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04 Jan 2021, 09:11
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C
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Difficulty:
65%
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Question Stats:
49% (01:14) correct
51%
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wrong
based on 206
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What is the Standard Deviation of the set containing values A, B, C, and D?
(1) A + B + C + D = 60
(2) A^2 + B^2 + C^2 + D^2 = 1050
Are You Up For the Challenge: 700 Level Questions
(1) A + B + C + D = 60
(2) A^2 + B^2 + C^2 + D^2 = 1050
Are You Up For the Challenge: 700 Level Questions
05 Jan 2021, 15:49
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IMO C
\(\sigma = \sqrt{\frac{\sum(x_i-x_{av})^2}{n}}\)
From 1: Avg. can be calculated, missing information for the squares of the difference (e.g., \((A-15)^2+(B-15)^2+(C-15)^2+(D-15)^2\)) --> Hence, insuf
From 2: We have the squares, but the sum of A + B + C + D is missing, and there is no way to calculate the average --> Hence, insuf
Joining 1 + 2 --> all pieces necessary to solve. Hence, C
\(\sigma = \sqrt{\frac{\sum(x_i-x_{av})^2}{n}}\)
From 1: Avg. can be calculated, missing information for the squares of the difference (e.g., \((A-15)^2+(B-15)^2+(C-15)^2+(D-15)^2\)) --> Hence, insuf
From 2: We have the squares, but the sum of A + B + C + D is missing, and there is no way to calculate the average --> Hence, insuf
Joining 1 + 2 --> all pieces necessary to solve. Hence, C
27 Jan 2021, 05:13
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frroldao
I don't disagree at all with frroldao
But I'll just make it easier for people to see a foiled equation so it can be more intuitive.
\(\sigma = \sqrt{\frac{\sum(x_i^2)}{n} - \mu}\)
Where \(\mu\) is the mean, \(n\) is the set count and \(\sum(x_i^2)\) is the sum of squares.
This essentially says, if you have the sum of squares and can find the average, you have the standard deviation.
16 Aug 2026, 11:36
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Whenever you're given the sum of the values, and the sum of the squares of the values, you can find the standard deviation. It's worth memorizing this if you don't feel like looking at the whole SD formula.
Bunuel
