Question

Suppose that 12% of test scores for the unit 1 test were above 85 and 8% were below 70. Assuming a normal

distribution, what is the mean and standard deviation of the distribution of test scores?

Answer 1

Mean = 78.2

Standard deviation = 5.8

Explanation:

Mathematically z-score;

= (x-mean)/SD

From the question;

12% of test scores were above 85

Thus;

P( x > 85) = 12%

P(x > 85) = 0.12

Now let’s get the z-score that has a probability of 0.12

This can be obtained from the standard normal distribution table and it is = 1.175

Thus;

1.175 = (85 - mean)/SD

let’s call the mean a and the SD b

1.175 = (85-a)/b

1.175b = 85 - a

a = 85 - 1.175b ••••••••(i)

Secondly 8% of scores were below 70

Let’s find the z-score corresponding to this proportion;

We use the standard normal distribution table as usual;

P( x < 70) = 0.08

z-score = -1.405

Thus;

-1.405 =( 70-a)/b

-1.405b = 70-a

a = 70 + 1.405b ••••••(ii)

Equate the two a

70 + 1.405b = 85 - 1.175b

85 -70 = 1.405b + 1.175b

15 = 2.58b

b = 15/2.58

b = 5.81

a = 70 + 1.405b

a = 70 + 1.405(5.81)

a = 78.16

So mean = 78.2 and Standard deviation is 5.8