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). \( C=10 \)