Question

5) Intelligence as measured by IQ tests is normally distributed with a mean 100 and standard deviation 15. Suppose groups of 10 people is selected at random. What percentage of these groups will have a mean IQ between 105 and 110?

Answer 1

Solution :

Given that ,

mean =  $\mu$ = 100

standard deviation = $\sigma$ = 15

n = 10

$\mu$ = 100

T

$\sigma$
T
=$\sigma$  / $\sqrt$ n =15 / $\sqrt$ 10=4.7434

#Mean IQ between 105 and 110 is P(105<<110)

T

P(105<
T
<110 ) = P[(105 -100) / 4.7434<(
T
- $\mu$
T
) / $\sigma$
T
< (110 -100) / 4.7434]

= P(1.05<z <2.11)

=P(z<2.11)-P(z<1.05)

value of z is obtin from standard normal table

=0.9825-0.8541

=0.1284

12.84% of these group will have mean IQ between 105 and 110