Question

The distribution of IQ scores can be modeled by a normal distribution with mean 100 and standard deviation 15.

Answer 1

Answer:

4.4% of the population with IQ between 120 and 125.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 15

We are given that the distribution of IQ scores is a bell shaped distribution that is a normal distribution.

a) Let X be a person's IQ score.

Then, density functions for IQ scores is given by:

$P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{z^2}{2}}\\\\\text{where,}\\\\z = \frac{x-\mu}{\sigma}\\\\P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\\\\P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-100)^2}{450}}$

b) P(population with IQ between 120 and 125.)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

$P(120 \leq x \leq 125) = P(\displaystyle\frac{120 - 100}{15} \leq z \leq \displaystyle\frac{125-100}{15}) = P(1.33 \leq z \leq 1.66)\\\\= P(z \leq 1.66) - P(z < 1.33)\\= 0.952 - 0.908 = 0.044 = 4.4\%$

$P(120 \leq x \leq 125) = 4.4\%$