A score of 58 was 2 standard deviations below the mean --> 58 = Mean - 2d
A score of 98 was 3 standard deviations above the mean --> 98 = Mean + 3d

Solving above for Mean = 74.

Answer: A.

For more about the Standard Deviation please check: Standard Deviation.

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Quick solution --> please see below
58=M-2D
98=M+3D
98-58=M+3D-M-2D
40=5D
D=8 so, One SD =8

58=M-2D
58+2D=M
58+(2x8)=M
58+16=M
So M = 74

apply the same for the other equation and you will get to M=74.

Thanks!

Welcome to GMAT Club. Below is a solution to your question.

A score of 58 was 2 standard deviations below the mean --> 58 = Mean - 2d
A score of 98 was 3 standard deviations above the mean --> 98 = Mean + 3d

Solving above for Mean = 74.

Answer: A.

Can someone please elaborate as how is this solved?

you can either solve as Bunuel did or exploit number properties and plug in from the answer choices:

58 = Mean - 2d -----> 2d = Mean - 58
98 = Mean + 3d -----> 3d = 98 - Mean

Thus your d must be both a multiple of 2 (2n) or a multiple of 3 (2n+1). Giving a fast glance at the answer choices, since E-E = E, the first equation is satisfied in every answer choice. The second equation, however, is satisfied in two answers only, respectively A and D.

Now, since d is a unique value for both equations, calculating d must yield the same result. It does but for one choice only and that choice is A.

\(d=\frac{74-58}{2}=\frac{98-74}{3}\)



This is intended to be a back-up method, for sure not as fast as that one Bunuel came up with but when you're under pressure everything that pops up in your mind is useful.
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These are concepts of normal distribution. You don't need to delve much into details but you must understand what 1 SD, 2 SD etc mean.

Say you have a distribution of weights in a class of children. The mean weight of the class is, say 46 kgs. If I tell you that 1 standard deviation is of 5 kgs, it means that 1 SD below mean is 5 kgs less than 46 kgs which is 41 kgs. This also means that 34% children have weight ranging from 46 kgs to 41 kgs but you don't have to worry about that.

So 2 SD below mean will be 2*5 = 10 kgs less than 46 i.e. 36 kgs.

Similarly, 3 SD above mean will be 3*5 = 15 kgs more than 46 i.e. 61 kgs.

In the original question, you are given that 58 is 2 SD below mean and 98 is 3 SD above mean. So from 58 to 98, we have 5 SDs.
\(\frac{{98 - 58}}{5} = 8\) = SD

Since SD is 8 so mean is at \(58 + 2*8 = 74\)
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Every list of numbers has a mean. every number on the list has a deviation from the mean: that is how far that number is from the mean.
deviation from the mean = (value) – (mean)
Or
Value= Mean - Deviation
58 = M - 2d
98 = M - 3d
Solve it
M = 74

For those confused like myself, 58=M-2d and 98=M+3d is a system of equations. We want the value of M, so isolate D in one of the equations. then plug that value of d into the other equation and solve for M.

I am getting m=-22. I will be glad is someone point out the mistake i have committed.

58=m-2d 1
98=m-3d 2

subtracting 2 from 1

d=-40.

substituting d in 1

m =-22.

hi manishbhusal,
98 is 3 deviation above ..
so 98=m+3d and not m-3d..

Mean -2 SD = 58 ..... eq 1
Mean +3SD =98 ........eq 2

Subtract eq 2 from eq1
5SD= 40
SD = 40/5--> 8

So the standard deviation is 8
now to find the mean
either add 2*SD to 58 = 2*8+58 = 18+58=74
Or subtract 3*SD from 98 = 98-3*8 = 98-24 = 74
THE ANSWER IS DEFINITELY A = 74
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The easiest approach ever:
98-58=40
there are 5 SD in between; hence,
40/5=8
58+2(SD) i.e. 58+16=74
or
98-3(SD) i.e. 98-24=74

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?

The range in given question is 98-58 = 40

58 is 2 SD below mean
90 is 3 SD above mean

So there is 5 SD in between 58 and 98. 5 SD=40 , SD = 8

The mean score is 74
Ans :A
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----------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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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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m -2d =58
m+3d = 98
5d = 40
d =8
So, mean = 98 - 3*8 = 74

m+3d=98---I

m-2d=58---II

Substracting II from I: m+3d-m+2d=98-58

=5d=40

d=5

Substituting the value in II
m-16=58

m=74

The answer is A.
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Thank you, I messed up while solving for these equations.(I know, such a silly mistake, but a costly one!) Your quick solution helped me so much, thanks.



Thanks!