Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105μ=105 and a standard deviation sigma equals 20σ=20. Find the probability that a randomly selected adult has an IQ between 8989 and 121121.

57.62% probability that a randomly selected adult has an IQ between 89 and 121.

Step-by-step explanation:

When the distribution is normal, we use 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:

Find the probability that a randomly selected adult has an IQ between 89 and 121.

This is the pvalue of Z when X = 121 subtracted by the pvalue of Z when X = 89. So

X = 121

has a pvalue of 0.7881

X = 89

has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

57.62% probability that a randomly selected adult has an IQ between 89 and 121.

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step-by-step explanation:

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answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column