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# Help! what is standard deviation for 89 93 92 88 89 87 95 94 91 92

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Help! what is standard deviation for 89 93 92 88 89 87 95 94 91 92

Guest May 20, 2017
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#1
+1807
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As of further review, my solution is wrong, so I'll just delete this garble that I wrote...

TheXSquaredFactor  May 20, 2017
edited by Guest  May 21, 2017
#3
+26547
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You have calculated the Mean Absolute Deviation Guest #1, not the standard deviation!

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Alan  May 21, 2017
#5
+1807
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I guess I should take my solution down then because my solution is wrong...

I feel bad for giving a user incorrect info...

TheXSquaredFactor  May 21, 2017
#2
+1

Find the (sample) standard deviation of the list:
(89, 93, 92, 88, 89, 87, 95, 94, 91, 92)

The standard deviation is given by:
sqrt((variance))

The (sample) variance of a list of numbers X = {X_1, X_2, ..., X_n} with mean μ = (X_1 + X_2 + ... + X_n)/n is given by:
(abs(X_1 - μ)^2 + abs(X_2 - μ)^2 + ... + abs(X_n - μ)^2)/(n - 1)

There are n = 10 elements in the list X = {89, 93, 92, 88, 89, 87, 95, 94, 91, 92}:
(abs(X_1 - μ)^2 + abs(X_2 - μ)^2 + abs(X_3 - μ)^2 + abs(X_4 - μ)^2 + abs(X_5 - μ)^2 + abs(X_6 - μ)^2 + abs(X_7 - μ)^2 + abs(X_8 - μ)^2 + abs(X_9 - μ)^2 + abs(X_10 - μ)^2)/(10 - 1)

The elements X_i of the list X = {89, 93, 92, 88, 89, 87, 95, 94, 91, 92} are:
X_1 = 89
X_2 = 93
X_3 = 92
X_4 = 88
X_5 = 89
X_6 = 87
X_7 = 95
X_8 = 94
X_9 = 91
X_10 = 92
(abs(89 - μ)^2 + abs(93 - μ)^2 + abs(92 - μ)^2 + abs(88 - μ)^2 + abs(89 - μ)^2 + abs(87 - μ)^2 + abs(95 - μ)^2 + abs(94 - μ)^2 + abs(91 - μ)^2 + abs(92 - μ)^2)/(10 - 1)

The mean (μ) is given by

μ = (X_1 + X_2 + X_3 + X_4 + X_5 + X_6 + X_7 + X_8 + X_9 + X_10)/10 = (89 + 93 + 92 + 88 + 89 + 87 + 95 + 94 + 91 + 92)/10 = 91:
(abs(89 - 91)^2 + abs(93 - 91)^2 + abs(92 - 91)^2 + abs(88 - 91)^2 + abs(89 - 91)^2 + abs(87 - 91)^2 + abs(95 - 91)^2 + abs(94 - 91)^2 + abs(91 - 91)^2 + abs(92 - 91)^2)/(10 - 1)

The values of the differences are:
89 - 91 = -2
93 - 91 = 2
92 - 91 = 1
88 - 91 = -3
89 - 91 = -2
87 - 91 = -4
95 - 91 = 4
94 - 91 = 3
91 - 91 = 0
92 - 91 = 1
10 - 1 = 9
(abs(-2)^2 + abs(2)^2 + abs(1)^2 + abs(-3)^2 + abs(-2)^2 + abs(-4)^2 + abs(4)^2 + abs(3)^2 + abs(0)^2 + abs(1)^2)/9

The values of the terms in the numerator are:
abs(-2)^2 = 4
abs(2)^2 = 4
abs(1)^2 = 1
abs(-3)^2 = 9
abs(-2)^2 = 4
abs(-4)^2 = 16
abs(4)^2 = 16
abs(3)^2 = 9
abs(0)^2 = 0
abs(1)^2 = 1
(4 + 4 + 1 + 9 + 4 + 16 + 16 + 9 + 0 + 1)/9

4 + 4 + 1 + 9 + 4 + 16 + 16 + 9 + 0 + 1 = 64:
64/9

The standard deviation is given by
sqrt((variance)) = sqrt(64/9) = 8/3:

Guest May 20, 2017
#4
+26547
0

No need for the "abs" Guest #2 as the terms are squared and will end up positive anyway.  You still get the correct result though.

.

Alan  May 21, 2017

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