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August 1st, 2017, 07:09 AM   #1
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From: Israel

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Can you help me please about the following exercise:

The variance of two numbers is 25, therefore, their range is:
a. 5
b. 10
c. 25
d. The range can not be calculated without knowing numbers.
e. None of the answers is correct.

I think the answer is d, but I'm not sure :\


If you didn't understand, I will explain again (I'm not from country where people speak english, and sorry if my english is bad).

Thanks a lot!

Last edited by IlanSherer; August 1st, 2017 at 07:11 AM.
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August 1st, 2017, 07:41 AM   #2
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A crucial point here is that the problem says "two number". If there were more than two numbers, then (d) would be correct.

But knowing that there are exactly two numbers, and letting those two numbers be "m" and "n" (we can take "n" to be the larger), then we know that their mean is $\displaystyle \frac{m+ n}{2}= \frac{m}{2}+ \frac{n}{2}$. The variance then is given by $\displaystyle (m- \frac{m}{2}- \frac{n}{2})^2+ (n- \frac{m}{2}- \frac{n}{2})^2$$\displaystyle = (\frac{m}{2}- \frac{n}{2})^2+ (\frac{n}{2}- \frac{m}{2})^2= 2(\frac{m}{2}- \frac{n}{2})^2= \frac{1}{2}(n- m)^2$

Now, suppose that is 25: then $\displaystyle (n- m)^2= 50$ so, since n> m, $\displaystyle n- m= 25\sqrt{2}$.

I would say the correct answer is (e).
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Last edited by Country Boy; August 1st, 2017 at 07:44 AM.
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August 1st, 2017, 07:53 AM   #3
Joined: Jan 2017
From: California

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$s\approx \frac{range}{4}$

Therefore, $range\approx 20$
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