Question 1: Page 303 Problem 11

A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out?

mean = lifetime / n

Answer: 10 hours

mean = 1000 / 100
print(mean)
## [1] 10

Question 2: Page 303 Problem 14

Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density

fZ(z) = \((1/2) \lambda e^{-\lambda |z|}\)

Answer:

The pdf of an exponential distribution is fx(x) = \(\lambda e^{- \lambda x}\) when x is greater than or equal to 0.

To find the pdf of Z, we will find the integral:

fZ(z) = \(\int_{-\infty}^{\infty} fX1(x) * fX2(z - x) \,dx\)

fX1(x) * fX2(z - x) = \(\lambda^2 e^{- \lambda x} e^{- \lambda(z-x)}\)

Since x must be greater than or equal to 0, we know the integral for fZ(z) must be 0 to infinity, and we can stop at z.

fZ(z) = \(\lambda^2 \int_{0}^{z} e^{- \lambda x} e^{\lambda x - \lambda z} \ dx\) + \(\lambda^2 \int_{z}^{\infty} e^{- \lambda x} e^{\lambda x + \lambda z} \ dx\)

Exponents can combine for simplification. Then, each term gets integrated.

fZ(z) = \(\frac{\lambda^2}{-2\lambda} e^{-2\lambda x + \lambda z} \Big|_{0}^{z} + \frac{\lambda^2}{-2\lambda} e^{-2\lambda x - \lambda z} \Big|_{z}^{\infty} \\\)

fZ(z) = \(-\frac{1}{2} e^{-\lambda z} + \frac{1}{2} e^{\lambda z}\)

That last equation can be reduced to fZ(z) = \(\frac{1}{2} e^{- \lambda |z|}\)

Question 3: Page 320-321 Problem 1

Let X be a continuous random variable with mean µ = 10 and variance \(\sigma ^2\) = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

  1. P(|X − 10| ≥ 2).

  2. P(|X − 10| ≥ 5).

  3. P(|X − 10| ≥ 9).

  4. P(|X − 10| ≥ 20).

Chebyshev’s Inequality = \(P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2}\)

Part a.

Answer: 8.3333

# Standard deviation
stdev <- (100 / 3)^0.5

# k * st.dev. = 2
k <- 2 / stdev

value <- 1 / (k*k)

value
## [1] 8.333333

Part b.

Answer: 1.3333

# Standard deviation
stdev <- (100 / 3)^0.5

# k * st.dev. = 5
k <- 5 / stdev

value <- 1 / (k*k)

value
## [1] 1.333333

Part c.

Answer: 0.4115

# Standard deviation
stdev <- (100 / 3)^0.5

# k * st.dev. = 9
k <- 9 / stdev

value <- 1 / (k*k)

value
## [1] 0.4115226

Part d.

Answer: 0.0833

# Standard deviation
stdev <- (100 / 3)^0.5

# k * st.dev. = 20
k <- 20 / stdev

value <- 1 / (k*k)

value
## [1] 0.08333333