Question

. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person selected at random will have an IQ of 110 or greater?

Answer 1

Answer:

2.28% probability that a person selected at random will have an IQ of 110 or greater

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 problem, we have that:

$\mu = 100, \sigma = 5$

What is the probability that a person selected at random will have an IQ of 110 or greater?

This is 1 subtracted by the pvalue of Z when X = 110. So

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

$Z = \frac{110 - 100}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that a person selected at random will have an IQ of 110 or greater