Question

calculate Corect mean. If €(fxi-5)=20 for of group size 10 then find arithmetic mean​

Answer 1

Arithmetic Mean =

∑f

i

∑x

i

f

i

⇒Arithmetic mean =

7+k+8+4+5

(5×7)+(10×k)+(15×8)+(20×4)+(25×5)

⇒arithmetic mean =

7+k+8+4+5

35+10k+120+80+125

=14

⇒336+14k=360+10k

⇒4k=24

⇒k=6

Hope it will help you

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Answer 2

Answer:

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Class 11

>>Maths

>>Statistics

>>Shortcut Method to find Variance and Standard Deviation

>>The mean and standard deviation of 20 ob

Question

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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 :

(i) If wrong item is omitted

(ii) If it is replaced by 12

Medium

Solution

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(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