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?

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.

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:

a) over 80?

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

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

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

has a pvalue of 0.9656.

X = 111

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

has a pvalue of 0.9505

1 - 0.9505 = 0.0495

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

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step-by-step explanation:

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