Question

according to a recent reporting on a standardized test, the average math score for students in a particular state was 558. assume the scores are normally distributed with a standard deviation of 104. answer parts (a) through (c) below including an appropriately labeled and shaded normal curve for each part.
the percentage that scored between 600 and 652 is 16.1% (round to one decimal place as needed.)
c. suppose students who scored in the top 5% of test takers in the state were eligible for a special scholarship program. what score would qualify students for this scholarship program?

Explanation:

Step1: Find the z - score for the top 5%

The top 5% means the area to the right of the z - score is \(0.05\). So the area to the left is \(1 - 0.05=0.95\). Using a standard normal table or calculator, the z - score corresponding to an area of \(0.95\) is approximately \(z = 1.645\).

Step2: Use the z - score formula to find the raw score

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 558\) (mean), \(\sigma=104\) (standard deviation), and \(z = 1.645\).
Rearrange the formula for \(x\): \(x=\mu+z\sigma\).
Substitute the values: \(x = 558+1.645\times104\).
First, calculate \(1.645\times104 = 1.645\times(100 + 4)=1.645\times100+1.645\times4=164.5+6.58 = 171.08\).
Then, \(x=558 + 171.08=729.08\approx729.1\).

Answer:

The score that would qualify students for the scholarship program is approximately \(729.1\). For the normal curve, option A is correct as it shows the shaded area (top 5%) on the right - hand side of the curve.