The test scores on a certain Society of Actuaries professonal examination are No

Question

The test scores on a certain Society of Actuaries professonal examination are Normally distributed with a mean score of =65 and a standard deviation of =7%.

(a) What is the probability that a random chosen person who is writing this test will score at most 68%?

(b) What proportion of all persons writing this test will score between 74% and 82% on the exam?

(c) 34% of all persons writing this Examination will not pass. What is the minimum mark needed to pass this test?

Please post your steps. Thanks!

Explanation / Answer

Mean is 65 and s is 7. z is given as (x-mean)/s

a) P(x<68)=P(z<(68-65)/7)=P(z<0.43), from normal distribution table we get 0.6664

b) P(74<x<82)=P((74-68)/7<z<(82-68)/7)=P(0.86<z<2) or P(z<2)-P(z<0.86), from normal distribution table we get 0.9772-0.8051=0.1721

c) we need to find the z value for 0.34 , which will be negative of 1-0.34 i.e negative of 0.66. thus z from the normal table is -0.41

thus answer is mean+z*s=65-0.41*7=62.13

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