The mean Algebra 2 test score is 81.5, with a standard deviation of 5.1

Question

The mean Algebra 2 test score is 81.5, with a standard deviation of 5.1. There are 130 students taking the course. Assuming a normally distributed curve, how many students scored between 76.4 and 86.6?

Answer 1

Hi Caleb

 

To solve this problem we need a z-table.

 

We need to values from the z table, one for 76.4 and one for 86.6.

 

Recall 

        X-μ

z = -------- 

          σ

 

For X = 86.6

z = (86.6-81.5)/5.1 = 1

 

For X = 76.4

z = (76.4-81.5)/5.1 = -1

 

Now we need to look these values up in a z table.

 

For z=1 we have 0.8413

For z=-1 we have 0.1587

 

Recall that these values represent the percentage from -∞ to the value shown.

 

To get the percentage between the two scores, we nee the difference between the two z values

0.8413-0.1587 = 0.6826

 

So 68.26% of the students scored between 76.4 and 86.6.

 

Since 130 students took the test, this gives us 

130(0.6826) = 88.738

 

88 students scored between 76.4 and 86.6.