Question

assume that adults have iq scores that are normally distributed with a mean of 95.2 and a standard deviation 22.1. find the first quartile ( q_1 ), which is the iq score separating the bottom 25% from the top 75%. (hint: draw a graph.)
the first quartile is
(type an integer or decimal rounded to one decimal place as needed.)

Explanation:

Step1: Find the z - score corresponding to the bottom 25%

Using the standard normal distribution table (or a calculator with a normal - distribution function), the z - score \(z\) such that \(P(Z\lt z)=0.25\). The z - score is approximately \(z=- 0.674\).

Step2: Use the z - score formula to find the IQ score \(x\)

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 95.2\) (mean) and \(\sigma = 22.1\) (standard deviation).
We want to solve for \(x\), so we can rewrite the formula as \(x=\mu+z\sigma\).
Substitute \(\mu = 95.2\), \(z=-0.674\), and \(\sigma = 22.1\) into the formula:
\(x=95.2+(-0.674)\times22.1\)
\(x = 95.2-0.674\times22.1\)
First, calculate \(0.674\times22.1\):
\(0.674\times22.1=(0.6 + 0.07+0.004)\times22.1=0.6\times22.1+0.07\times22.1 + 0.004\times22.1\)
\(=13.26+1.547+0.0884 = 14.8954\)
Then \(x=95.2-14.8954=80.3046\approx80.3\)

Answer:

\(80.3\)