Question

The grades on the last math exam had a mean of 88%. Assume the population of grades on math exams is known to be normally distributed with a standard deviation of 6. What percent of students earn a score between 72% and 90%? (5 points)

0.6255

0.5843

0.6754

0.5346

Answer 1

explain also how to do it

Answer 2

This can be done using a standard normal distribution table. First you need to find the z-scores for 72 and 90. Can you do that?

Answer 3

Nope

Answer 4

I want to know how to do this, the teacher just gives us homework and tells us to figure it out by ourselves.

Answer 5

@wio

Answer 6

You can use the standard normal distribution table here if you need it
http://lilt.ilstu.edu/dasacke/eco148/ztable.htm
The z-scores are found using the formula:
\[z=\frac{X-\mu}{\sigma}\]
The z-score for 90 is:
\[\frac{90-88}{6}=0.33\]
And the z-score for 72 is:
\[\frac{72-88}{6}=-2.67\]
Now you need to use a standard normal distribution table to find the cumulative probabilities for z-scores of 0.33 and -2.67. Please do that and post your results.

Answer 7

Is the answer C?

Answer 8

You need to find the two values that I asked for. Then we can quickly find the solution.

Answer 9

.0038 and .6293

Answer 10

Good work! Now just do this calculation to find the percent of students who earn a score between 72% and 90%:
\[(0.6293-0.0038) \times100=\ ?\ percent\]

Answer 11

A

Answer 12

Actually the question asks for a percentage, but the answer choices are decimal fractions.
Yes, A is correct :)