Question

scores on a standardized intelligence test are normally distributed with a mean of 100 and a standard deviation of 12 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 \(\sigma = 12\).

$$ z=\frac{76 - 100}{12}=\frac{-24}{12}=- 2 $$

Step2: Find the percentage using the standard normal distribution table

We want to find \(P(X\lt76)\), which is equivalent to \(P(Z\lt - 2)\) in the standard normal distribution (\(X\) is the original normal random variable with mean \(\mu = 100\) and standard deviation \(\sigma=12\), and \(Z\) is the standard normal random variable \(Z=\frac{X - \mu}{\sigma}\)).
Looking up the value of \(P(Z\lt - 2)\) in the standard normal distribution table, we know that \(P(Z\lt - 2)=0.0228\)

Answer:

\(2.28\)