2021-06-25T09:52:31-04:00
Calculate the standard deviation of the following data;
45, 50, 67, 68, 75, 80, 70, 65, 70,90
1
2021-06-28T03:53:25-0400
n = 10 n=10 n = 10
μ = ∑ i = 1 n x i n = 1 10 ( 45 + 50 + 67 + 68 + 75 + 80 \mu=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{1}{10}(45+50+67+68+75+80 μ = n i = 1 ∑ n x i = 10 1 ( 45 + 50 + 67 + 68 + 75 + 80
+ 70 + 65 + 70 + 90 ) = 68 +70+65+70+90)=68 + 70 + 65 + 70 + 90 ) = 68
σ 2 = ∑ i = 1 n ( x i − μ ) 2 n = 1 10 ( ( 45 − 68 ) 2 + ( 50 − 68 ) 2 \sigma^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\mu)^2}{n}=\dfrac{1}{10}((45-68)^2+(50-68)^2 σ 2 = n i = 1 ∑ n ( x i − μ ) 2 = 10 1 (( 45 − 68 ) 2 + ( 50 − 68 ) 2
+ ( 67 − 68 ) 2 + ( 68 − 68 ) 2 + ( 75 − 68 ) 2 +(67-68)^2+(68-68)^2+(75-68)^2 + ( 67 − 68 ) 2 + ( 68 − 68 ) 2 + ( 75 − 68 ) 2
+ ( 80 − 68 ) 2 + ( 70 − 68 ) 2 + ( 65 − 68 ) 2 +(80-68)^2+(70-68)^2+(65-68)^2 + ( 80 − 68 ) 2 + ( 70 − 68 ) 2 + ( 65 − 68 ) 2
+ ( 70 − 68 ) 2 + ( 90 − 68 ) 2 ) = 154.8 +(70-68)^2+(90-68)^2)=154.8 + ( 70 − 68 ) 2 + ( 90 − 68 ) 2 ) = 154.8
σ = σ 2 = 154.8 ≈ 12.4419 \sigma=\sqrt{\sigma^2}=\sqrt{154.8}\approx12.4419 σ = σ 2 = 154.8 ≈ 12.4419
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