Question

reading improvement program to help students improve their reading, a school district decides to implement a reading program. it is to be administered to the bottom 6% of the students in the district, based on the scores on a reading achievement exam. if the average score for the students in the district is 116.5, find the cutoff score that will make a student eligible for the program. the standard deviation is 23. assume the variable is normally distributed. use the standard normal distribution table. round z - value calculations to 2 decimal places and the final answer to the nearest whole number. cutoff score =

Explanation:

Step1: Find the z - score

We want the bottom \(6\%\) of the normal distribution. So, we need to find the \(z\) - score \(z\) such that \(P(Z<z)=0.06\).
Looking up in the standard normal distribution table (the \(z\) - table), the \(z\) - score corresponding to a left - tail area of \(0.06\) is approximately \(z=-1.56\).

Step2: Use the z - score formula

The \(z\) - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 116.5\) (mean), \(\sigma = 23\) (standard deviation), and \(z=-1.56\).
We want to solve for \(x\). Rearranging the formula gives \(x=\mu+z\sigma\).
Substitute the values: \(x = 116.5+(-1.56)\times23\).
First, calculate \((-1.56)\times23=-35.88\).
Then, \(x=116.5 - 35.88=80.62\).

Answer:

\(81\)