Answer:
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 15
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]
a) P(IQ greater than 131.5)
P(x > 131.5)
[tex]P( x > 131.5) = P( z > \displaystyle\frac{131.5 - 100}{15}) = P(z > 2.1)[/tex]
[tex]= 1 - P(z \leq 2.1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 131.5) = 1 - 0.982 = 0.018 = 1.8\%[/tex]
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5
