I really need help with these 3 questions! 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. Assuming a normally distributed curve, what percent of students scored between 81.5 and 91.7? Remember to include your % sign.

Answer 1

Hello,

 

When trying to calculate probabilities using distributions, it is usually easier to compute z-scores. Any z-score can be computed using the following formula:

 

z = (X - M)/S, where X is a score, M is the mean and S is the standard deviation. In your case, we have two:

 

(81.5 - 81.5)/5.1 = 0

and

(91.7 - 81.5)/5,1 = 2

 

Now we need to determine what percentage of scores are between z=0 and z=2. To do this, we need to use a normal distribution table (your teacher should have given you one of there should be one in your textbook). When we look up these z-scores, the value we find is 0.47725. This means that 47.725% of scores are between 81.5 and 91.7