How to find variance of first 50 even natural numbers?
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First 50 even natural numbers are
2,4,6,8,.......,100
Now Mean (average) of first 50 even natural numbers
μ=2+4+6+8+....+10050μ=2(1+2+3+....+50)50μ=(1+2+3+....+50)25μ=50(50+12)25μ=51
Now Sum of squares of natural even natural numbers
∑x2n=2100=22+42+62+........+1002=22⎛⎝⎜12+22+32+.....+502⎞⎠⎟=4×50(50+1)(2×50+1)6=4×50×51×1016Now variance=∑x2n=2100n−μ2=4×50×51×101650−512=2×17×101−2601=3434−2601=833
2,4,6,8,.......,100
Now Mean (average) of first 50 even natural numbers
μ=2+4+6+8+....+10050μ=2(1+2+3+....+50)50μ=(1+2+3+....+50)25μ=50(50+12)25μ=51
Now Sum of squares of natural even natural numbers
∑x2n=2100=22+42+62+........+1002=22⎛⎝⎜12+22+32+.....+502⎞⎠⎟=4×50(50+1)(2×50+1)6=4×50×51×1016Now variance=∑x2n=2100n−μ2=4×50×51×101650−512=2×17×101−2601=3434−2601=833
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