Question

Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16. If 100 people are randomly selected, find the probability that their mean IQ score is greater than 103. (rounded to the four decimal places)

Answer 1

0.03 is the probability that for the sample mean IQ score is greater than 103.

Explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 16

Sample size, n = 100

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling =

`(\sigma)/(√(n)) = (16)/(√(100)) = 1.6`

P( mean IQ score is greater than 103)

P(x > 103)

`P( x > 103) = P( z > \displaystyle(103 - 100)/(1.6)) = P(z > 1.875)`

`= 1 - P(z \leq 1.875)`

Calculation the value from standard normal z table, we have,

`P(x > 103) = 1 - 0.970 =0.03= 3\%`

0.03 is the probability that for the sample mean IQ score is greater than 103.