Question 30

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 and b?

Solution

We are given that :
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80
Now taking mean
we get $$30 = a+b+8+5+10$$
so we get $$a+b = 7$$ (1)
Now variance = 6.80
so $$6.80 =( (a-6)^2 +(b-6)^2+(8-6)^2+(5-6)^2+(10-6)^2)/5$$
we get $$6.8(5) = (a-6)^2+(b-6)^2+1+16+4$$
So we get $$13 = (a-6)^2 +(b-6)^2$$
$$a=7-b$$
we get $$13 = (1-b)^2+(b-6)^2$$
we get $$b =4$$
and so $$a= 3$$

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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 and b?

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