Question 1 :
The mean and standard deviation of 20 items are found to be 10 and 2 respectively. At the time of checking it was found that an item 12 was wrongly entered as 8. Calculate the correct mean and standard deviation.
Solution :
Mean of 20 items = 10
x̄ = 10
Standard deviation of 20 items = 2
σ = 2, wrong value = 8, correct value = 12
Mean (x̄) = (Σx/n)
10 = Σx/20
10(20) = Σx
Σx = 200
200 represents sum of 20 items which contain the wrong value.
correct Σx = 200 - wrong value + correct value
= 200-8+12
= 204
Corrected mean(x̄) = (Σx/n)
= 204/20
= 10.2
Variance (σ2) = (Σx2/n)-(Σx/n)2
4 = (Σx2/n)-(10)2
4 = (Σx2/n) - 100
4+100 = Σx2/20
104(20) = Σx2
Σx2 = 2080
But this is also wrong Σx2, to find the corrected Σx2 we use the formula.
Correct Σx2 = 2080-82+122
= 2080-64+144
= 2160
Now, let us find correct variance
= corrected (Σx2/n)-corrected (Σx/n)2
= (2016/20) - (10.2)2
= 108-104.04
= 3.96
Correct(σ) = √3.96
= 1.99
Correct mean = 10.2
Correct standard deviation = 1.99.
Question 2 :
Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the items and the sum of the squares of all the items.
Solution :
Mean of 100 items (x̄) = 48
Standard deviation (σ) = 10
here we have to answer for two questions that is sum of all items (Σx) and sum of squares all items (Σx2).
x̄ = Σx/n
48 = Σx/100
Σx = 100(48)
= 4800
Sum of all items = 4800
To find sum of squares of all items, we have to find variance (σ²).
σ2 = (Σx2/n) - (Σx/n)2
102 = (Σx2/100)-(4800/100)2
102 = (Σx2/100)-482
100 = (Σx2/100)-2304
100+2304 = (Σx2/100)
2404 = Σx2/100
Σx2 = 2404(100)
Σx2 = 240400
Sum of squares of all items = 240400.
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