Question

1. A continuous random variable is uniformly distributed between 20 and 120.

1. Find the following probabilities –

1. P(X<70)
2. P(X>50)
3. P(X=50)
4. P(30<X<90)

2. What are the mean and standard deviation of this distribution?

Answer 1

TOPIC:Uniform distribution.

- Define, xr Onif (20, 120) So, the palf of xis- (2) { 120-20 16:20 X$120. ( 0 ; otherwise. ; otherwise. af we need O P(X< 79. - 1 *° fex) dx . jex) dx. AH 20 - 70-20 10o 50 100 -10.50(Am).

③ we needs XP ( x 750). TX 750 120 dx 50 a 20 dx . 100 = to do. [2] 120 50 120-50 100 - 120-5 10.40 i we need, P ( x = 50). I (Ams). takes any X is a continuous random variable, the proobability that particular value, is zero.

We need =) P (30 < x <90). Jxdx J 30 30 - Too [x700 go-30 100 210.60 / (An). by we need, mean (x). = E(X). 120 x Jx) dx. = 1 20 = 100 1120 xdx 20 doc" (120"-20") | Holland

Again, E (x2). 1120 x² jeni dr. 20 2.0 too [2 = 5733. 3333 i var (x). = E(X2) - E²(X). = 833. 3333. standard deviation (x). = sd (x). a Juare (x? 833.3333 m[28.8675 (Aus)

(Rounded to four decimal places).