Question

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.Out of a randomly selected 1650 people from the population, how many of themwould have an IQ between 79 and 100, to the nearest whole number?

Answer 1

The probability density for a selected person to be in this interval of IQ is:

P(79< x < 100)

Here, x is a variable representing the IQ of that person. Now, to evaluate this probability density, we first need to find the z-score of this variable. Then, we can use a z-score table to solve it. Finally, we multiply the probability density by the number of people, that is, by 1650.

The z-score is found this way:

z = (x-mean)/(stardard deviation)

P(79< x < 100) = P( [79-100]/15 < z < [100-100]/15) = P(-1.4 < z < 0)

Now, the probability of finding z in this interval equals the probability that z is less than 0, minus the probability that z is less or equal to -1.4:

P(79 < x < 100) = P(-1.4 < z < 0) = P(z < 0) - P(z ≤ -1.4)

These values can be found in a z-score table, so we get:

P(z < 0 ) = 0.5

P(z ≤ -1.4) = 0.0808

Then:

P(79 < x < 100) = 0.5 - 0.0808 = 0.4192

Finally, multiplying this probability density by the number of people, we find:

0.4192 * 1650 = 691.68 ~ 692

Therefore, 692 would have an IQ in that interval.