Question

suppose that scores on a particular test are normally distributed with a mean of 120 and a standard deviation of 16. what is the minimum score needed to be in the top 15% of the scores on the test? carry your intermediate computations to at least four decimal places, and round your answer to one decimal place.

Explanation:

Step1: Find the z - score

We want the top 15% of the scores. So the area to the left of the z - score is $1 - 0.15=0.85$. Looking up in the standard normal distribution table (or using a calculator with a normal - distribution function), the z - score corresponding to an area of 0.85 is approximately $z = 1.0364$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the original normal distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know $\mu = 120$, $\sigma = 16$, and $z = 1.0364$. Rearranging the formula for $x$ gives $x=\mu+z\sigma$.

Step3: Calculate the value of x

Substitute the values into the formula: $x = 120+1.0364\times16$. First, calculate $1.0364\times16 = 16.5824$. Then, $x=120 + 16.5824=136.5824$.

Answer:

$136.6$