Question

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What percentage of people have an IQ score between 69 and 132, to the nearest tenth?

Answer 1

Answer:

96.4% of people have an IQ score between 69 and 132

Step-by-step explanation:

When the distribution is normal, we use 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:

$\mu = 100, \sigma = 15$

What percentage of people have an IQ score between 69 and 132, to the nearest tenth?

The proportion is the pvalue of Z when X = 142 subtracted by the pvalue of Z when X = 69. The percentage is the proportion multiplied by 100. So

X = 132

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

$Z = \frac{132 - 100}{15}$

$Z = 2.13$

$Z = 2.13$ has a pvalue of 0.983

X = 69

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

$Z = \frac{69 - 100}{15}$

$Z = -2.07$

$Z = -2.07$ has a pvalue of 0.019

0.983 - 0.019 = 0.964

0.964*100 = 96.4%

96.4% of people have an IQ score between 69 and 132