Question

If X is uniformly distributed with mean 1 and variance 4/3, find P(X<0).​

Answer 1

Answer:

1/4

Step-by-step explanation:

mean=1 and. variance=4/3

(a+b)/2=1. ( (a-b) ^2)/12=4/3

a+b=2. (a-b) ^2=16

a-b=4

equating both we get a= -1 and b=3

p(X<0)=1/(3+1) x (0+1)=1/4

Answer 2

Given:

X is uniformly distributed.

Mean = 1 ; Variance = 4/3.

To find:

P(X<0)

Solution:

We can solve this mathematical problem using the following mathematical concept.

Assume X follows uniform distribution in (a,b).

We know that for uniform distribution:

Mean = $\frac{b+a}{2}$  and variance = $\frac{(b-a)^{2} }{12}$

According to the question:

Mean = $\frac{b+a}{2}$ = 1 and variance = $\frac{(b-a)^{2} }{12}$ = $\frac{4}{3}$

We get,

• $a+b = 2$
• $(b-a) ^{2} = 16\\b - a = 4, -4$
• On solving we get, a = 1 and b = 3.

∴ f(x) = $\frac{1}{4}$ , -1 < x < 3

∴ P[X<0] = $\int_{-1}^{0} f(x) d x = \frac{1}{4}$

Therefore, P(X<0) = $\frac{1}{4}$.