Question

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 . ​

Answer 1

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