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:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling =
[tex]\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{100}} = 1.6[/tex]
P( mean IQ score is greater than 103)
P(x > 103)
[tex]P( x > 103) = P( z > \displaystyle\frac{103 - 100}{1.6}) = P(z > 1.875)[/tex]
[tex]= 1 - P(z \leq 1.875)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 103) = 1 - 0.970 =0.03= 3\%[/tex]
0.03 is the probability that for the sample mean IQ score is greater than 103.
