Question

Let\({X_1},{X_2},....{X_{30}}\)be independent random variables each having a discrete distribution with p.f.

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{4}\;\;\;\;\;\;if\;\;x = 0\;or\;2\\\frac{1}{2}\;\;\;\;\;\;if\;\;x = 1\\0\;\;\;\;\;\;\;otherwise\end{array} \right.\)

Use the central limit theorem and the correction for continuity to approximate the probability that\({X_1} + \cdots + {X_{30}}\)is at most 33.

Short answer

Approximate value of the probability that \({X_1} + \cdots + {X_{30}}\) is at most 33 is \(0.8169\).

Step-by-step solution

Given information

Let\({X_1}......{X_{30}}\)be independent random variables with p.f.

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{4}\;\;\;\;\;\;if\;\;x = 0\;or\;2\\\frac{1}{2}\;\;\;\;\;\;if\;\;x = 1\\0\;\;\;\;\;\;\;otherwise\end{array} \right.\)

Finding mean and variance of random variable X

For a discrete random variable,

\(E\left( {{X_i}} \right) = \sum {{X_i}{\rm P}\left( {{X_i}} \right)} \)

Hence from the p.f., mean of\({X_i}\)is:

\(\begin{array}{l}E\left( {{X_i}} \right) = 0 \times {\rm P}\left( {{X_0}} \right) + 1 \times {\rm P}\left( {{X_1}} \right) + 2 \times {\rm P}\left( {{X_2}} \right)\\ = 0\left( {\frac{1}{4}} \right) + 1\left( {\frac{1}{2}} \right) + 2\left( {\frac{1}{4}} \right)\\ = 0 + \frac{1}{2} + \frac{2}{4}\end{array}\)

\(\begin{array}{c}E\left( {{X_i}} \right) = \frac{1}{2} + \frac{1}{2}\\ = 1\end{array}\)’

And\(E{\left( {{X_i}} \right)^2} = 1.5\)

So the variance of\({X_i}\)is

\(\begin{array}{l}Var\left( {{X_i}} \right) = E{\left( {{X_i}} \right)^2} - {\left[ {E\left( {{X_i}} \right)} \right]^2}\\Var\left( {{X_i}} \right) = 1.5 - 1\\Var\left( {{X_i}} \right) = 0.5\end{array}\)

The central limit theorem says that\(Y = {X_1} + {X_2} + \cdots + {X_{30}}\) has approximately the normal distribution with mean

\(30\left( 1 \right) = 30\)

Variance is

\(\left[ {30\left( {0.5} \right)} \right] = 15\)

Finding the probability of \({X_1} +   \cdots  + {X_{30}}\) is at most 33

Using the correction for continuity, we would assume that Y has the normal distribution with mean 30 and variance 15 and compute the probability that\(Y \le 33.5\)

\(\begin{array}{c}\phi \left( {\frac{{\left[ {33.5 - 30} \right]}}{{\sqrt {15} }}} \right) = \phi \left( {0.904} \right)\\ = 0.8169\end{array}\)

By using the standard normal table.

Therefore,approximate value of probability that\({X_1}...... + {X_{30}}\)is at most 33 is\(0.8169\).