Question

Need Help ASAP

Answer 1

Given that mean of quiz scores = 6.4 and standard deviation = 0.7

And we need to use Chebyshev's theorem to find the range in which 88.9% of data will reside.

Chebyshev's theorem states that "Specifically, no more than $\frac{1}{k^{2} }$ of the distribution's values can be more than k standard deviations away from the mean".

That is $1-\frac{1}{k^{2} } = 0.889$

$\frac{1}{k^{2} } = 1-0.889 = 0.111$

$k^{2} = \frac{1}{0.111}$

k = 3

So, we want the range of values within 3 standard deviations of mean.

Hence range is [mean -3*standard deviation, mean +3*standard deviation]

                      = [6.4 - 3*0.7 , 6.4+3*0.7]

                      = [6.4 - 2.1 , 6.4+2.1] = [4.3,8.5]