Question

Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Find the probability that a randomly selected adult has an IQ greater than 131.5. Group of answer choices

Answer 1

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:

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

a) P(IQ greater than 131.5)

P(x > 131.5)

$P( x > 131.5) = P( z > \displaystyle\frac{131.5 - 100}{15}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

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

$P(x > 131.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is  the probability that a randomly selected adult has an IQ greater than 131.5