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.

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 problem, we have that:

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

has a pvalue of 0.7881

X = 88

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.

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d

a_n = a_1 + (n - 1)d

the first term is 6.

the common difference d = ?

when you find d, replace n with 7 and d with the common difference to find your answer.