Data 605 discussion week 8 - # 8.2.1

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

\[mean μ = 10\] \[variance σ^2 = 100/3\] \[Standard deviation σ = sqrt (100/3)\]

\[P(|X-\mu|\geq k\sigma)\leq\frac{1}{k^2}\]

(a) P( | X - 10 | >= 2)

\[ k \sigma=2 \] \[ k= \frac{2}{\sqrt{100/3}} \] \[ P(|X - 10|\geqslant 2) = \frac{1}{k^{_{2}}} = 8.3333 \approx 1\]

(b) P( | X - 10 | >= 5)

\[ k \sigma=5 \] \[ k= \frac{5}{\sqrt{100/3}} \] \[ P(|X - 10|\geqslant 5) = \frac{1}{k^{_{2}}} = 1.3333 \approx 1$ \]

(c) P( | X - 10 | >= 9)

\[ k \sigma=9 \] \[ k= \frac{9}{\sqrt{100/3}} \] \[ P(|X - 10|\geqslant 9) = \frac{1}{k^{_{2}}} = 0.4115 \]

(d) P( | X - 10 | >= 20)

\[ k \sigma=20 \] \[ k= \frac{20}{\sqrt{100/3}} \] \[ P(|X - 10|\geqslant 20) = \frac{1}{k^{_{2}}} = 0.0833$ \]

Test:

u <- 10
o <- sqrt(100/3)

x1 <-c(2, 5, 9, 20)

k <- (x1 )/o

p1 <-  round(1/(k^2), 4)
p1
## [1] 8.3333 1.3333 0.4115 0.0833