For a certain exam, a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above

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

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

Answer: A
Source: official guide

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BTGModeratorVI wrote:
Wed Jan 06, 2021 8:07 am
For a certain exam, a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean. What was the mean score for the exam?

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

Answer: A
Source: official guide
----------ASIDE---------------------
A little extra background on standard deviations above and below the mean

If, for example, a set has a standard deviation of 4, then:
1 standard deviation = 4
2 standard deviations = 8
3 standard deviations = 12
1.5 standard deviations = 6
0.25 standard deviations = 1
etc


So, if the mean of a set is 9, and the standard deviation is 4, then:
2 standard deviations ABOVE the mean = 17 [since 9 + 2(4) = 17]
1.5 standard deviations BELOW the mean = 3 [since 9 - 1.5(4) = 3]
3 standard deviations ABOVE the mean = 21 [since 9 + 3(4) = 21]
etc.
-------------------------------------------------------------

For this question, we can let M=mean and let D=the standard deviation

So, 58 is 2 standard deviations below the mean translates into M - 2D = 58
and 98 is 3 standard deviations above the mean translates into M + 3D = 98

When we solve this system of equations, we get M=74 and D=8

So the answer is A

Cheers,
Brent
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BTGModeratorVI wrote:
Wed Jan 06, 2021 8:07 am
For a certain exam, a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean. What was the mean score for the exam?

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

Answer: A
Source: official guide
Solution:

If we let x = mean and s = standard deviation, then a score of 58 that was 2 standard deviations below the mean can be translated as 58 = x – 2s. Similarly, a score of 98 that was 3 standard deviations above the mean can be translated as 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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