MEAN AND STANDARD DEVIATION FROM INCORRECT MEAN AND STANDARD DEVIATION

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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