Question

the iq scores on the weschler adult intelligence scale (wais) follow a normal distribution with a mean of 100 and a standard deviation of 15. using a z - score to standardize the given values, what proportion of iq scores are lower than 120? (all problems in this edla were done using the formulas and the \by hand\ method. if you are using a calculator then round your answers to 3 decimal places and you should be able to match the correct answer.) 0.9082 0.0918 0.0000 1.0000

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 120\), \(\mu=100\), and \(\sigma = 15\).

$$ z=\frac{120 - 100}{15}=\frac{20}{15}\approx1.33 $$

Step2: Find the proportion using the standard normal table

We want to find \(P(X < 120)\), which is equivalent to \(P(Z<1.33)\) in the standard normal distribution.
Looking up the value of \(z = 1.33\) in the standard normal table (the cumulative - distribution function of the standard normal distribution), we find that \(P(Z < 1.33)=0.9082\)

Answer:

0.9082