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

Answer 1

Answer:

a) $E(X) = \frac{a+b}{2}=\frac{670+750}{2}= 710$

The variance is given by:

$Var(X) =\frac{(b-a)^2}{12}= \frac{(750-670)^2}{12}= 5333.3333$

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

$Sd(X) = \sqrt{5333.333}= 23.09$

b) $P(X$

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

$F(X) = \frac{x-a}{b-a} , a \leq x \leq b$

And replacing we got:

$P(X$

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:

$X \sim Unif(a=670, b = 750)$

Part a

For this case the expected value is given by:

$E(X) = \frac{a+b}{2}=\frac{670+750}{2}= 710$

The variance is given by:

$Var(X) =\frac{(b-a)^2}{12}= \frac{(750-670)^2}{12}= 5333.3333$

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

$Sd(X) = \sqrt{5333.333}= 23.09$

Part b

For this case we want to find this probability:

$P(X$

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

$F(X) = \frac{x-a}{b-a} , a \leq x \leq b$

And replacing we got:

$P(X$