scores above 13.1.

Answer 1

Given that, mean = 11.9

And P(10.1 < X < 13.7) = 0.452 ( or 45.2%)

a) First we find, z-score such that,

P(-z < Z < z) = 0.452

=> 2 * P(Z < z) - 1 = 0.452

=> 2 * P(Z < z) = 1.452

=> P(Z < z) = 0.726

Using standard normal z-table we get, z-score corresponds to probability 0.726 is z = 0.60

We know,

z = ( X - mean) / sd

=> 0.60 = (13.7 - 11.9) / sd

=> sd = 1.8/0.60

=> sd = 3

Therefore, standard deviation is 3

b) We want to find, P(X > 13.1)

z = (13.1 - 11.9) /3 = 0.4

=> P(X > 13.1)

= P(Z > 0.4)

= 1 - P(Z < 0.4)

= 1 - 0.6554

= 0.3446

=> P(X > 13.1) = 0.3446

In percentage: 0.3446 * 100% = 34.46%