Question

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The grades on the last history exam had a mean of 75%. Assume the population of grades on history exams is known to be normally distributed with a standard deviation of 7. What percent of students earn a score between 70% and 82%?

A. 0.6285

B. 0.6101

C. 0.6024

D. 0.5892

Answer 1

I said C for this one

Answer 2

@ParthKohli can u help with stats?

Answer 3

Have you considered calculating the Z-scores?

Answer 4

\(\dfrac{70-75}{7} = -5/7 = -0.714285\)

You do the other one.

Answer 5

okay, so is it going to be
70-82
-------
7
?

Answer 6

so how do you know which numbers to use?

Answer 7

No good. Use the mean. We need to know how many standard deviations above the mean (75). (82-75)/7 = ??

Answer 8

oh, the mean is 75 .... OH!!!!
okay, so its 7/7 which is 1

Answer 9

No, we must determine how much Standard Normal probability lies between Z = -0.714285 and Z = +1. This can be done many ways. You probably have a calculator that will do it.

Answer 10

ok, so is it averaging?
I'm really sorry, i'm kinda new to this thing

Answer 11

You should have some sort of cumulative distribution table or calculator function.

In MS Excel 2013 it is this =NORM.DIST(-0.714285,0,1,1) = 0.237525. This suggests that 23.7525% of the probability of the Standard Normal Distribution lies to the left of -0.714285 standard deviations from the mean.

In a table, and there are many...

https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf

We read the values: Z = -0.71 ==> 0.23885 and Z = -0.72 ==> 0.23576. Interpolating might give Z = -0.714285 ==> 0.23753. This is very close to the value MS Excel gave us.

You have to find a way that you understand and that is available to you.

Answer 12

ok, thank you for your help :)

Answer 13

After that, you must do the same thing for Z = +1. Tables and calculators are different. You have to understand your output. In my examples, I calculated probability to the left of the Z-Score. Another table may be to the right. In this case, the answer is:

Using MS Excel
P(Z < 1) - P(Z < -0.714285) = 0.841345 - 0.237525 = 0.603819
Using the table and interpolation
0.603813

Make sure you understand both the problem statement and the tools. Either one lacking can make it all go horribly wrong. :-)

Answer 14

ok :)
I'll look into that :D