Question

A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit of 750.a. Calculate the mean and the standard deviation for the distribution. (Round intermediate calculation for standard deviation to 4 decimal places and final answer to 2 decimal places.)

Answer 1

You can compute both the mean and second moment directly using the density function; in this case, it's

$f_X(x)=\begin{cases}\frac1{750-670}=\frac1{80}&\text{for }670\le x\le750\\0&\text{otherwise}\end{cases}$

Then the mean (first moment) is

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x\,\mathrm dx=710$

and the second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x^2\,\mathrm dx=\frac{1,513,900}3$

The second moment is useful in finding the variance, which is given by

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2=\dfrac{1,513,900}3-710^2=\dfrac{1600}3$

You get the standard deviation by taking the square root of the variance, and so

$\sqrt{V[X]}=\sqrt{\dfrac{1600}3}\approx23.09$