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.) b. What is the probability that X is less than 730

Respuesta :

Answer:

a) [tex] E(X) = \frac{a+b}{2}=\frac{670+750}{2}= 710[/tex]

The variance is given by:

[tex] Var(X) =\frac{(b-a)^2}{12}= \frac{(750-670)^2}{12}= 5333.3333[/tex]

And the deviation is just the square root of the variance and we got:

[tex] Sd(X) = \sqrt{5333.333}= 23.09[/tex]

b) [tex] P(X<730)[/tex]

And for this case we can use the cumulative distribution given by:

[tex] F(X) = \frac{x-a}{b-a} , a \leq x \leq b[/tex]

And replacing we got:

[tex] P(X<730)= F(730) = \frac{730-670}{750-670}=0.75[/tex]

Step-by-step explanation:

For this case we assume that X is our random variable and we know that the distribution for X is given by:

[tex] X \sim Unif(a=670, b = 750)[/tex]

Part a

For this case the expected value is given by:

[tex] E(X) = \frac{a+b}{2}=\frac{670+750}{2}= 710[/tex]

The variance is given by:

[tex] Var(X) =\frac{(b-a)^2}{12}= \frac{(750-670)^2}{12}= 5333.3333[/tex]

And the deviation is just the square root of the variance and we got:

[tex] Sd(X) = \sqrt{5333.333}= 23.09[/tex]

Part b

For this case we want to find this probability:

[tex] P(X<730)[/tex]

And for this case we can use the cumulative distribution given by:

[tex] F(X) = \frac{x-a}{b-a} , a \leq x \leq b[/tex]

And replacing we got:

[tex] P(X<730)= F(730) = \frac{730-670}{750-670}=0.75[/tex]

Q&A Education