Solution
Let X X X represent the marks of 7 7 7 students in a test
X = 66 , 73 , 59 , 62 , 69 , 65 , 70 X=66,73,59,62,69,65,70 X = 66 , 73 , 59 , 62 , 69 , 65 , 70
(a) mean X ˉ = ∑ X i n \bar X=\dfrac{\sum X_i}n X ˉ = n ∑ X i
X ˉ = 66 + 73 + 59 + 62 + 69 + 65 + 70 7 \bar X=\frac{66+73+59+62+69+65+70}{7} X ˉ = 7 66 + 73 + 59 + 62 + 69 + 65 + 70
X ˉ = 464 7 = 66.288 \bar X=\dfrac{464}7=66.288 X ˉ = 7 464 = 66.288
(b) Variance
X i X i − X ˉ ( X i − X ˉ ) 2 59 − 7.288 53.115 62 − 4.288 18.387 65 − 1.288 1.659 66 − 0.288 0.083 69 2.712 7.355 70 3.712 13.779 73 6.712 45.051 ∑ 139.429 ~~~~~~~\def\arraystretch{1.5}\begin{array}{c:c:c}X_i
&X_i -\bar X& (X_i -\bar X)^2\\\hline59&-7.288&53.115\\\hdashline62&-4.288&18.387\\\hdashline65&-1.288&1.659\\\hdashline66&-0.288&0.083\\\hdashline69&2.712&7.355\\\hdashline70&3.712&13.779\\\hdashline73&6.712&45.051\\\hline\sum&&139.429\\\hline\end{array} X i 59 62 65 66 69 70 73 ∑ X i − X ˉ − 7.288 − 4.288 − 1.288 − 0.288 2.712 3.712 6.712 ( X i − X ˉ ) 2 53.115 18.387 1.659 0.083 7.355 13.779 45.051 139.429
σ 2 = 139.429 7 = 19.92 \sigma^2=\dfrac{139.429}7=19.92 σ 2 = 7 139.429 = 19.92
(c) Standard deviation
σ = v a r i a n c e \sigma=\sqrt{variance} σ = v a r ian ce
σ = 19.92 \sigma=\sqrt{19.92} σ = 19.92
σ = 4.463 \sigma=4.463 σ = 4.463
Comments