Question

Adult intelligence scores are distributed approximately normally with μ = 100 and σ =
15. Therefore, means of simple random samples of n intelligence scores are distributed
approximately normally with with μ = 100 and σ = 15/Ön. In each part of this question, carry out any
calculations using two places after the decimal point for z scores and four places after the decimal point for
proportions.

a) What proportion of intelligence scores is lower than 94?

b)  What proportion of means of random samples of 9 intelligence scores is lower than 94?

c) What proportion of means of random samples of 25 intelligence scores is lower
than 94?

d) How low must an intelligence score be to be in the lowest 25?

e) How low must the mean of a random sample of 9
intelligence scores be to be in the lowest 2%?

f) How low must the mean of a random sample of 25
intelligence scores be to be in the lowest 2%?

Answer 1

Solution :

Given that ,

mean = = 100

standard deviation = =15

P(x <94 ) = P(( x -) / (94-100) / 15)

= P(z < -0.4)

Using z table

= 0.3446

b.

n = 9

=100

=  / n = 15/ 9=5

P( < 94) = P[( - ) / < (94-100) / 5]

= P(z < -1.2)

Using z table

= 0.1151  

c.

n=25

=100

=  / n = 15/ 25=3

P( < 94) = P[( - ) / < (94-100) / 3]

= P(z < -2)

Using z table

= 0.0228

d.

Using standard normal table,

P(Z < z) = 25%

=(Z < z) = 0.25  

= P(Z < z ) = 0.25  

z = -0.67

Using z-score formula  

x = z +

x = -0.67*15+100

x = 89.95

x=90

e.

n = 9

= 100

= / n = 15 /9=5

Using standard normal table,

P(Z < z) = 2%

= P(Z < z) = 0.02  

= P(Z < -2.05) = 0.02

z = -2.05

Using z-score formula  

= z * +

= -2.05*5+100

= 89.75

f.

n = 25

= 100

= / n = 15 /25=3

Using standard normal table,

P(Z < z) = 2%

= P(Z < z) = 0.02  

= P(Z < -2.05) = 0.02

z = -2.05

Using z-score formula  

= z * +

= -2.05*3+100

= 93.85