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I'm stuck on this question, can someone help me, many thanks.

If $U$ is uniformly distributed with mean $5$ and variance $3$, what is $P(U<4)$?

update(this problem has been solved): I made a mistake when calculating, the result should be under the condition : x ranges from a to b. The final result is 1/3.

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    $\begingroup$ uniformly distributed over what set? $\endgroup$ – mathworker21 Jan 17 at 12:44
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    $\begingroup$ Where did you get stuck with solving this? What have you tried already? Questions without context and visible effort often get downvoted and closed. $\endgroup$ – postmortes Jan 17 at 12:58
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Hint:- Let us suppose $U$ follows Uniform distribution with parameter $a$ and $b$.

Mean=$E(U)=\frac{b+a}{2}=5$ and Variance $=V(U)=\frac{(b-a)^2}{12}=3$.

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    $\begingroup$ You are assuming that $U$ is continuous Uniform. All that is stated is that it is Uniform (which might be continuous or discrete) $\endgroup$ – wolfies Jan 17 at 12:47
  • $\begingroup$ Yes I know this formula, the result is 1/3 right? I just realized I made a mistake, x should be in the range from a to b $\endgroup$ – wawawa Jan 17 at 12:48
  • $\begingroup$ @ Cecilia Yes, you are correct. I forgot to mention one thing that you solve the two equations under the restriction that $a<b$. $\endgroup$ – user440191 Jan 17 at 12:50
  • $\begingroup$ @ wolfies you are absolutely right. I should have mentioned that I am assuming continuous uniform distribution. However, if one assume that it is the case of discrete uniform, I think thee will be a issue regarding the consideration of the set over which it is uniformly distributed ( as mentioned by @mathworker above) $\endgroup$ – user440191 Jan 17 at 12:58
  • $\begingroup$ I urge @Cecilia to consider the discrete case when the set is $\{1,2,\dots N\}$ and see what happens. $\endgroup$ – user440191 Jan 17 at 12:59
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As the mean is directly in the middle between $a$ and $b$ you can set

  • $a = 5-x$ and $b=5+x$ for $x>0$

Now solve $$\sigma^2 = \frac{(b-a)^2}{12}= \frac{(2x)^2}{12}=\frac{x^2}{3} = 3\stackrel{x>0}{\Rightarrow} x= 3$$ So, $U$ "lives" on $[a,b]=[2,8]$. It follows $$P(U<4) = \frac{4-2}{8-2}=\frac{2}{6}=\frac{1}{3}$$

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  • $\begingroup$ Thank you so much for the answer, however @Bhargob answers the question before you haha and I've already chosen his answer, thanks for your time and I really appreciated it. $\endgroup$ – wawawa Jan 17 at 14:59

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