In: Statistics and Probability
The average MCAT score follows a Normal distribution, with a mean of μ = 508 and a standard deviation of σ = 8. What is the probability that the mean IQ score of 100 randomly selected people will be more than 510?
Solution :
Given that ,
mean =
= 508
standard deviation =
= 8
n = 100
=
= 508
=
/
n = 8 /
100 = 0.8
P(
> 510) = 1 - P(
< 510)
= 1 - P[(
-
) /
< (510 - 508) / 0.8 ]
= 1 - P(z < 2.50)
Using z table,
= 1 - 0.9938
= 0.0062