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Statistics Part 1

Section: Mathematics

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Question : 91 of 100

Marks: +1, -0

The mean of the numbers

a, b, 8,5,10

is 6 and the variance is

6.80

. Then which one of the following gives possible values of

a

b

$a=0, b=7$
$a=5, b=2$
$a=1, b=6$
$a=3, b=4$

Solution:

Given that,

Mean of

a, b, 8,5,10=6

\therefore \; \; \; \frac{a+b+8+5+10}{5}=6

\Rightarrow a+b+23=30

\Rightarrow a+b=7 . . . . . .(1)

Variance

=\; \frac{\sum(x_i-A)^2}{n}=6.8

\Rightarrow \; \frac{(6-a)^2+(6-b)^2+(6-8)^2+(6-5)^2+(6-10)^2}{5}=6.5

\Rightarrow (6-a)^2+(6-b)^2+4+1+16=34

\Rightarrow (6-a)^2+(6-b)^2=13

\Rightarrow a^2+b^2=25

By using (1) we get,

\; a^2+(7-a)^2=25

\; \Rightarrow \; \; a^2-7 a+12=0

\; \therefore \; \; a=3,4

then

b=4,3

.

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