Esme noted the test marks, x, of 16 people in a class. She found that Σ x = 824 and that the standard deviation of x was 6.5.?

Σ(x - 50)² => Σ(x² - 100x + 2500)

i.e. Σx² - 100Σx + 16(2500)

or, Σx² - 82400 + 40000

Hence, Σ(x - 50)² = Σx² - 42400....(1)

Now, Var(x) = (Σx²)/n - (Σx/n)²

so, 6.5² = (Σx²)/16 - (824/16)²

i.e. 42.25 = (Σx²)/16 - 2652.25

Then, (Σx²)/16 = 2694.5

=> Σx² = 43112

So, from (1) we have 43112 - 42400 = 712

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Mean, ¯x = Σx/16 = 824/16 = 51.5

Standard deviation, s = √[Σ(x - ¯x)²/16] = 6.5
Variance, s² = Σ(x - 51.5)²/16 = 6.5²
Σ(x² - 103x + 51.5²) = 676
Σx² - 103Σx + Σ51.5² = 676
Σx² - 103*824 + 51.5²*16 = 676
Σx² = 43112

Σx² - 2*50*Σx + Σ50² = 43112 - 2*50*Σx + Σ50²
Σ(x² - 50)² = 43112 - 100*824 + 50²*16
Σ(x² - 50)² = 712