Scores on a 100-point final exam administered to all applied calculus classes at a large university are normally distributed with a mean of 69.5 and a standard deviation of 26.45.

Question

Scores on a 100-point final exam administered to all applied calculus classes at a large university are normally distributed with a mean of 69.5 and a standard deviation of 26.45.

What percentage of students taking the test had scores between 60 and 80? (Round your answer to one decimal place.)

Answer 1

Hi Aevenir,

This problem requires you to find Z scores and look up a distribution table.

In the context of this problem, you need two z-scores.

Z = (Score on Test - Average Score) / Standard Dev

Z₆₀ = (60 - 69.5) / 26.45 = -0.3592

Z₈₀ = (80 - 69.5) / 26.45 = 0.3970

Pull up a table that determines the area beneath the bell curve from negative infinity to your value.

For your Z₆₀ of -0.3592, your area is 0.3594.

For you Z₈₀ of 0.3970, your area is 0.6517.

You're interested in the people who are between these two scores, so you need to subtract the area for Z₆₀ from the area for Z₈₀.

0.6517 - 0.3594 = 0.2923

Approximately 29.2% of students scored between 60 and 80 on this exam. Note that this number may be slightly off based on the number of decimals used on the z-score chart.

Answer 2

Are you sure about the numbers? A standard deviation of 26.45 is very large. It puts a substantial proportion of quiz scores at greater than 100 and even some less than 0. Not realistic.

Have a look at this desmos graph I made: https://www.desmos.com/calculator/ydl2erxcab

You can use it to play around with any normal distribution. a is the standard deviation, b is the mean. c is the lower limit and d is the upper limit.

At the end of the day you will be using some sort of technology to get an answer. It is important to understand the fundamentals first. Consider scheduling a lesson to help you get a handle on using the normal distribution.