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Question
The mean and S.D. of 200 items are found to be 60 and 20 respectively. At the time of calculation, two items were wrongly taken as 3 and 67 instead of 13 and 17. Find the correct mean and variance.
Solution
Here, n = 200, `bar"x"` = Mean = 60, S.D. = 20
Wrongly taken items are 3 and 67.
Correct items are 13 and 17.
Now, `bar"x"` = 60
∴ `1/"n" sum_("i" = 1)^"n" "x"_"i"` = 60
∴ `1/200 sum_("i" = 1)^"n" "x"_"i"` = 60
∴ `sum_("i" = 1)^"n" "x"_"i"` = 200 × 60
∴ `sum_("i" = 1)^"n" "x"_"i"` = 12000
Correct value of `sum_("i" = 1)^"n" "x"_"i" = sum_("i" = 1)^"n" "x"_"i"` – (sum of wrongly taken items) + (sum of correct items)
= 12000 – (3 + 67) + (13 + 17)
= 12000 – 70 + 30
= 11960
Correct value of mean = `1/"n" xx "correct value of" sum_("i" = 1)^"n" "x"_"i" = 1/200 xx 11960` = 59.8
Now, S.D. = 20
Variance = (S.D.)2 = 202
∴ Variance = 400
∴ `1/"n" sum_("i" = 1)^"n" "x"_"i"^2 - (bar"x")^2` = 400
∴ `1/200 sum_("i" = 1)^"n" "x"_"i"^2 - (60)^2` = 400
∴ `1/200 sum_("i" = 1)^"n" "x"_"i"^2` = 400 + 3600
∴ `sum_("i" = 1)^"n" "x"_"i"^2` = 4000 × 200
∴ `sum_("i" = 1)^"n" "x"_"i"^2` = 800000
Correct value of `sum_("i" = 1)^"n" "x"_"i"^2`
= `sum_("i" = 1)^"n" "x"_"i"^2` – (Sum of squares of wrongly taken items) + (Sum of squares of correct items)
= 800000 – (32 + 672) + (132 + 172)
= 800000 – (9 + 4489) + (169 + 289)
= 800000 – 4498 + 458 = 795960
∴ Correct value of Variance
`=(1/"n" xx "correct value of " sum_("i" = 1)^"n" "x"_"i"^2) - ("correct value of" bar"x")^2`
= `1/200 xx 795960 - (59.8)^2`
= 3979.8 – 3576.04
= 403.76
∴ The correct mean is 59.8 and correct variance is 403.76.
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