Need asap suppose that quiz scores in a beginning statistics class have a mean of 6.4 with a standard deviation of 0.7. using chebyshev's theorem, state the range in which at least 88.9% of the data will reside. do not round your answers.

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 of the distribution's values can be more than k standard deviations away from the mean".

That is

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]

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: