Solve Chebychev's Theorem: Mean 50, Standard Deviation 5

[robasc]

Could you please explain and break the steps down in solving the answer to this question for me?



In a distribution of 200 values, the mean is 50 and the standard deviation is 5. Use Chebychev's theorem.



a. at least what percentage of the values will fall between 10 and 30?



50 - 30 = 20



k = 20 / 5 = 4



1 - 1 / k^2 = 1 - 1 / 4^2 = 1 - 1 / 16 = .0625 = 1 - .0625 = .9375 or
%93.75



I got this part right but now this is the part I am having trouble with:



b. At least what percentage of the values will fall between 12 and 28?





Also, does at least mean to subtract?

 

[0rthodontist]

Chebychev's theorem states that at least 1 - 1/k^2 of the distribution lies within k standard deviations of the mean. Personally I don't see how you can use it here, maybe there's a trick. For part (a) your answer of 93% is the minimum percentage of values that fall between 30 and 70, not between 10 and 30.

 

[tacman]

Chebychev's theorem states that for any distribution, the proportion of values that fall within k standard deviations of the mean is at least (1-1/k^2). In this case, k is calculated as (x-mean)/standard deviation.

So for part b, we need to find the proportion of values that fall within 2 standard deviations of the mean (12 and 28 are both 2 standard deviations away from the mean of 50). To find this proportion, we can use the same formula as in part a, but with k = 2.

k = 2 standard deviations = (28-50)/5 = -22/5 = -4.4

1- 1/k^2 = 1 - 1/(-4.4)^2 = 1 - 1/19.36 = 0.948 = 94.8%

Therefore, at least 94.8% of the values will fall between 12 and 28.

No, "at least" does not mean to subtract. In this context, it means that the minimum percentage of values that will fall within the given range is the calculated percentage. It is possible that more than 94.8% of the values will actually fall within this range, but we can be confident that at least 94.8% will fall within it.