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.)

Respuesta :

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

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

Then the mean (first moment) is

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

and the second moment is

[tex]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[/tex]

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

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

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

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