Question

A student obtained the mean and the standard deviation of 100
observations as 40 and 5.1. It was later found that one observation
wrongly copied as 50, the correct figure being 40. Find the correctn
and the S.D.​

Answer 1

Answer:

Here is ur answer

Step-by-step explanation:

Correct option is

B

Mean =39.9, S.D. =5

We have, n=100, mean

X

ˉ

=40 and standard deviation

(σ)=5.1

∵

X

ˉ

=

n

∑xi

⇒n

X

ˉ

=xi

⇒10×40=∑xi

Now, corrected mean (

X

ˉ

)=

n

Incorrect∑xi−(Sumofincorrectvalues)+(sum of correct values)

where Incorrect ∑xi− (sum of incorrect values)+(sum of correct values)

equal is correct ∑xi

⇒ corrected ∑xi=4000−(50)+40

=3990

∴ correct mean

X

ˉ

=

100

3990

=39.9

Now, σ=5.1,∴ variance =(5.1)

2

=26.01

26.01=

100

1

(∑xi

2

)−(

100

1

∑xi)

2

=

100

1

×(∑xi

2

)−(

100

4000

)

2

⇒∑xi

2

=162601

⇒ Incorrect ∑xi

2

=162601

Now correct ∑xi

2

= (incorrect ∑xi

2

)-(Sum of squares of incorrect values)+(sum of squares of correct values)

correct xi

2

=162601−(50)

2

+(40)

2

=161701

Corrected (σ)=

n

1

corrected∑xi

2

−(

n

1

(corrected∑xi))

2

=

100

161701

−(

100

3990

)

2

=

25

=5

⇒ correct mean =39.9 and correct standard deviation =5.

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