A continuous random variable X has a uniform distribution over some interval [L, R] with mean $7$ and variance $3$ (i.e. $E(X) = 7$ and $Var(X)=3$).

Question

A continuous random variable X has a uniform distribution over some interval [L, R] with mean $7$ and variance $3$ (i.e. $E(X) = 7$ and $Var(X)=3$).

(a) Compute $P(X \leq 5$ or $X \geq 9)$

Answer:

$E(X) = \frac{L + R}{2} = 7 \to L + R = 14$

and

$Var(X) = \frac{(R-L)^2}{12} = 3 \to R - L = 6$

so $L = 4$, $R = 10$

thus,

$P(X \leq 5$ or $X \geq 9) = ?$

How do I compute above ?

Answer 1

Using the properties of a uniform distribution:

P(X≤5)=(5-L)/(R-L)=1/6 and

P(X≥9)=(R-9)/(R-L)=1/6

P(X≤5 or X≥9)= 1/6 + 1/6 =1/3