Question

scores on a standardized intelligence test are normally distributed with a mean of 100 and a standard deviat the test who score below 76. the percentage of people taking the test who score below 76 is % (type an integer or a decimal.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 76\), \(\mu=100\), and assume \(\sigma = 16\) (a common standard deviation for IQ tests).

$$z=\frac{76 - 100}{16}=\frac{-24}{16}=-1.5$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X < 76)\), which is equivalent to \(P(Z<-1.5)\) in the standard normal distribution (\(Z\sim N(0,1)\)).
Looking up the value of \(z = - 1.5\) in the standard normal table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: normalcdf(-1000,-1.5,0,1)), we find that \(P(Z < - 1.5)=0.0668\)

Answer:

\(6.68\)