Question

Consider the random variable X with following probability distribution: f (x )space space equals space fraction numerator 2 x space minus 1 over denominator 16 end fraction comma space space space space space space x space equals space 1 comma space 2 comma space 3 comma space 4. a) Find the probability that X is at most 2. b) What is the expected value of X

Answer 1

a) The probability that X is at most 2

= P(X ≤ 2) = 0.25

b) The expected value of X = 3.125

Explanation:

The probability distribution is given as

f(x) = (2x - 1)/16, for x = 1,2,3,4

a) The probability that X is at most 2

= P(X ≤ 2) = P(X = 1) + P(X = 2)

P(X = 1) = f(x=1) = [(2×1) - 1]/16 = 0.0625

P(X = 2) = f(x=2) = [(2×2) - 1]/16 = 0.1875

P(X ≤ 2) = P(X = 1) + P(X = 2) = 0.0625 + 0.1875 = 0.25

b) The expected Value of X

Expected value is given as

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

x = 1,2,3,4

But we need the corresponding probabilities for each variable

P(X = 1) = f(x=1) = 0.0625

P(X = 2) = f(x=2) = 0.1875

P(X = 3) = f(x=3) = [(2×3) - 1]/16 = 0.3125

P(X = 4) = f(x=4) = [(2×4) - 1]/16 = 0.4375

The probability mass function can then be given in tabular form as

X | p(x)

1 | 0.0625

2 | 0.1875

3 | 0.3125

4 | 0.4375

E(X) = Σ xᵢpᵢ = (1×0.0625) + (2×0.1875) + (3×0.3125) + (4×0.4375)

= 0.0625 + 0.375 + 0.9375 + 1.75

= 3.125

Hope this Helps!!!