Question

Let X be a random variable with mean 11 and variance 9. Using Tchebysheff's theorem,find

(a) a lower bound for  $P(6 < X < 16)$

(b) the value of C such that $P(|X-11|\geq C)\leq 0.09$

 

 

Answer 1

Solution

Let X be random variable with $E(X)=11\hspace{2mm},V(X)=9$

(a) a lower bound for  $P(6 < X < 16)$

Tchebysheff's  theorem :

$\implies P(E(X)-k\delta< X< k\delta+E(X))\geq 1-\frac{1}{k^2}$

since $E(X)-k\delta=6\implies k=(11-6)\times \frac{1}{3}=\frac{5}{3}$

Then  $P(6 < X < 16)\geq 1-\frac{9}{25}=\frac{16}{25}$

Therefore. Lower bound is $\frac{16}{25}$

(b) Find the value C such that

$\implies P(|X-11|\geq C)\leq 0.09$

By chebysheff's theorem  $P(|X-11|\geq C)\leq \frac{V(X)}{C^2}$

then  $\frac{V(X)}{C^2}=0.09\implies C^2=\frac{V(X)}{0.09}\implies C=10$

Therefore.$C=10$

Therefore.

a). Lower bound is $\frac{16}{25}$

b). $C=10$