Question

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 8. Only 5% of the students taking the test scored higher than what grade? (Ch 6) answer is 83.81 but please show and explain how, what z table you used and the numbers. thanks

Answer 1

Answer

Given that

mean = 69

standard deviation (sd) = 8

we have to find test score corresponding to top 5%, here top 5% means test score higher than bottom 95%

We will use percentile to z score table, this table is the z critical value table for one tailed hypothesis test

using this table, z score for 95 percentile is 1.645

So, required test score = mean +z*sd

setting mean = 69, sd = 8 and z = 1.645

test score = 69 + (1.645*8)

= 69+13.16

= 82.16

(please check your answer because the correct answer is 82.16, I have checked it using TI 84 calculator as well)