Question

Suppose that \({X_1},...,{X_n}\)form a random sample of size n from a distribution for which the mean is 6.5 and the variance is 4. Determine how large the value of n must be in order for the following relation to be satisfied:

\({\bf{P}}\left( {{\bf{6}} \le {{{\bf{\bar X}}}_{\bf{n}}} \le {\bf{7}}} \right) \ge {\bf{0}}{\bf{.8}}\)

Short answer

80

Step-by-step solution

Given information

Mean = 6.5

Variance = 4

We need to calculate the sample size such that \(P\left( {6 \le {{\bar X}_n} \le 7} \right) \ge 0.8\)

Calculation of sample size

Chebyshev inequality

Let X be the random variable for which \({\rm{Var}}\left( X \right)\) exists. Then every for every number \(t > 0\) we have the following inequality

\(P\left( {\left| {X - E\left( X \right)} \right| \ge t} \right) \le \frac{{{\rm{Var}}\left( X \right)}}{{{t^2}}}\)

Using Chebyshev inequality

\(P\left( {6 \le {{\bar X}_n} \le 7} \right) \ge 0.8\)

\(\begin{array}{c}P\left( {\left| {{{\bar X}_n} - \mu } \right| < \frac{1}{2}} \right) \ge 1 - \frac{4}{{n \times \frac{1}{4}}}\\ = 1 - \frac{{16}}{n}\end{array}\)

Now

\(\begin{array}{c}1 - \frac{{16}}{n} \ge 0.8\\n - 16 \ge 0.8n\\n \ge 80\end{array}\)

So, the required sample size is 80.