Thread: Uniform Distribution Question

Hi, I need help with this question

What is the probability that a random variable V which is uniformly
distributed on the interval [0; 1] lies within 2 standard deviation units of its mean?

I was told to standardise this variable by taking away its mean which is $\displaystyle 0.5$ and dividing by it's standard deviation which is $\displaystyle 1/\sqrt12$

With that I end up with z-values -2 and 2
$\displaystyle P(-2 \leq z \leq 2) = 0.9545$

Is it correct to use a normal here?

Originally Posted by alan4cult

Hi, I need help with this question

What is the probability that a random variable V which is uniformly
distributed on the interval [0; 1] lies within 2 standard deviation units of its mean?

I was told to standardise this variable by taking away its mean which is $\displaystyle 0.5$ and dividing by it's standard deviation which is $\displaystyle 1/\sqrt12$

With that I end up with z-values -2 and 2
$\displaystyle P(-2 \leq z \leq 2) = 0.9545$

Is it correct to use a normal here?

No. You can use that IFF the variable follows Standard Normal Distr.

Originally Posted by alan4cult

Hi, I need help with this question

What is the probability that a random variable V which is uniformly
distributed on the interval [0; 1] lies within 2 standard deviation units of its mean?

I was told to standardise this variable by taking away its mean which is $\displaystyle 0.5$ and dividing by it's standard deviation which is $\displaystyle 1/\sqrt12$

With that I end up with z-values -2 and 2
$\displaystyle P(-2 \leq z \leq 2) = 0.9545$

Is it correct to use a normal here?

$\displaystyle \mu = \frac{1}{2}$ and $\displaystyle \sigma = \frac{1}{\sqrt{12}}$ (see http://en.wikipedia.org/wiki/Uniform...n_(continuous)). It is simple to calculate $\displaystyle \int_{\frac{1}{2} - \frac{2}{\sqrt{12}}}^{\frac{1}{2} + \frac{2}{\sqrt{12}}} \, dv$.