Question

if the stanford - binet iq test is a normal curve with a mean of 100 and a standard deviation of 20, what percent of the population has an iq between 90 and 110?
54.6%
12.1%
71.4%
38.3%

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 100\) (mean) and \(\sigma=20\) (standard deviation).
For \(x = 90\), \(z_1=\frac{90 - 100}{20}=\frac{- 10}{20}=-0.5\)
For \(x = 110\), \(z_2=\frac{110 - 100}{20}=\frac{10}{20}=0.5\)

Step2: Use the standard normal distribution table

The probability \(P(-0.5<Z<0.5)\) can be found using the property \(P(-a < Z < a)=2\Phi(a)-1\), where \(\Phi(a)\) is the cumulative distribution function of the standard normal distribution.
From the standard normal table, \(\Phi(0.5)=0.6915\)
Then \(P(-0.5 < Z < 0.5)=2\times0.6915-1\)
\(P(-0.5 < Z < 0.5)=1.383 - 1=0.383\)

Answer:

38.3%