Question

For a group of 200 candidates the arithmetic mean was found to be 40. Later on it was
employees.
7.
found that score 43 was misread as 34. Find the correct arithmetic mean.
of 20 observations is found to be 10. On rechecking, it was found​

Answer 1

Step-by-step explanation:

We have

n=200.

x

ˉ

40,σ=15

Now,

x

ˉ

=

n

1

∑xi

∴

200

1

∑xi=40

⇒∑xi=40×200=8000

Since, the score was miss read this sum is incorrect

⇒ corrected ∑xi=8000−34−53+43+35

=8000−7

=7993

∴ corrected mean =

200

∑xi

=

200

7993

=39.955

S.D.=σ=15

⇒ variance =15

2

=225

Now,

Variance =(

n

1

∑xi

2

)−(

n

1

∑xi)

2

∴

200

1

∑xi

2

–(40)

2

=225

⇒∑(xi)

2

=200×1825=365000

Now,

This is an incorrect reading.

∴ corrected ∑xo

2

=365000−34

2

−53

2

+43

2

+35

2

=365000−1156−2809+1849+1225

=364109

Corrected variance =(

n

1

corrected ∑xi)–(corrected m)

2

=(

200

1

×364109)–(39.955)

2

=1820.545−1596.402

Corrected variance =224.14

Corrected S.D=

corrected variance

=

224.14

Corrected S.D=14.97

Hence, this is the answer.

Answer 2

Answer:

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