Probability Distribution

In probability theory and the statistics, a probability distribution identifies either the probability of each value of a random variable (when the variable is discrete), or the probability of the value falling within a particular interval (when the variable is continuous). The probability distribution describes the range of the possible values that a random variable can get and the probability that the value of the random variable is within any subset of that range.

I like to share this Non Central T Distribution with you all through my article.

The Normal distribution is often called as the "bell curve", when the random variable takes the values in the set of real numbers. Let us see some sample problems on probability distribution statistics.

Examples

Given below are some of the examples on Probability Distribution Statistics.

Example 1:

A continuous random variable X has probability distribution function f(x) = 3x2, 0 ≤ x ≤ 1, Find a and b such that. (i) P(X ≤ a) = P(X > a) and (ii) P(X > b) = 0.05. Calculate probability distribution for this function.

Solution:

(i) Since the total probability is 1, [Given that P(X ≤ a) = P (X > a)

P(X ≤ a) + P(X > a) = 1

i.e., P(X ≤ a) + P(X ≤ a) = 1

⇒ P(X ≤ a) =1/2⇒ `int`3x2dx = 12

i.e.,[3x3/3]a0=1/2 ⇒ a3 =1/2

i.e., a = 1/213

(ii) P(X > b) = 0.05

∴ `int` f(x) dx = 0.05 ∴ `int` b1 3x2  dx = 0.05

[3x3]31b= 0.05 ⇒ 1 − b3 = 0.05

b3 = 1 − 0.05 = 0.95 =95

100 ⇒ b = 19/2013

Example 2:  A random variables X has probability mass function as in the probability distribution tables given below
X
0 1 2 3 4 5 6
P(X = x) k 3k 5k 7k 9k 11k 13k



(1) Find k.

(2) Evaluate P(X < 4), P(X ≥ 5) and P(3< X ≤ 6)

Solution:

(1) Since P(X = x) is a probability mass function `sum_(n=0)^6` P(X = x) = 1

ie.,P(X=0) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)+P(X = 6) = 1.

⇒ k + 3k + 5k + 7k + 9k + 11k + 13k = 1 ⇒ 49 k = 1 ⇒ k =1/49

(2) P(X < 4) = P(X = 0) + P(X = 1 ) + P(X = 2) + P(X = 3) =1/49 +3/49 +5/49 +7/49 =16/49

P(X ≥ 5) = P(X = 5) + P(X = 6) =11/49 +13/49 =24/49

P(3 < X ≤ 6) = P(X = 4) + P(X = 5) + P(X = 6) =9/49 +11/49 +13/49 =33/49

∴ The smallest value of x for which P(X ≤ x) > 1/2 is 4.

Example 3:

In a class, the average number of marks obtained by student in Physics is 0.52, chemistry is 0.48 and in both Physics and chemistry they obtained 0.37. Find the total average obtained in either Physics or chemistry.

Solution:

P(A) = Average number of marks in Physics = 0.52

P(B) = Average number of marks in Chemistry = 0.48

P(A and B) = Average number of marks in both physics and chemistry = 0.37

P(A or B)    = P(A) + P(B) – P(A-B)

= 0.52 + 0.48 – 0.37

= 0.63