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?
Expected Value for Exponential Distribution \[ E[X] = \frac{1}{\lambda} \]
Step 1: Calculate Rate parameter lambda for one bulb \[ \lambda = \frac{1}{1000} \ \]
Step 2: Calculate the Rate Parameter for the Minimum Lifetime for 100 bulbs \[ n\lambda = 100 \times \frac{1}{1000} = \frac{1}{10} \ \]
Step 3: Expected time for first bulb burn out \[ E[X_{\text{min}}] = \frac{1}{n\lambda} = \frac{1}{\frac{1}{10}} = 10 \, \text{hours} \]
# Rate parameter lambda
lambda_single_bulb <- 1 / 1000 # failure event rate
# Calculate rate for 100 bulbs
n_bulbs <- 100
lambda_combined <- n_bulbs * lambda_single_bulb
# Expected value
expected_time_first_bulb_out <- 1 / lambda_combined
# Results
cat("The expected time for the first lightbulb to burn out is:", expected_time_first_bulb_out, "hours\n")## The expected time for the first lightbulb to burn out is: 10 hoursLet X be a continuous random variable with mean μ = 10 and variance σ2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
Given:
Step 1: Calculate \(k\)
\[ k = \frac{2}{\sigma} = \frac{2}{\frac{10}{\sqrt{3}}} = \frac{2\sqrt{3}}{10} \approx 0.346 \]
Step 2: Chebyshev’s Inequality
\[ P(|X - 10| \geq 2) \leq \frac{1}{k^2} \approx \frac{1}{0.346^2} \] The calculation yielded an upper bound of 833%, Chebyshev’s Inequality cannot provide a meaningful upper bound for deviations greater than 2 (above 100%)
Step 1: Calculate \(k\)
\[ k = \frac{5}{\sigma} = \frac{5}{\frac{10}{\sqrt{3}}} = \frac{5\sqrt{3}}{10} \approx 0.866 \]
Step 2: Chebyshev’s Inequality
\[ P(|X - 10| \geq 5) \leq \frac{1}{k^2} \approx \frac{1}{0.866^2} \] The calculation yielded an upper bound of 133%, Chebyshev’s Inequality cannot provide a meaningful upper bound for deviations greater than 5 (above 100%)
Step 1: Calculate \(k\)
\[ k = \frac{9}{\sigma} = \frac{9}{\frac{10}{\sqrt{3}}} = \frac{9\sqrt{3}}{10} \approx 1.559 \]
Step 2: Chebyshev’s Inequality
\[ P(|X - 10| \geq 9) \leq \frac{1}{k^2} \approx \frac{1}{1.559^2} \] For a deviation of 9, the upper bound on the probability is approximately 41.15% (no more than 41.15% of the values of \(X\) are 9 units or more away from its mean)
Step 1: Calculate \(k\)
\[ k = \frac{20}{\sigma} = \frac{20}{\frac{10}{\sqrt{3}}} = \frac{20\sqrt{3}}{10} \approx 3.464 \]
Step 2: Chebyshev’s Inequality
\[ P(|X - 10| \geq 20) \leq \frac{1}{k^2} \approx \frac{1}{3.464^2} \]
For a deviation of 20, the upper bound on the probability is approximately 8.33% (no more than 8.33% of the values of \(X\) are 20 units or more away from its mean)
# Given data
mu <- 10
variance <- 100 / 3
sigma <- sqrt(variance)
# (a) Calculations for deviation = 2
k_a <- 2 / sigma
upper_bound_a <- 1 / (k_a^2)
cat("For (a) P(|X-10| >= 2):\n", "k =", k_a, "\n", "Upper Bound =", upper_bound_a, "\n\n")## For (a) P(|X-10| >= 2):
## k = 0.3464102
## Upper Bound = 8.333333# (b) Calculations for deviation = 5
k_b <- 5 / sigma
upper_bound_b <- 1 / (k_b^2)
cat("For (b) P(|X-10| >= 5):\n", "k =", k_b, "\n", "Upper Bound =", upper_bound_b, "\n\n")## For (b) P(|X-10| >= 5):
## k = 0.8660254
## Upper Bound = 1.333333# (c) Calculations for deviation = 9
k_c <- 9 / sigma
upper_bound_c <- 1 / (k_c^2)
cat("For (c) P(|X-10| >= 9):\n", "k =", k_c, "\n", "Upper Bound =", upper_bound_c, "\n\n")## For (c) P(|X-10| >= 9):
## k = 1.558846
## Upper Bound = 0.4115226# (d) Calculations for deviation = 20
k_d <- 20 / sigma
upper_bound_d <- 1 / (k_d^2)
cat("For (d) P(|X-10| >= 20):\n", "k =", k_d, "\n", "Upper Bound =", upper_bound_d, "\n\n")## For (d) P(|X-10| >= 20):
## k = 3.464102
## Upper Bound = 0.08333333