QUESTION 1:
Show that f(x)=x^2/55 for x=1,2,3,4,5 is a probability distribution function of a discrete random variable X. Hence, find
(a) P(X=3)
(b) P(X>2)
(c) P(X≤4)
(d) P(1<X≤5)
(e) P(1≤X≤3)
QUESTION 2:
Given the discrete random variable X has the following probability distribution function
g(x)={(2px ,x=0,1,2)/(px^2 ,x=3,4)}
Where p is a constant. Show that p=1/31. Hence, find
(a) P(X>4)
(b) P(X≤3)
(c) P(1≤X<4)
(d) P(|X-3|<2)
Discrete random variables- Cumulative Distribution Function
QUESTION 3:
The discrete random variable X has the following probability distribution function.
f(x)={■(1/6&x=0,2@1/3&x=1,3@0&otherwise)}
(a) Find the cumulative distribution function F(x)
(b) P(X≤2)
(c) P(X<1)
(d) P(0<X≤3)
(e) P(X>2)
(f) P(X<-1)
QUESTION 4:
The discrete random variable X has the following probability distribution function.
f(x)={■(1/8&x=0,3@3/8&x=1,2@0&otherwise)}
(a) Find the cumulative distribution function F(x)
(b) P(X≤-0.5)
(c) P(X≤2.5)
(d) P(X≤4)
QUESTION 5:
The cumulative distribution function F(x) of a discrete random variable X is given by:
F(x)=kx,x=1.2,3,4
Find
(a) the value of the constant k
(b) the probability distribution of X
(c) P(X=2)
(d) P(X>3)
(e) P(1≤X≤3)
To show that a given function is a probability distribution function (PDF) of a discrete random variable X, we need to check three conditions:
1) Non-negativity: The function should always be non-negative for any value of x.
2) Sum of probabilities: The sum of all probabilities should be equal to 1.
3) Probabilities of individual events: The probability of any event X=x should be between 0 and 1 (inclusive).
We will apply these conditions to the given functions in each question and then answer the subsequent parts of the questions.
QUESTION 1:
The given probability distribution function is:
f(x) = x^2/55 for x = 1, 2, 3, 4, 5
Let's check the conditions:
1) Non-negativity: For any value of x, f(x) = x^2/55 is non-negative.
2) Sum of probabilities: Let's calculate the sum of all probabilities:
f(1) + f(2) + f(3) + f(4) + f(5) = (1^2/55) + (2^2/55) + (3^2/55) + (4^2/55) + (5^2/55)
= 1/55 + 4/55 + 9/55 + 16/55 + 25/55
= 55/55
= 1
The sum of all probabilities is equal to 1.
3) Probabilities of individual events: The probability of any event X=x should be between 0 and 1.
(a) P(X=3) = f(3) = 3^2/55 = 9/55
(b) P(X>2) = P(X=3) + P(X=4) + P(X=5) = 9/55 + 16/55 + 25/55 = 50/55
(c) P(X≤4) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1/55 + 4/55 + 9/55 + 16/55 = 30/55
(d) P(1<X≤5) = P(X=2) + P(X=3) + P(X=4) + P(X=5) = 4/55 + 9/55 + 16/55 + 25/55 = 54/55
(e) P(1≤X≤3) = P(X=1) + P(X=2) + P(X=3) = 1/55 + 4/55 + 9/55 = 14/55
So, to summarize:
(a) P(X=3) = 9/55
(b) P(X>2) = 50/55
(c) P(X≤4) = 30/55
(d) P(1<X≤5) = 54/55
(e) P(1≤X≤3) = 14/55