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.

Let \({\rm{X}}\) be the consumer's call duration. It is given that the call duration is exponentially distributed with parameter\({\rm{\lambda }}\). Which means that it's pdf is given as:

(a) The first one charges a flat rate of \({\rm{10}}\) cents per minute, whereas the second charges a flat rate of \({\rm{99}}\) cents for calls up to \({\rm{20}}\) minutes in duration and then \({\rm{10}}\) cents for each additional minute exceeding \({\rm{20}}\).

If the call duration is less than \({\rm{10}}\) minutes than plan-\({\rm{1}}\) is better since it charges will be less than \({\rm{99}}\) cents.

If the call duration is more than or equal to \({\rm{10}}\) minutes than plan-\({\rm{2}}\) is better because for the next \({\rm{10}}\) minute plan-\({\rm{2}}\) doesn't charge anything while plan-\({\rm{1}}\) continues to charge \({\rm{10}}\) cents per minute. Hence for call duration of \({\rm{20}}\) or more, plan-\({\rm{2}}\) gives \({\rm{10}}\) free minutes.

Here we talked about call duration for a specific call. We can tell that plan-l is better when expected call duration is small but we will have to mathematically calculate the exact minutes at which both plans are equally economical. But it will surely be greater than \({\rm{10}}\) minutes.

(b) When call duration is \({\rm{x}}\) minutes then let \({{\rm{h}}_{\rm{1}}}{\rm{(x)}}\) and \({{\rm{h}}_{\rm{2}}}{\rm{(x)}}\)be the cost functions (in cents) for the first and second plan respectively.

Now we recall the following proposition:

Proposition: If \({\rm{Y}}\) is a continuous rv with pdf \({\rm{f(y)}}\) and \({\rm{g(y)}}\) is any function of\({\rm{y}}\), then

\({\rm{E(g(y)) = }}\int_{{\rm{ - \yen }}}^{\rm{\yen}} {\rm{g}} {\rm{(y) \times f(y) \times dy}}\)

Since \({{\rm{h}}_{\rm{1}}}{\rm{(x)}}\) and \({{\rm{h}}_{\rm{2}}}{\rm{(x)}}\) are functions of \({\rm{X}}\) (call duration) and \({\rm{X}}\) is exponentially distributed, hence it's pdf \({\rm{f(x)}}\)is given as :

Expected value of \({{\rm{h}}_{\rm{1}}}{\rm{(x)}}\) is given as :

\(\begin{aligned}{\rm{E}}\left( {{{\rm{h}}_{\rm{1}}}{\rm{(x)}}} \right) & = \int_{{\rm{ - \yen}}}^{\rm{\yen }} {{{\rm{h}}_{\rm{1}}}} {\rm{(x) \times f(x) \times dx}}\\ & =\int_{\rm{0}}^{\rm{\yen }} {{\rm{(10x)}}} \left( {{\rm{\lambda \times }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right){\rm{ \times dx}}\\ &= \left( {{\rm{(10x)}}\left( {{\rm{ - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)} \right)_{\rm{0}}^{\rm{\yen}}{\rm{ - }}\int_{\rm{0}}^{\rm{\yen }} {{\rm{(10)}}} {\rm{ \times }}\left( {{\rm{ - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\\&= 0 - 10 \left( {\int_{\rm{0}}^{\rm{\yen}} {\rm{ - }} {{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\\ &= - 10\left( {\frac{{{{\rm{e}}^{{\rm{ - \lambda x}}}}}}{{\rm{\lambda }}}} \right)_{\rm{0}}^{\rm{\yen}}\\ &= - 10 \left( {{\rm{0 - }}\frac{{\rm{1}}}{{\rm{\lambda }}}} \right)\\{\rm{E}}\left( {{{\rm{h}}_{\rm{1}}}{\rm{(x)}}} \right) &= \frac{{{\rm{10}}}}{{\rm{\lambda }}}\end{aligned}\)

(Use partial fraction)

Expected value of \({{\rm{h}}_{\rm{2}}}{\rm{(x)}}\) is given as:

