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Standard Deviation Problem

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by adi » Sat Mar 08, 2008 2:28 pm

Timer

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Answers

A

B

C

D

E

Stats

Difficulty—

For a certain examination, a score of 58 was 2 standard deviations below the mean, and a score of 98 was 3 standard deviations above the mean. What was the mean score for the examination?

A) 74
B) 76
C) 78
D) 80
E) 82

The correct answer should be A.

Thanks in advance! :D
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Re: Standard Deviation Problem

by Stuart@KaplanGMAT » Sat Mar 08, 2008 3:36 pm
adi wrote:For a certain examination, a score of 58 was 2 standard deviations below the mean, and a score of 98 was 3 standard deviations above the mean. What was the mean score for the examination?

A) 74
B) 76
C) 78
D) 80
E) 82

The correct answer should be A.

Thanks in advance! :D
We can set up two equations with two unknowns to solve this problem.

Let M = the mean
Let SD = 1 standard deviation

M + 3SD = 98
M - 2SD = 58

if we subtract we get:

5SD = 40
SD = 8

and subbing back in:

M - 2(8) = 58
M = 58 + 16
M = 74

We also could have used logic rather than equations:

The spread between 58 and 98 is a total of 5 standard deviations (one is 2 below, one is 3 above), so those 5 SDs = 40, and 1 SD = 40/5 = 8.

If 58 is 2 SDs below the mean, then 58 + 2(8) = 74 is the mean.
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by yalephd2007 » Sun Apr 13, 2008 4:56 pm
thank you for your detailed explonation
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by dalwow » Wed Apr 16, 2008 12:13 pm
Stuart-

For this problem I just took the total variance (98-58 ) which was 40 and divided that by 5 standard deviations. That told me one standard deviation was 8 pts. Then we could easily figure it out from that. Is that just a coincidence for this problem or is that a viable way to work this out in the future? Thanks

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by Natalllllia » Mon Dec 01, 2008 8:30 pm
thank you
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by Jeff@TargetTestPrep » Thu Aug 30, 2018 4:46 pm
adi wrote:For a certain examination, a score of 58 was 2 standard deviations below the mean, and a score of 98 was 3 standard deviations above the mean. What was the mean score for the examination?

A) 74
B) 76
C) 78
D) 80
E) 82
If we let x = mean and s = standard deviation, then a score of 58 that was 2 standard deviations below the mean would mean 58 = x - 2s. Similarly, a score of 98 that was 3 standard deviations above the mean would mean 98 = x + 3s.

We can use the two equations to determine a value of x.

x - 2s = 58

x + 3s = 98

If we subtract the first equation from the second equation, we have:

5s = 40

s = 8

Now we can determine x:

x + 3s = 98

x + 3(8) = 98

x + 24 = 98

x = 74

Answer: A

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by [email protected] » Mon Sep 03, 2018 7:52 pm
Hi All,

We're told that for a certain examination, a score of 58 was 2 standard deviations BELOW the mean and a score of 98 was 3 standard deviations ABOVE the mean. We're asked for the mean score for the examination. The math behind this question is just basic Algebra, but depending on how you organize your information, you can save some time on the math that you'll need to do.

To start, the difference between a 98 and a 58 is 5 total Standard Deviations (since a 98 is '3 above' and a 58 is '2 below'). Thus, 98 - 58 = 40 is equal to 5 Standard Deviations.
40/5 = 8... meaning that 1 Standard Deviation is 8 'away' from the mean. At this point, we can either "go up" from 58 or "go down" from 98, but we'll get to the Mean either way as long as we're counting up the proper number of Standard Deviations.

From 58, we add 2 S.D.s --> 58 + 2(8) = 74
From 98, we subtract 3 S.D.s --> 98 - 3(8) = 74

Final Answer: A

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