Question 1168213: To help students improve their reading, a school district decides to implement a reading program. It is to be administered to the bottom 9% 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 115.8, find the cutoff score that will make a student eligible for the program. The standard deviation is 23. Assume the variable is normally distributed. Round z-value calculations to 2 decimal places and the final answer to the nearest whole number.
Cutoff score
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $X$ be the random variable representing the score on the reading achievement exam. We are given that $X$ follows a normal distribution with a mean ($\mu$) of $115.8$ and a standard deviation ($\sigma$) of $23$.
We want to find the cutoff score below which the bottom 9% of the students fall. This means we need to find the score $x$ such that $P(X < x) = 0.09$.
First, we need to find the z-score corresponding to the 9th percentile (0.09) of the standard normal distribution. We look up the z-value in a standard normal distribution table or use a calculator such that the cumulative probability to the left of $z$ is 0.09.
From the standard normal distribution table, the z-value corresponding to a cumulative probability of approximately 0.09 is around -1.34.
Now, we use the z-score formula to convert this z-value back to the original score scale:
$z = \frac{x - \mu}{\sigma}$
We have $z = -1.34$, $\mu = 115.8$, and $\sigma = 23$. We need to solve for $x$:
$-1.34 = \frac{x - 115.8}{23}$
Multiply both sides by 23:
$-1.34 \times 23 = x - 115.8$
$-30.82 = x - 115.8$
Add 115.8 to both sides:
$x = 115.8 - 30.82$
$x = 84.98$
We need to round the final answer to the nearest whole number.
$x \approx 85$
Therefore, the cutoff score that will make a student eligible for the program is approximately 85.
Final Answer: The final answer is $\boxed{85}$
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