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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{100}} = 1.6$

P( mean IQ score is greater than 103)

P(x > 103)

$P( x > 103) = P( z > \displaystyle\frac{103 - 100}{1.6}) = P(z > 1.875)$

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

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.