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Miscellaneous · Q21
Q.

Two batsmen A and B, in their last 5 innings, scored the following runs:

Batsman A4050604555
Batsman B3070456540

Find the mean and standard deviation of the runs scored by each batsman, and state, using standard deviation, which batsman is more consistent.

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For A: 40,50,60,45,5540,50,60,45,55, sum =250=250, mean =250/5=50=250/5=50. Deviations: −10,0,10,−5,5-10,0,10,-5,5; squares: 100,0,100,25,25100,0,100,25,25; sum =250=250; variance =250/5=50=250/5=50; SDA=50=52≈7.07\text{SD}_A=\sqrt{50}=5\sqrt2\approx7.07. For B: 30,70,45,65,4030,70,45,65,40, sum =250=250, mean =250/5=50=250/5=50. Deviations: −20,20,−5,15,−10-20,20,-5,15,-10; squares: 400,400,25,225,100400,400,25,225,100; sum =1150=1150; variance =1150/5=230=1150/5=230; SDB=230≈15.17\text{SD}_B=\sqrt{230}\approx15.17. Both batsmen have the same mean of 5050 runs, so the mean alone gives no way to compare them — but SDA≈7.07\text{SD}_A\approx7.07 is much smaller than SDB≈15.17\text{SD}_B\approx15.17, showing A's scores stay much closer to the common mean of 5050 across the 5 innings, while B's scores swing far more widely. Since a smaller standard deviation means less variability (more consistency) around the mean, batsman A is the more consistent player. [!ANSWER] Mean =50=50 runs for both; SDA≈7.07\text{SD}_A\approx7.07, SDB≈15.17\text{SD}_B\approx15.17; batsman A is more consistent.

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