Introduction to Probability, Grinstead, C. Snell, J., 1997
Page 320
Question 1

Question
1) Let 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.

  1. P(|X − 10| ≥ 2).
  2. P(|X − 10| ≥ 5).
  3. P(|X − 10| ≥ 9).
  4. P(|X − 10| ≥ 20).

Chebyshev’s inequality provides an estimate of how a distribution varies from it’s mean.
The formula results in an upper bound of probability that a random variable varies from the mean by a certain number of standard deviations.

Below is Chebyshev’s inequality formula:
\(P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}\)

Because we are provided variance we can calculate standard deviation, which is equal to the square root of the variance. We can then plug the standard deviations into the formular.

#Given
mean <- 10
variance <- 100/3

#SD
sd <- sqrt(variance)

#k the number of standard deviations away from the mean
k_1 <- 2 / sd
k_2 <- 5 / sd
k_3 <- 9 / sd
k_4 <- 20 / sd

#upper bounds 
upper_bound_1 <- 1 / (k_1^2)
upper_bound_2 <- 1 / (k_2^2)
upper_bound_3 <- 1 / (k_3^2)
upper_bound_4 <- 1 / (k_4^2)

#results
u_b_1<-bound<-cat("Upper bound for P(|X - 10| >= 2):", upper_bound_1, "\n")
## Upper bound for P(|X - 10| >= 2): 8.333333
u_b_2<-cat("Upper bound for P(|X - 10| >= 5):", upper_bound_2, "\n")
## Upper bound for P(|X - 10| >= 5): 1.333333
u_b_3<-cat("Upper bound for P(|X - 10| >= 9):", upper_bound_3, "\n")
## Upper bound for P(|X - 10| >= 9): 0.4115226
u_b_4<-cat("Upper bound for P(|X - 10| >= 20):", upper_bound_4, "\n")
## Upper bound for P(|X - 10| >= 20): 0.08333333