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.

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step-by-step explanation:

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3