Question

Let X be a random variable for which \({\bf{E}}\left( {\bf{X}} \right){\bf{ = \mu }}\)and\({\bf{Var}}\left( {\bf{X}} \right){\bf{ = }}{{\bf{\sigma }}^{\bf{2}}}\).Construct a probability distribution for X such that \({\bf{P}}\left( {\left| {{\bf{X - \mu }}} \right| \ge {\bf{3\sigma }}} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{9}}}\)

Short answer

X	P(X=x)
\(\mu - 3\sigma \)	\(\frac{1}{{18}}\)
\(\mu \)	\(\frac{8}{9}\)
\(\mu + 3\sigma \)	\(\frac{1}{{18}}\)
Total	1

Step-by-step solution

Given information

We need to construct a probability distribution for a random variable X such that \(P\left( {\left| {X - \mu } \right| \ge 3\sigma } \right) = \frac{1}{9}\) .

Step-2: Construction of probability distribution

Let us assume the random variable X assumes three values \(\mu - 3\sigma ,\mu ,\mu + 3\sigma \)

It is given that \(P\left( {\left| {X - \mu } \right| \ge 3\sigma } \right) = \frac{1}{9}\) .

\( \Rightarrow P\left( {X = \mu - 3\sigma } \right) = P\left( {X = \mu + 3\sigma } \right) = \frac{1}{{18}}\)

Hence \(\begin{aligned}{}P\left( {X = \mu } \right) &= 1 - \left( {\frac{1}{{18}} + \frac{1}{{18}}} \right)\\ &= \frac{8}{9}\end{aligned}\)

Hence the probability distribution

X	P(X=x)
\(\mu - 3\sigma \)	\(\frac{1}{{18}}\)
\(\mu \)	\(\frac{8}{9}\)
\(\mu + 3\sigma \)	\(\frac{1}{{18}}\)
Total	1