Question

A standardized test in language facility has been administered in a district for many years and had a bell shaped distribution with mean 135 and standard deviation 16. An administrator wants to use the current classes test results to assign students to different classes. The upper 1/7 of students will be assigned to the advanced literature course.

What test score will separate the top 1/7 of students from the lower 6/7 of students?
a. 126
b. −0.57
c. 139
d. 152
e. 479
f. +1.07

Answer 1

To determine the test score that separates the top 1/7 of students from the lower 6/7, we can use the z-score associated with a cumulative probability of 6/7. By rearranging the formula, we can calculate the test score to be approximately 149.

Step-by-step explanation:

To calculate the test score that will separate the top 1/7 of students from the lower 6/7, we need to find the z-score associated with the upper 1/7.

We can use the formula z = (x - µ) / σ, where x is the desired test score, µ is the mean, and σ is the standard deviation.

Since we want the upper 1/7, we can use the z-score associated with a cumulative probability of 1 - 1/7 = 6/7.

Using a z-table or a calculator, we can find that the z-score corresponding to a cumulative probability of 6/7 is approximately 0.8416.

Now we can rearrange the formula to solve for x: x = z * σ + µ. Plugging in the values, we get x = 0.8416 * 16 + 135 = 148.6656. Rounded to the nearest whole number, the test score that separates the top 1/7 from the lower 6/7 is 149.