Question

If the inter-quartile range is the distance between the first and third quartiles, then the inter-decile range is the distance between the first and ninth decile. (Deciles divide a distribution into ten equal parts.) If IQ is normally distributed with a mean of 100 and a standard deviation of 16, what is the inter-decile range of IQ

Answer 1

Answer:

The inter-decile range of IQ is 40.96.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 question, we have that:

$\mu = 100, \sigma = 16$

First decile:

100/10 = 10th percentile, which is X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 100}{16}$

$X - 100 = -1.28*16$

$X = 79.52$

Ninth decile:

9*(100/10) = 90th percentile, which is X when Z has a pvalue of 0.9. So it is X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 100}{16}$

$X - 100 = 1.28*16$

$X = 120.48$

Interdecile range:

120.48 - 79.52 = 40.96

The inter-decile range of IQ is 40.96.