Question Help Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100μ=100 and a standard deviation sigma equals 15σ=15. Find the probability that a randomly selected adult has an IQ between 8888 and 112112.

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Answer:

Step-by-step explanation:

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 problem, we have that:

[tex]\mu = 100, \sigma = 15[/tex]

Find the probability that a randomly selected adult has an IQ between 88 and 112.

This is the pvalue of Z when X = 112 subtracted by the pvalue of Z when X = 88.

X = 112

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{112 - 100}{15}[/tex]

[tex]Z = 0.8[/tex]

[tex]Z = 0.8[/tex] has a pvalue of 0.7881

X = 88

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{88 - 100}{15}[/tex]

[tex]Z = -0.8[/tex]

[tex]Z = -0.8[/tex] has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

57.62%  probability that a randomly selected adult has an IQ between 88 and 112.