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x is a random variable with range (2,3), mean=9/4 what is the variance? I need to know if what the variance is by only having the mean and range. Please help

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  • $\begingroup$ which is the distribution of $x$? $\endgroup$ Commented Jan 19, 2021 at 22:58
  • $\begingroup$ These are the only details, the question goes as follows: x is a random variable with range consisting of {2,3}, mean=9/4. Find the variance. $\endgroup$ Commented Jan 19, 2021 at 23:02
  • $\begingroup$ yes, i think so $\endgroup$ Commented Jan 19, 2021 at 23:16
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    $\begingroup$ @integratethis a continuous uniform distribution has mean in the center, which would be 5/2 (fixed typo.. point was its not 9/4), so there seems to be a problem here. $\endgroup$
    – Mark
    Commented Jan 19, 2021 at 23:24
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    $\begingroup$ @Mark you are right. I guess op meant the probabilities are not known for the two values, but you are given $X = \{2, 3\}$ , not a "range", and the mean is $9/4$. $\endgroup$ Commented Jan 20, 2021 at 0:27

1 Answer 1

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You know $1 = p_2 + p_3$ and $9/4 = 2 p_2 + 3 p_3$, so you can figure out that $p_2 = 3/4$. Then it's a matter of computing the variance as $(2 - 9/4)^2p_2 + (3 - 9/4)^2p_3$.

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  • $\begingroup$ so the answer is 5/8, however in my book it says its 3/16 $\endgroup$ Commented Jan 19, 2021 at 23:35
  • $\begingroup$ The answer is $3/16$. $\endgroup$ Commented Jan 19, 2021 at 23:40
  • $\begingroup$ I'm sorry, but could you go over it again $\endgroup$ Commented Jan 19, 2021 at 23:48
  • $\begingroup$ $(-1/4)^2(3/4) + (3/4)^2(1/4) ...$ $\endgroup$ Commented Jan 19, 2021 at 23:51
  • $\begingroup$ thank you so much @aruralreader $\endgroup$ Commented Jan 19, 2021 at 23:52

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