Question

LetXbe a continuous random variable with densityf(x) ={3e−3x,whenx >0,0,elsewhere.(a) Verify thatfis a density function.(b) CalculateP(−1< X <1).(c) CalculateP(X <5).(d) CalculateP(2< X <4|X <5).(e) Find the cumulative distribution function ofX.(f) Find theEXandEe2X.(g) Find the variance ofX.

Answer 1

In this question, we verify that f(x) is a density function, calculate probabilities for the given random variable X, find the cumulative distribution function, calculate the expected value and variance, and determine the probability density function.

Step-by-step explanation:

To verify that f(x) is a density function, we need to check two conditions:

1. For any x, f(x) is greater than or equal to 0.
2. The integral of f(x) over the entire range is equal to 1.

a) The first condition is satisfied since 3e^(-3x) is always positive for x > 0.

b) To calculate P(-1 < X < 1), we integrate f(x) from -1 to 1:
P(-1 < X < 1) = ∫(from -1 to 1) 3e^(-3x) dx
= [-e^(-3x)](from -1 to 1) = -e^(-3)+e^3

c) To calculate P(X < 5), we integrate f(x) from 0 to 5:
P(X < 5) = ∫(from 0 to 5) 3e^(-3x) dx
= [-e^(-3x)](from 0 to 5) = -e^(-15)+1

d) To calculate P(2 < X < 4 | X < 5), we need to find the conditional probability:
P(2 < X < 4 | X < 5) = P(2 < X < 4 and X < 5) / P(X < 5)
We calculate the numerator and denominator separately:

• Numerator: ∫(from 2 to 4) 3e^(-3x) dx = [-e^(-3x)](from 2 to 4) = -e^(-12)+e^(-6)
• Denominator: P(X < 5) = -e^(-15)+1

Therefore, P(2 < X < 4 | X < 5) = (-e^(-12)+e^(-6))/(-e^(-15)+1)