\(\begin{aligned}{\rm{E}}\left( {{{\rm{h}}_{\rm{2}}}{\rm{(x)}}} \right)&= \int_{{\rm{ - \yen }}}^{\rm{\yen }} {{{\rm{h}}_{\rm{2}}}} {\rm{(x) \times f(x) \times dx}}\\&= 99 + \int_{{\rm{20}}}^{\rm{\yen}} {{\rm{(10(}}} {\rm{x - 20))}}\left( {{\rm{\lambda \times }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right){\rm{ \times dx}}\\&= 99 + \left( {{\rm{(10(x - 20))}}\left( {{\rm{ - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)} \right)_{{\rm{20}}}^{\rm{\yen}}{\rm{ - }}\int_{{\rm{20}}}^{\rm{\yen }} {{\rm{(10)}}} {\rm{ \times }}\left( {{\rm{ - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\\ &= 99 + (0 - 0) - 10\left( {\int_{{\rm{20}}}^{\rm{\yen }} {\rm{ - }} {{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)\\&= 99 - 10\left( {\frac{{{{\rm{e}}^{{\rm{ - \lambda x}}}}}}{{\rm{\lambda }}}} \right)_{{\rm{20}}}^{\rm{\yen}}{\rm{ (Use partial fraction) }}\\& = 99 - 10\left( {{\rm{0 - }}\frac{{{{\rm{e}}^{{\rm{ - 20\lambda }}}}}}{{\rm{\lambda }}}} \right)\\{\rm{E}}\left( {{{\rm{h}}_{\rm{2}}}{\rm{(x)}}} \right)&= 99 + \frac{{{\rm{10}}{{\rm{e}}^{{\rm{ - 20\lambda }}}}}}{{\rm{\lambda }}}\end{aligned}\)

Let the expected call duration be \({\rm{\mu }}\) minutes.

It is already given that the call duration has exponential distribution. We know that the expected value for exponential

distribution is given as\({\rm{1/\lambda }}\). Since we have denoted it as \({\rm{\mu }}\), hence

\({\rm{\lambda = }}\frac{{\rm{1}}}{{\rm{\mu }}}\)

Expected call duration is \({\rm{10}}\) minutes:

\(\begin{array}{c}{\rm{\lambda = }}\frac{{\rm{1}}}{{\rm{\mu }}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10}}}}\\{\rm{ = 0}}{\rm{.1}}\\{\rm{E}}\left( {{{\rm{h}}_{\rm{1}}}{\rm{(x)}}} \right){\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{0}}{\rm{.1}}}}{\rm{ = 100}}\\{\rm{E}}\left( {{{\rm{h}}_{\rm{2}}}{\rm{(x)}}} \right){\rm{ = 99 + }}\frac{{{\rm{10}}{{\rm{e}}^{{\rm{ - 20 \times 0}}{\rm{.1}}}}}}{{{\rm{0}}{\rm{.1}}}}\\{\rm{ = 112}}{\rm{.53}}\end{array}\)

Since the expected cost is lower for plan-\({\rm{1}}\), hence plan-\({\rm{1}}\) is better when expected call duration is \({\rm{10}}\) minutes.

Expected call duration is \({\rm{15}}\) minutes:

\(\begin{array}{c}{\rm{\lambda = }}\frac{{\rm{1}}}{{\rm{\mu }}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{15}}}}\\{\rm{E}}\left( {{{\rm{h}}_{\rm{1}}}{\rm{(x)}}} \right){\rm{ = }}\frac{{{\rm{10}}}}{{\frac{{\rm{1}}}{{{\rm{15}}}}}}{\rm{ = 150}}\\{\rm{E}}\left( {{{\rm{h}}_{\rm{2}}}{\rm{(x)}}} \right){\rm{ = 99 + }}\frac{{{\rm{10}}{{\rm{e}}^{{\rm{ - 20/15}}}}}}{{\frac{{\rm{1}}}{{{\rm{15}}}}}}\\{\rm{ = 138}}{\rm{.54}}\end{array}\)

Since the expected cost is lower for plan-\({\rm{2}}\), hence plan-\({\rm{2}}\) is better when expected call duration is \({\rm{15}}\) minutes.