Assume that adults have IQ scores that are normally distributed with a mean of 95.9 and a standard deviation of 16.4. Find the following probabilities. Round your answers to three decimal places; add trailing zeros as needed. The probability that a randomly selected adult has an IQ greater than 122.6 is 0.052 . The probability that a randomly selected adult has an IQ lower than 92.0 is 0.413 . The probability that a randomly selected adult has an IQ between 80.0 and 110.0 is 0.639 .

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.

Step-by-step 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!!!

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

the answer is; biodiversity resulting from few ancestors

adaptive radiation occurs when a population is subjected to a large natural environment with spatial ecological variations. the parts of the population in the different ecologies begin to adapt to their local environments. due to non-random mating across the large populations, this causes the different populations from the large population to speciate within their local environments. they, therefore, diverge to different species sharing a common ancestor.