1. (b) What proportion of the population would have IQ above 115? (express as a decimal, to 3 decimal places)

To find the proportion of the population with an IQ above 115, we first need to calculate the z-score for an IQ of 115. The z-score is a measure of how many standard deviations an element is from the mean.

The formula for the z-score is (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation.

So, for an IQ of 115, the z-score would be (115 - 100) / 15 = 1.

In a normal distribution, about 84% of the data falls below 1 standard deviation above the mean. This is a known value from the standard normal distribution table.

So, if 84% of people have an IQ below 115, then 100% - 84% = 16% of people have an IQ above 115.

Expressed as a decimal to three decimal places, this is 0.160.