If X is normally distributed, with mean \(\mu\) and variance \(\sigma^2\) , find an upper bound for the following probabilities, using Chebyshev’s Inequality:

a) \(P(|X - \mu| \geq \sigma)\)

b) \(P(|X - \mu| \geq 2 \sigma)\)

c) \(P(|X - \mu| \geq 3 \sigma)\)

d) \(P(|X - \mu| \geq 4 \sigma)\)

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

Answer: 1

k <- 1

value <- 1 / (k*k)

value
## [1] 1

Answer: 1/2

k <- 2

value <- 1 / (k*k)

value
## [1] 0.25

Answer: 1/9

k <- 3

value <- 1 / (k*k)

value
## [1] 0.1111111

Answer: 1/16

k <- 4

value <- 1 / (k*k)

value
## [1] 0.0625

Now find the exact value using the program NormalArea or the normal table in Appendix A, and compare.

To find the exact values, I used a table of Z-scores. I found the area in between the values Z = \(\pm\) 1.00, Z = \(\pm\) 2.00, Z = \(\pm\) 3.00, and Z = \(\pm\) 4.00 for the appropriate question. For example, I found the area in between Z = \(\pm\) 1.00 for part a.

a) Exact value

\(P(|X - \mu| \geq \sigma) = 1 - P(-\sigma \leq Z \leq \sigma)\)

\(P(Z \leq \sigma)\) = 0.8413

\(P(Z \leq -\sigma)\) = 0.1587

Answer: 0.3174

1 - (0.8413 - 0.1587)
## [1] 0.3174

b) Exact value

\(P(|X - \mu| \geq 2 \sigma) = 1 - P(-2 \sigma \leq Z \leq 2 \sigma)\)

\(P(Z \leq 2 \sigma)\) = 0.9772

\(P(Z \leq -2 \sigma)\) = 0.0228

Answer: 0.0456

1 - (0.9772 - 0.0228)
## [1] 0.0456

c) Exact value

\(P(|X - \mu| \geq 3 \sigma) = 1 - P(-3 \sigma \leq Z \leq 3 \sigma)\)

\(P(Z \leq 3 \sigma)\) = 0.9987

\(P(Z \leq -3 \sigma)\) = 0.0013

Answer: 0.0026

1 - (0.9987 - 0.0013)
## [1] 0.0026

d) Exact value

\(P(|X - \mu| \geq 4 \sigma) = 1 - P(-4 \sigma \leq Z \leq 4 \sigma)\)

\(P(Z \leq 4 \sigma)\) = 0.9999

\(P(Z \leq -4 \sigma)\) = 0.0000

Answer: 0.0001

1 - (0.9999 - 0.0000)
## [1] 1e-04