Question

suppose sat mathematics scores are normally distributed with a mean of 518 and a standard deviation of 113. a university plans to recruit students whose scores are in the top 9%. what is the minimum score required for recruitment? round your answer to the nearest whole number, if necessary.

Explanation:

Step1: Find the z - score

We want the top \(9\%\) of the normal distribution. So the area to the left of the z - score is \(1 - 0.09=0.91\).
Using a standard normal table (or a calculator with a normal - distribution function, such as the inverse of the cumulative distribution function for a standard normal distribution \(Z\sim N(0,1)\)), we find that the z - score \(z\) corresponding to an area of \(0.91\) is approximately \(z = 1.34\).

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 \(X\sim N(\mu,\sigma^{2})\), \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
We know \(\mu = 518\), \(\sigma=113\), and \(z = 1.34\).
Substitute these values into the formula \(1.34=\frac{x - 518}{113}\).
First, multiply both sides of the equation by \(113\): \(1.34\times113=x - 518\).
Calculate \(1.34\times113 = 151.42\).
Then, solve for \(x\): \(x=151.42 + 518\).

Step3: Calculate the value of \(x\)

\(x=151.42+518=669.42\)

Answer:

\(669\)