Question

A random variable X follows the uniform distribution with a lower limit of 720 and an upper limit of 920. a. Calculate the mean and the standard deviation of this distribution. (Round intermediate calculation for standard deviation to 4 decimal places and final answer to 2 decimal places.) Mean Standard deviation b. What is the probability that X is less than 870? (Do not round intermediate calculations. Round your answer to 2 decimal places.) Probability

Answer 1

(a) The support has length 920 - 720 = 200, so X has probability density

`f_X(x)=\begin{cases}\frac1{200}&\text{for }720\le x\le920\\0&\text{otherwise}\end{cases}`

X has mean

`E[X]=\displaystyle\int_(-\infty)^\infty xf_X(x)\,\mathrm dx=\frac1{200}\int_(720)^(920)x\,\mathrm dx=\boxed{820}`

and variance

`V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2`

The second moment is

`E[X^2]=\displaystyle\int_(-\infty)^\infty x^2f_X(x)\,\mathrm dx=\frac1{200}\int_(720)^(920)x^2\,\mathrm dx=\frac{2,027,200}3`

and so

`V[X]=\frac{2,027,200}3-820^2=\frac{10,000}3`

The standard deviation is the square root of the variance, so

`SD[X]=√(V[X])=\sqrt{\frac{10,000}3}\approx\boxed{57.73}`

(b)

`P(X<870)=\displaystyle\int_(-\infty)^(870)f_X(x)\,\mathrm dx=\frac1{200}\int_(720)^(870)\mathrm dx=(870-720)/(200)=\boxed{0.75}`