Uniform Distribution Question

**alan4cult** · November 18th 2009, 05:57 AM

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 $0.5$ and dividing by it's standard deviation which is $1/\sqrt12$

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

Is it correct to use a normal here?

**aman_cc** · November 18th 2009, 08:42 AM

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 $0.5$ and dividing by it's standard deviation which is $1/\sqrt12$

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

Is it correct to use a normal here?

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

**mr fantastic** · November 18th 2009, 03:34 PM

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 $0.5$ and dividing by it's standard deviation which is $1/\sqrt12$

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

Is it correct to use a normal here?

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

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