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.

pdf of \({\rm{X}}\) is given, which can also be written as:

Now we first calculate standard deviation of given \({\rm{r}}{\rm{.v}}{\rm{.}}\)

Expected value of \({\rm{f(x)}}\) is given as:

\(\begin{aligned}E(x) &= \int_{{\rm{ - \yen }}}^{\rm{\yen }} {\rm{x}} {\rm{ \times f(x) \times dx}}\\& = \int_{{\rm{ - \yen }}}^{\rm{0}} {{\rm{(0}}{\rm{.5x)}}} \left( {{\rm{\lambda \times }}{{\rm{e}}^{{\rm{\lambda x}}}}} \right){\rm{ \times dx + }}\int_{\rm{0}}^{\rm{\yen }} {{\rm{(0}}{\rm{.5x)}}} \left( {{\rm{\lambda \times }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right){\rm{ \times dx}}\end{aligned}\)

If we replace \({\rm{x}}\) by \({\rm{ - t}}\)in the first integral then it is equal to negative of second one. Hence

\({\rm{E(x) = 0}}\)

\(\begin{aligned}{\rm{E}}\left( {{{\rm{x}}^{\rm{2}}}} \right) &= \int_{{\rm{ - \yen }}}^{\rm{\yen }} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times f(x) \times dx}}\\ &= \int_{{\rm{ - \yen }}}^{\rm{0}} {\left( {{\rm{0}}{\rm{.5}}{{\rm{x}}^{\rm{2}}}} \right)} \left( {{\rm{\lambda \times }}{{\rm{e}}^{{\rm{\lambda x}}}}} \right){\rm{ \times dx + }}\int_{\rm{0}}^{\rm{\yen }} {\left( {{\rm{0}}{\rm{.5}}{{\rm{x}}^{\rm{2}}}} \right)} \left( {{\rm{\lambda \times }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right){\rm{ \times dx}}\end{aligned}\)

If we replace \({\rm{x}}\) by \({\rm{ - t}}\) in the first integral then it is equal to second integral. Hence

\(\begin{aligned} &= 2\int_{\text{0}}^{\text{¥}} {\left( {{\text{0}}{\text{.5}}{{\text{x}}^{\text{2}}}} \right)} \left( {{\text{λ x }}{{\text{e}}^{{\text{ - λ x}}}}} \right){\text{ x dx}} \\ &= \int_{\text{0}}^{\text{¥}} {{{\text{x}}^{\text{2}}}} \left( {{\text{λ x }}{{\text{e}}^{{\text{ - λ x}}}}} \right){\text{ x dx}} \\ &= \left[ {\left( {{{\text{x}}^{\text{2}}}} \right)\left( {{\text{ - }}{{\text{e}}^{{\text{ - λ x}}}}} \right)} \right]_{\text{0}}^{\text{¥}}{\text{ - }}\int_{\text{0}}^{\text{¥}} {{\text{(2x)}}} \left( {{\text{ - }}{{\text{e}}^{{\text{ - λ x}}}}} \right) \\ &= 0 + 2\left( {\int_{\text{0}}^{\text{¥}} {{{\text{e}}^{{\text{ - λx}}}}} } \right) \\& = 2\left[ {{\text{x}}\left( {{\text{ - }}{{\text{e}}^{{\text{ -λ x}}}}} \right)} \right]_{\text{0}}^{\text{¥}}{\text{ - 2}}\int_{\text{0}}^{\text{¥}} {\left( {{\text{ - }}{{\text{e}}^{{\text{ - λ x}}}}} \right)} \\ & = 0 - 2{\left[ {{\text{0 - }}\frac{{\text{1}}}{{\text{λ }}}} \right]^{{\text{(use partial fraction) }}}} \\ {\text{E}}\left( {{{\text{x}}^{\text{2}}}} \right) &= \frac{{\text{2}}}{{\text{λ}}} \\ \end{aligned} \)

(Use partial fraction again)

Hence the standard deviation can be given as:

\(\begin{aligned}\sigma &= \sqrt {{\rm{E}}\left( {{{\rm{x}}^{\rm{2}}}} \right){\rm{ - (E(X)}}{{\rm{)}}^{\rm{2}}}} \\&= \sqrt {\frac{{\rm{2}}}{{\rm{\lambda }}}{\rm{ - 0}}} {\rm{\sigma }}\\&= \frac{{\sqrt {\rm{2}} }}{{\rm{\lambda }}}\end{aligned}\)

The standard deviation is given to us as\({\rm{40}}{\rm{.9\;km}}\). Hence

\(\begin{aligned}\frac{{\sqrt {\rm{2}} }}{{\rm{\lambda }}} &= 40{\rm{.9\lambda }}\\&= \frac{{\sqrt {\rm{2}} }}{{{\rm{40}}{\rm{.9}}}}{\rm{\lambda }}\\ &= 0 {\rm{.0348}}\end{aligned}\)

(b)

The probability that the extent of daily sea-ice change is within\({\rm{1}}\)standard deviation of the mean value is given as\({\rm{P(X£ 40}}{\rm{.9)}}\). Hence

\(\begin{aligned}{\text{P(X£40}}\text.9) &= \int_{\text{0}}^{{\text{40}}{\text{.9}}} {\text{f}} {\text{(x) × dx}} \\ \ &= \int_{\text{0}}^{{\text{40}}{\text{.9}}} {\text{0}} {\text{.5 λ× }}{{\text{e}}^{{\text{ - λ x}}}}{\text{ × dx}} \\ &= 0{\text{.5 λ}}\left[ {\frac{{{\text{ - }}{{\text{e}}^{{\text{ - λ x}}}}}}{{\text{ λ}}}} \right]_{\text{0}}^{{\text{40}}{\text{.9}}} \\ &= - 0{\text{.5}}\left[ {{{\text{e}}^{{\text{ - λ (40}}{\text{.9)}}}}{\text{ - 1}}} \right] \\ & = 0{\text{.5}}\left[ {{\text{1 - }}{{\text{e}}^{{\text{ - (0}}{\text{.0348)(40}}{\text{.9)}}}}} \right]{\text{P(X£40}}{\text{.9)}} \\ &= 0{\text{.3784}} \\ \end{aligned} \)