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For a certain exam,a score of 58 was 2 standard deviations b
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06 Mar 2012, 03:06
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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
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For a certain exam,a score of 58 was 2 standard deviations b
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06 Mar 2012, 03:11




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Re: For a certain exam,a score of 58 was 2 standard deviations b
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14 Aug 2013, 09:37
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. Answer: A. 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 b
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17 Apr 2014, 07:59
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. 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 EE = 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. \(d=\frac{7458}{2}=\frac{9874}{3}\) This is intended to be a backup 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 b
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17 Apr 2014, 18:19
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 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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Re: For a certain exam,a score of 58 was 2 standard deviations b
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13 Sep 2014, 18:38
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 b
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08 Dec 2014, 15:44
For those confused like myself, 58=M2d 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 b
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14 Mar 2015, 23:18
I am getting m=22. I will be glad is someone point out the mistake i have committed.
58=m2d 1 98=m3d 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 b
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14 Mar 2015, 23:33
manishbhusal wrote: I am getting m=22. I will be glad is someone point out the mistake i have committed.
58=m2d 1 98=m3d 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 m3d..
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For a certain exam,a score of 58 was 2 standard deviations b
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23 Jul 2016, 01:46
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 = 983*8 = 9824 = 74 THE ANSWER IS DEFINITELY A = 74
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Re: For a certain exam,a score of 58 was 2 standard deviations b
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10 Dec 2017, 09:20
Quick solution > please see below 58=M2D 98=M+3D 9858=M+3DM2D 40=5D D=8 so, One SD =8
58=M2D 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.
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For a certain exam,a score of 58 was 2 standard deviations b
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23 Jan 2018, 13:44
The easiest approach ever: 9858=40 there are 5 SD in between; hence, 40/5=8 58+2(SD) i.e. 58+16=74 or 983(SD) i.e. 9824=74



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Re: For a certain exam,a score of 58 was 2 standard deviations b
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25 Jul 2018, 17:55
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 9858 = 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 b
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28 Aug 2018, 14:52
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, Brent
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Re: For a certain exam,a score of 58 was 2 standard deviations b
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30 Aug 2018, 16:45
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 Answer: A
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