Question

scores on a test have a mean of 74.6 and 5 percent of the scores are above 89. the scores have a distribution that is approximately normal. find the standard deviation. round your answer to the nearest tenth, if necessary.

Explanation:

Step1: Find the z - score corresponding to the upper 5%

Since 5% of the scores are above 89, the area to the left of the z - score is \(1 - 0.05=0.95\). Looking up in the standard normal distribution table (or using a calculator with a normal - distribution function), the z - score \(z\) such that \(P(Z<z)=0.95\) is approximately \(z = 1.645\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x = 89\), \(\mu=74.6\), and \(z = 1.645\).
Substitute the values into the formula: \(1.645=\frac{89 - 74.6}{\sigma}\).

Step3: Solve for \(\sigma\)

First, simplify the numerator: \(89 - 74.6=14.4\).
Then, from \(1.645=\frac{14.4}{\sigma}\), we can solve for \(\sigma\) by cross - multiplying: \(\sigma=\frac{14.4}{1.645}\).
\(\sigma=\frac{14.4}{1.645}\approx8.8\)

Answer:

\(8.8\)