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# For a certain exam, a score of 58 was 2 standard deviations below mean

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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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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!
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
Bunuel wrote:
MaithiliGokarn wrote:
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

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.

Can someone please elaborate as how is this solved?
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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patternpandora wrote:
Bunuel wrote:
MaithiliGokarn wrote:
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

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.

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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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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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
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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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.
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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.
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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manishbhusal wrote:
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..
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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MaithiliGokarn wrote:
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

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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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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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
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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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?

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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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
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MaithiliGokarn wrote:
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

----------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,
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
MaithiliGokarn wrote:
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

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

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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
MaithiliGokarn wrote:
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

m -2d =58
m+3d = 98
5d = 40
d =8
So, mean = 98 - 3*8 = 74
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
Top Contributor
MaithiliGokarn wrote:
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

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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For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
aabougabal wrote:
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.

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!
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
JeffTargetTestPrep wrote:
MaithiliGokarn wrote:
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

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

Hello, my understanding is that in system of equations, when applying elimination, we need to add the two equations. Why are you subtracting the first equation from the second?
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Re: For a certain exam, a score of 58 was 2 standard deviations below mean [#permalink]
bamfromtowerofgod wrote:
JeffTargetTestPrep wrote:
MaithiliGokarn wrote:
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

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

Hello, my understanding is that in system of equations, when applying elimination, we need to add the two equations. Why are you subtracting the first equation from the second?

When you are dealing with equations ( where sign ‘=‘ is used), you can add/subtract/multiply.
But when you are dealing with inequalities ( where sign ‘<,>’ are used), you can only add the same sides of inequality.
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