Question

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

(a) Let X be a person's IQ score. Write the formula for the density function of IQ scores.
p(x)=

(b) Estimate the fraction of the population with IQ between 120 and 125.
fraction =

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:

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

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

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

Answer 2

Using the normal distribution, it is found that:

a)

The density function is:

$P(x) = \frac{1}{2\sqrt{450\pi}}e^{-\frac{(x - 100)^2}{450}}$

b)

0.0443 = 4.43% of the population has IQ between 120 and 125.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem:

• Mean of 100 inches, thus $\mu = 100$.
• Standard deviation of 15 inches, thus $\sigma = 15$.

Item a:

The density function for the normal distribution is:

$P(x) = \frac{1}{2\sqrt{2\pi\sigma^2}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}$

Considering $\mu = 100, \sigma = 15$.

$P(x) = \frac{1}{2\sqrt{450\pi}}e^{-\frac{(x - 100)^2}{450}}$

Item b:

The fraction is the p-value of Z when X = 125 subtracted by the p-value of Z when X = 120, thus:

X = 125:

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

$Z = \frac{125 - 100}{15}$

$Z = 1.67$

$Z = 1.67$ has a p-value of 0.9525.

X = 120:

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

$Z = \frac{120 - 100}{15}$

$Z = 1.33$

$Z = 1.33$ has a p-value of 0.9082.

0.9525 - 0.9082 = 0.0443.

0.0443 = 4.43% of the population has IQ between 120 and 125.

A similar problem is given at https://brainly.com/question/16873840