the mean and standard deviation of 20 observation are found to be 10 and 2 respectively. on rechecking, it was found that an observation 8 was incorrect. calculate the correct mean and standard deviation in each of the following cases:(1) if wrong item is omitted. (2) if it is replaced by 12 .
Answers
Answer:
Answer
(i) Given, number of observations n=20
Incorrect mean =10
Incorrect standard deviation =2
x
ˉ
=
n
1
i=1
∑
20
x
i
10=
20
1
i=1
∑
20
x
i
⇒
i=1
∑
20
x
i
=200
So, the incorrect sum of observations =200
Correct sum of observation =200−8=192
⇒ Correct mean =
19
Correctsum
=
19
192
=10.1
Standard deviation σ=
n
1
i=1
∑
n
x
i
2
−
n
2
1
(
i=1
∑
n
x
i
)
2
=
n
1
i=1
∑
n
x
i
2
−(
x
ˉ
)
2
⇒2=
20
1
Incorrrect
i=1
∑
n
x
i
2
−(10)
2
⇒4=
20
1
Incorrect
i=1
∑
n
x
i
2
−100
⇒Incorrect
i=1
∑
n
x
i
2
=2080
∴ Correct
i=1
∑
n
x
i
2
=Incorrect
i=1
∑
n
x
i
2
−(8)
2
=2080−64
=2016
∴ Correct Standard deviation =
n
Correct∑x
i
2
−(CorrectMean)
2
=
19
2016
−(10.1)
2
106.1
−102.01
=
4.09
= 2.02
(ii) When 8 is replaced by 12
Incorrect sum of observation =200
∴ Correct sum of observations =200−8+12=204
∴ Correct mean =
20
Correctsum
=
20
204
=10.2
Standard deviation σ=
n
1
i=1
∑
n
x
i
2
−
n
2
1
(
i=1
∑
n
x
i
)
2
=
n
1
i=1
∑
n
x
i
2
−(
x
ˉ
)
2
⇒2=
20
1
Incorrect
i=1
∑
n
x
i
2
−(10)
2
⇒ 4=
20
1
Incorrect
i=1
∑
n
x
i
2
−100
⇒Incorrect
i=1
∑
n
x
i
2
=2080
∴Correct
i=1
∑
n
x
i
2
=Incorrect
i=1
∑
n
x
i
2
−(8)
2
+(12)
2
=2080−64+144
=2160
Correctstandarddeviation=
n
Correct∑x
i
2
−(Correctmean)
2
=
20
2160
−(10.2)
2
=
108−104.04
=
3.96
=1.98