Suppose the results of an IQ test are normally distributed, with a mean of 10 and a standard deviation of 16. If 150 people took the test, how many do you expect would score between 95 and 130?

To find the number of people who would score between 95 and 130, we need to use the concept of the standard normal distribution. We can convert the given values into z-scores and then find the proportion of values between these z-scores.

First, let's convert the scores of 95 and 130 into z-scores using the formula: z = (x - mean) / standard deviation.

For score 95:
z1 = (95 - 10) / 16 = 5.9375

For score 130:
z2 = (130 - 10) / 16 = 7.5

Next, we can use a standard normal distribution table or a calculator to find the proportions associated with these z-scores.

The proportion between 95 and 130 can be calculated as P(z1 < z < z2) or P(z < z2) - P(z < z1).

Using a standard normal distribution table, we find:
P(z < 5.9375) ≈ 1 (Note: The value is so large that it is essentially 1.)
P(z < 7.5) ≈ 1

Then, the proportion between these two z-scores is:
P(z1 < z < z2) = P(z < 7.5) - P(z < 5.9375) ≈ 1 - 1 ≈ 0

This implies that the proportion of people scoring between 95 and 130 in this normal distribution is almost zero. Therefore, we would expect a very small number, if any, to fall into this range.

Specifically, since 150 people took the test, we wouldn't expect any of them to score between 95 and 130, based on the given distribution parameters.