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by gmatnmein2010 » Mon Feb 15, 2010 8:40 pm
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
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Source: — Problem Solving |

by shashank.ism » Mon Feb 15, 2010 8:52 pm
gmatnmein2010 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
Let M be the mean & S = 1 standard deviation

M + 3S = 98 ---------(i)
M - 2S = 58 ---------(ii)

solving above 2 equations , we get:
S = 8
--> M = 2S+58 = 74
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by thephoenix » Mon Feb 15, 2010 9:55 pm
x - 2sd = 58
x + 3sd = 98

SD = 8 and Mean (x) = 74
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by komal » Tue Feb 16, 2010 10:53 am
gmatnmein2010 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
x - 2sd = 58
x + 3sd = 98

SD = 8 and Mean (x) = 74 in A.
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by sadullaevd » Tue Feb 16, 2010 8:33 pm
2sd=58

3sd =98

to find one sd=> (98-58)/5 =8

mean - 2sd = 58

mean 74

hope it helps
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