Question

Based on the Normal model ​N(100​,17​) describing IQ​ scores, what percent of individuals would be expected to have scores ​a) over 80​? ​b) under 89​? ​c) between 111 and 131​? ​d) over 128​?

Answer 1

a) 88.10% of students would be expected to score over 80

b) 25.78% of students would be expected to score under 89.

c) 22.34% of students would be expected to score between 111 and 131.

d) 4.95% of students would be expected to score over 128.

Explanation:

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

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

`Z = (X - \mu)/(\sigma)`

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:

`\mu = 100, \sigma = 17`

a) over 80​? ​

1 subtracted by the pvalue of Z when X = 80.

`Z = (X - \mu)/(\sigma)`

`Z = (80 - 100)/(17)`

`Z = -1.18`

`Z = -1.18` has a pvalue of 0.1190

1 - 0.1190 = 0.8810

88.10% of students would be expected to score over 80

b) under 89​? ​

Pvalue of Z when X = 89. So

`Z = (X - \mu)/(\sigma)`

`Z = (89 - 100)/(17)`

`Z = -0.65`

`Z = -0.65` has a pvalue of 0.2578.

25.78% of students would be expected to score under 89.

c) between 111 and 131​? ​

Pvalue of Z when X = 131 subtracted by the pvalue of Z when X = 111.

X = 131

`Z = (X - \mu)/(\sigma)`

`Z = (131 - 100)/(17)`

`Z = 1.82`

`Z = 1.82` has a pvalue of 0.9656.

X = 111

`Z = (X - \mu)/(\sigma)`

`Z = (111 - 100)/(17)`

`Z = 0.65`

`Z = 0.65` has a pvalue of 0.7422.

0.9656 - 0.7422 = 0.2234

22.34% of students would be expected to score between 111 and 131.

d) over 128​?

1 subtracted by the pvalue of Z when X = 128. So

`Z = (X - \mu)/(\sigma)`

`Z = (128 - 100)/(17)`

`Z = 1.65`

`Z = 1.65` has a pvalue of 0.9505

1 - 0.9505 = 0.0495

4.95% of students would be expected to score over 128.