Answer:
a) The probability that a randomly selected adult has an IQ greater than 122.6 is 0.052.
b) The probability that a randomly selected adult has an IQ lower than 92.0 is 0.405.
c) The probability that a randomly selected adult has an IQ between 80.0 and 110.0 is 0.639.
Explanation:
This is a normal distribution problem with
Mean = μ = 95.9
Standard deviation = σ = 16.4
a) The probability that a randomly selected adult has an IQ greater than 122.6
= P(x > 122.6)
We first normalize or standardize 122.6
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (122.6 - 95.9)/16.4 = 1.63
To determine the required probability
P(x > 122.6) = P(z > 1.63)
We'll use data from the normal distribution table for these probabilities
P(x > 122.6) = P(z > 1.63) = 1 - P(z ≤ 1.63)
= 1 - 0.94845
= 0.05155 = 0.052 to 3 d.p
b) The probability that a randomly selected adult has an IQ lower than 92.0 = P(x < 92)
We first normalize or standardize 92.0
z = (x - μ)/σ = (92 - 95.9)/16.4 = - 0.24
To determine the required probability
P(x < 92.0) = P(z < -0.24)
We'll use data from the normal distribution table for these probabilities
P(x < 92.0) = P(z < -0.24) = 0.40517 = 0.405 to 3 d.p
c) The probability that a randomly selected adult has an IQ between 80.0 and 110.0.
P(80 ≤ x ≤ 110)
We first normalize or standardize 80.0 and 110.0
For 80.0
z = (x - μ)/σ = (80 - 95.9)/16.4 = - 0.97
For 110.0
z = (x - μ)/σ = (110 - 95.9)/16.4 = 0.86
The required probability
P(80 ≤ x ≤ 110) = P(-0.97 ≤ z ≤ 0.86)
We'll use data from the normal distribution table for these probabilities
(80 ≤ x ≤ 110) = P(-0.97 ≤ z ≤ 0.86)
= P(z ≤ 0.86) - P(z ≤ -0.97)
= 0.80511 - 0.16602
= 0.63909 = 0.639 to 3 d.p
Hope this Helps!!!