Assume that adults have IQ scores that are normally distributed with a mean of 96 and a standard deviation of 15.7. Find the probability that randomly selected adults has and IQ greater than 123.4

4.05% probability that a randomly selected adult has an IQ greater than 123.4.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

Probability that a randomly selected adult has an IQ greater than 123.4.

This is 1 subtracted by the pvalue of Z when X = 123.4. So

has a pvalue of 0.9595

1 - 0.9595 = 0.0405

4.05% probability that a randomly selected adult has an IQ greater than 123.4.

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b