What is the surface area of the solid that this net can form?
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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]