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How large a random sample must be taken from a given distribution in order for the probability to be at least 0.99 that the sample mean will be within 2 standard deviations of the mean of the distribution?

Short Answer

Expert verified

25

Step by step solution

01

Given information

If X be the random variable then it is given that \(P\left( {\left| {{X_n} - \mu } \right| < 2\sigma } \right) \ge 0.99\). We need to calculate minimum sample size.

02

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

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

Using Chebyshev inequality

\(\begin{array}{c}P\left( {\left| {{X_n} - \mu } \right| < 2\sigma } \right) \ge 0.99\\1 - \frac{1}{{4n}} \ge 0.99\\n \ge 25\end{array}\)

\(\)So the minimum sample size required is 25.

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