(y - y1) = m (x -x1)
( y - 5/2) = -1/2 (x - -2)
(y - 5/2) = -1/2 ( x + 2)
y - 5/2 = -1/2 x - 1
y = -x/2 - 1 + 5/2
y = -x/2 - 2/2 + 5/2
y = -x/2 + 3/2
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]
90/135 = 0.666666667
0.666 x 100 = 66.6 %
Answer:
the answer to that would be y=3x-9