Answer:
0.1587 is the probability that the sample mean of 100 randomly selected adult's IQ is greater than 120.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 118
Standard Deviation, σ = 20
n = 100
We are given that the distribution of IQ scores is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
P( IQ is greater than 120)
P(x > 120)
[tex]P( x > 120) = P( z > \displaystyle\frac{120 - 118}{\frac{20}{\sqrt{100}}}) = P(z > 1)[/tex]
[tex]= 1 - P(z \leq 1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 120) = 1 - 0.8413 = 0.1587 = 15.87\%[/tex]
0.1587 is the probability that the sample mean of 100 randomly selected adult's IQ is greater than 120.
