Question

Suppose that F is a continuous c.d.f. on the real line, and let \({{\bf{\alpha }}_{\bf{1}}}\,{\bf{and}}\,{{\bf{\alpha }}_{\bf{2}}}\)be numbers such that \({\bf{F}}\left( {{{\bf{\alpha }}_{\bf{1}}}} \right)\,{\bf{ = 0}}{\bf{.3}}\,{\bf{and}}\,{\bf{F}}\left( {\,{{\bf{\alpha }}_{\bf{2}}}} \right){\bf{ = 0}}{\bf{.8}}{\bf{.}}\). Suppose 25 observations are selected at random from the distribution for which the c.d.f. is F. What is the probability that six of the observed values will be less than \({{\bf{\alpha }}_{\bf{1}}}\), 10 of the observed values will be between \({{\bf{\alpha }}_{\bf{1}}}\) and \({{\bf{\alpha }}_{\bf{2}}}\), and nine of the observed values will be greater than \({{\bf{\alpha }}_{\bf{2}}}\)?

Short answer

0.0060.

Step-by-step solution

Given information

A cdf F is given. Two points are located on it \({\alpha _1}\) and\({\alpha _2}\). Their respective cdf values is\(F\left( {{\alpha _1}} \right)\, = 0.3\,and\,F\left( {\,{\alpha _2}} \right) = 0.8.\)

Defining the data points

The elements of the population lie on this cdf graph. Let there be k distinct data points.

Let \({X_1}\) denote the elements from the ith group that lies below \({\alpha _1}\). The probability of a point lying here is denoted by \({P_1}\).

Similarly, let \({X_2}\) denote the elements from the 2nd group that lie between \({\alpha _1}\) and \({\alpha _2}\). The probability of a point lying here is denoted by \({P_2}\).

And lastly, let \({X_3}\) denote the elements from the 3^rd group that lie above \({\alpha _2}\).

The probability of a point lying here is denoted by \({P_3}\).

Thus, the points \({P_1} + {P_2} + {P_3} = 1\) defined by observing \({X_1} = {x_1},{X_2} = {x_2},{X_3} = {x_3}\).

Since, \(F\left( {{\alpha _1}} \right)\, = 0.3\,and\,F\left( {\,{\alpha _2}} \right) = 0.8.\)

Therefore,

\(\begin{aligned}{c}{P_1} &= 0.3\\{P_2} &= 0.8 - 0.3\\ &= 0.5\\{P_3}& = 1 - 0.8\\ &= 0.2\end{aligned}\)

Defining the pdf

This is an example of a multinomial distribution. Here are the data points \({X_i} \sim {\rm{Multinomial}}\left( {n,{p_i}} \right),\,i = 1,2,3\)

We have total 25 data points. Therefore, n=25.

Therefore, the pdf is

\(\left( {\frac{{25!}}{{{x_1}!{x_2}!{x_3}!}}} \right){p_1}^{{x_1}}{p_2}^{{x_2}}{p_3}^{{x_3}}\)

Substituting the values

It is given that \({x_1} = 6,{x_2} = 10,{x_3} = 9\)

Therefore,

\(\begin{aligned}{}p\left( x \right) &= \left( {\frac{{25!}}{{6!10!9!}}} \right){\left( {0.3} \right)^6}{\left( {0.5} \right)^{10}}{\left( {0.2} \right)^9}\\ &= 0.0060\end{aligned}\)

Therefore, the probability that six of the observed values are less than\({{\bf{\alpha }}_{\bf{1}}}\)10 are between\({{\bf{\alpha }}_{\bf{1}}}\), and\({{\bf{\alpha }}_{\bf{2}}}\)9 are above\({{\bf{\alpha }}_{\bf{2}}}\)is 0.0060.