91影视

A discrete random variable is one that can only take on a finite number of different values.

Given: the average period between load incidences is \({\rm{0}}{\rm{.5}}\) year.

(a) We may then expect one cargo every six months.

\(\begin{aligned} \lambda &= \mu \\ &= 1 per half year \\ & = 1t \end{aligned}\)

Because a two-year period is divided into four half-years, we may expect four loads:

\(\begin{aligned} lambda &= \mu \\ &= 1(4) \\ &= 4 \end{aligned}\)

Therefore, \({\rm{4}}\) loads are expected.

(b) Poisson probability formula:

\({\rm{P(X = k) = }}\frac{{{{\rm{\lambda }}^{\rm{k}}}{{\rm{e}}^{{\rm{ - \lambda }}}}}}{{{\rm{k!}}}}\)

At \({\rm{k = 0,1,}}...{\rm{,4}}\), evaluate the formula.

\(\begin{array}{c}{\rm{P(X = 0) = }}\frac{{{{\rm{4}}^{\rm{0}}}{{\rm{e}}^{{\rm{ - 4}}}}}}{{{\rm{0!}}}}\\ \approx {\rm{0}}{\rm{.0183}}\\{\rm{P(X = 1) = }}\frac{{{{\rm{4}}^{\rm{1}}}{{\rm{e}}^{{\rm{ - 4}}}}}}{{{\rm{1!}}}}\\ \approx {\rm{0}}{\rm{.0733}}\\{\rm{P(X = 5) = }}\frac{{{{\rm{4}}^{\rm{5}}}{{\rm{e}}^{{\rm{ - 4}}}}}}{{{\rm{5!}}}}\\ \approx {\rm{0}}{\rm{.1563}}\end{array}\)

For discontinuous or mutually exclusive occurrences, use the following addition rule:

\({\rm{P(A or B) = P(A) + P(B)}}\)

Fill in the blanks with the corresponding probabilities:

\(\begin{aligned}{\rm{P(X}} \le 5) &= P(X = 0) + P(X = 1) + \ldots + P(X = 5) \\ &= 0 {\rm{.0183 + 0}}{\rm{.0733 + \ldots + 0}}{\rm{.1563}}\\ &= 0 {\rm{.7851}}\end{aligned}\)

\({\rm{T183/84}}\)-calculator command: poissoncdf \({\rm{(4,5)}}\)

Rule of complements:

\({\rm{P(not A) = 1 - P(A)}}\)

Use the complement rule to help you:

\(\begin{aligned}P(X > 5) &= 1 - P(X \leqslant 5) \\ &= 1 - 0.7851 \\ &= 0.2149 \\ \end{aligned} \)

Therefore, the value is: \({\rm{P(X > 5) = 0}}{\rm{.2149}}\).

(c) Poisson probability formula:

\({\rm{P(X = k) = }}\frac{{{{\rm{\lambda }}^{\rm{k}}}{{\rm{e}}^{{\rm{ - \lambda }}}}}}{{{\rm{k!}}}}\)

At \({\rm{k = 0}}\), evaluate the formula:

\(\begin{aligned}{\rm{P(X = 0) = }}\frac{{{{\rm{\lambda }}^{\rm{0}}}{{\rm{e}}^{{\rm{ - \lambda }}}}}}{{{\rm{0!}}}}\\{\rm{ = }}{{\rm{e}}^{{\rm{ - \lambda }}}}\end{aligned}\)

This probability must be less than \({\rm{0}}{\rm{.1}}\):

\(\begin{aligned}{\rm{P(X = 0) = }}{{\rm{e}}^{{\rm{ - \lambda }}}}\\{\rm{ = 0}}{\rm{.1}}\end{aligned}\)

Calculate each side's logarithm:

\(\begin{array}{c}{\rm{ - \lambda = ln}}{{\rm{e}}^{{\rm{ - \lambda }}}}\\{\rm{ = ln0}}{\rm{.1}}\end{array}\)

Divide 鈥�1 from each side:

\(\begin{aligned}{\rm{\lambda = - ln0}}{\rm{.1}}\\ \approx {\rm{2}}{\rm{.3}}\end{aligned}\)

Therefore, the time is approximately \({\rm{2}}{\rm{.3}}\) half years or \({\rm{1}}{\rm{.15}}\) years (or longer).