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pb_india
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I just decided m=a+b+c+d to ease the calculation

In fact there is a flaw above

(a-m/4)^2+(b-m/4)^2+(c-m/4)^2+(d-m/4)^2

implies std^2 = 200+4*(m/4)^2-2m*4m/4
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1) a+b+c+d=50
We can get the mean 'm', but not enough

2) a^2+b^2+c^2+d^2=200
Not enough

1) and 2)

Std deviation = SQRT[ (m-a)^2 + (m-b)^2 + (m-c)^2 + (m-d)^2)/4]
Using 1 and 2, this can be calculated
C)
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pb_india
What is the standard deviation of the set: a, b, c, d

1) a+b+c+d=50
2) a^2+b^2+c^2+d^2=200


if the information about mean sq deviation were given, the value of SD could have found/calculated. but the sum of the component and sum of the sq of each component are not sufficient to answer the question. therefore, OA is E, but at the same time the question is flawed.

if a+b+c+d=50, then the sum of the square of each component can not be 200. this is because if we make a+b+c+d=50, sum of the square of these individual component would be more thyan 200. if we suppose sum of the square of each component be 200, the sum would be different from 50. for example, lets make a+b+c+d=50,

suppose, a =1, b = 7, c = 15, d = 27, sum = 50. but sum of the square is more than 200.

Now, lets make a^2+b^2+c^2+d^2=200
suppose, a =0, b = 6, c = 8, d = 10, sum = 22
a^2 =0, b^2 = 36, c^2 = 64, d^2 = 100, sum of the sq=200............1

even the sum of the sq of each component could be achieved by varying the value of each component. for example, if

a =0, b = 6, c = 8, d = -10, sum = 4
a^2 =0, b^2 = 36, c^2 = 64, d^2 = 100, sum of the sq=200 ........ 2

in case 1 and case 2, the value of SD varies. therefore, statements are not only insufficient to answer the question but also falwed.



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