91Ó°ÊÓ

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

(a) The pdf \({\rm{f(x)}}\) is given to us as:

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\rm{1}}}{{\rm{9}}}\left( {{\rm{4 - }}{{\rm{x}}^{\rm{2}}}} \right)}&{{\rm{ - 1 < x < 2}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

It is plotted in the figure given below:

(b) We recall the definition of cdf of a continuous variable.

Definition: The cumulative distribution function \({\rm{F(x)}}\)for a continuous rv \({\rm{X}}\) is defined for every number \({\rm{x}}\) by

\(F(x) = P(X£x) = \int_{ - \yen}^x f (y) \times dy\)

Since \({\rm{f}}\left( {\rm{x}} \right)\) is non-zero only for \({\rm{x}}\) between \({\rm{ - 1}}\) and \({\rm{2}}\) . Hence for any \({\rm{x}}\) between \({\rm{ - 1}}\)and \({\rm{2}}\), we can write:

\(\begin{aligned}F(X)&=\int_{{\rm{ - 1}}}^{\rm{x}} {\frac{{\rm{1}}}{{\rm{9}}}} \left( {{\rm{4 - }}{{\rm{y}}^{\rm{2}}}} \right){\rm{ \times dy}}\\&=\frac{{\rm{4}}}{{\rm{9}}}\int_{{\rm{ - 1}}}^{\rm{x}} {\rm{d}} {\rm{y - }}\frac{{\rm{1}}}{{\rm{9}}}\int_{{\rm{ - 1}}}^{\rm{x}} {{{\rm{y}}^{\rm{2}}}} {\rm{ \times dy}}\\&=\frac{{\rm{4}}}{{\rm{9}}}{\rm{[y]}}_{{\rm{ - 1}}}^{\rm{x}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{9}}}\left[ {\frac{{{{\rm{y}}^{\rm{3}}}}}{{\rm{3}}}} \right]_{{\rm{ - 1}}}^{\rm{x}}\\&=\frac{{\rm{4}}}{{\rm{9}}}{\rm{[x - ( - 1)] - }}\frac{{\rm{1}}}{{\rm{9}}}\left[ {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ - }}\frac{{{{{\rm{( - 1)}}}^{\rm{3}}}}}{{\rm{3}}}} \right]\\F(X)&=-\frac{{{{\rm{x}}^{\rm{3}}}}}{{{\rm{27}}}}{\rm{ + }}\frac{{{\rm{4x}}}}{{\rm{9}}}{\rm{ + }}\frac{{{\rm{11}}}}{{{\rm{27}}}}\end{aligned}\)

Thus overall \({\rm{F(X)}}\)can be given as:

\({\rm{F(x) = }}\left\{ {\begin{aligned}{{}{}}{\rm{0}}&{{\rm{x < - 1}}}\\{{\rm{ - }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{{\rm{27}}}}{\rm{ + }}\frac{{{\rm{4x}}}}{{\rm{9}}}{\rm{ + }}\frac{{{\rm{11}}}}{{{\rm{27}}}}}&{{\rm{ - 1£x£2}}}\\{\rm{1}}&{{\rm{x > 2}}}\end{aligned}} \right.\)

\({\rm{ The cdf F(x) is plotted in the figure given below : }}\)

(c) Let us denote the median as \({\rm{\tilde \mu }}\) know that at median: \({\rm{F(\tilde \mu ) = 0}}{\rm{.5}}\)

From the cdf that we derived in the last part:

\({\rm{F(0) = }}\frac{{{\rm{11}}}}{{{\rm{27}}}}{\rm{\gg 0}}{\rm{.41}}\)

Hence 0 is not the median temperature at which the reaction takes place.

And since: \({\rm{F(0) < 0}}{\rm{.5}}\)

Hence the median temperature is larger than\({\rm{0}}\).

(d)

It is given that \({\rm{Y}}\) is the number among the ten labs at which the temperature exceeds\({\rm{1}}\). Then

\({\rm{Y\gg Bin(n,p)}}\)

Here: \({\rm{n = 10}}\)

And \({\rm{p}}\)is the probability that the temperature in a lab exceeds\({\rm{1}}\). Hence

\(\begin{aligned}P(X > 1) &= 1 - F(1)\\&= 1 -\left( {{\rm{ - }}\frac{{{{{\rm{(1)}}}^{\rm{3}}}}}{{{\rm{27}}}}{\rm{ + }}\frac{{{\rm{4(1)}}}}{{\rm{9}}}{\rm{ + }}\frac{{{\rm{11}}}}{{{\rm{27}}}}} \right)\\&=\frac{{\rm{5}}}{{{\rm{27}}}}\end{aligned}\)

Proposition: Let \({\rm{X}}\) be a continuous rv with \({\rm{pdf(x)}}\) and cdf\({\rm{F(x)}}\). Then for any number a,

\({\rm{P(X > a) = 1 - F(a)}}\)