91影视

Probability is the likelihood that an event will occur and is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails.

(a)

Given:

\(p(x) = k\frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}\)

\(x = 1,2,3, \ldots \)

With a mean of \(\mu \), we have a Poisson distribution.

\(P(X = x) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\)

\(x = 0,1,2,3, \ldots \)

Because we know the Poisson distribution is a valid probability distribution, the sum of these probabilities must equal

\(\sum\limits_{x = 0}^{ + \infty } {\frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}} = 1\)

Replace \(\mu \)by \(\theta \):

\(\sum\limits_{x = 0}^{ + \infty } {\frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}} = 1\)

Remove the term with \(x = 0\)from the equation:

\(\frac{{{e^{ - \theta }}{\theta ^0}}}{{0!}} + \sum\limits_{x = 1}^{ + \infty } {\frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}} = 1\)

Simplify:

\({e^{ - \theta }} + \sum\limits_{x = 1}^{ + \infty } {\frac{{{e^{ - \theta }}{{\bf{\theta }}^x}}}{{x!}}} = 1\)

Subtract \({e^{ - \theta }}\)from each side:

\(\sum\limits_{x = 1}^{ + \infty } {\frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}} = 1 - {e^{ - \theta }}\)

Next, calculate the total probability of the given distribution:

\(\sum\limits_{x = 1}^{ + \infty } k \frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}} = k\sum\limits_{x = 1}^{ + \infty } {\frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}} = k\left( {1 - {e^{ - \theta }}} \right)\)

For the given function \(p(x)\)to be a valid pmf, the probability must be equal to \(1\):

\(k\left( {1 - {e^{ - \theta }}} \right) = 1\)

Divide each side of the equation by \(1 - {e^{ - \theta }}\):

\(k = \frac{1}{{1 - {e^{ - \theta }}}}\)

Hence, the required value of k is\(\frac{1}{{1 - {e^{ - \theta }}}}\).

(b)

Given:

\(p(x) = k\frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}\)

\(x = 1,2,3, \ldots \)

\(\theta = \mu = 2.313035\)

Result part (a):

\(k = \frac{1}{{1 - {e^{ - \theta }}}}\)

At \(k = 1,2,3,4,5\), evaluate the following probability:

\(p(1) = \frac{1}{{1 - {e^{ - 2.313035}}}}\frac{{{e^{ - 2.313035}}{{2.313035}^1}}}{{1!}} \approx 0.254\)

\(p(2) = \frac{1}{{1 - {e^{ - 2.313035}}}}\frac{{{e^{ - 2.010003}}{{2.313035}^2}}}{{2!}} \approx 0.294\)

\(p(3) = \frac{1}{{1 - {e^{ - 2.313035}}}}\frac{{{e^{ - 2.313035}}{{2.313035}^3}}}{{3!}} \approx 0.227\)

\(p(4) = \frac{1}{{1 - {e^{ - 2.313035}}}}\frac{{{e^{ - 2.313035}}{{2.313035}^4}}}{{4!}} \approx 0.131\)

\(p(5) = \frac{1}{{1 - {e^{ - 2.313035}}}}\frac{{{e^{ - 2.313035}}{{2.313035}^5}}}{{5!}} \approx 0.061\)

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

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

Add the corresponding probabilities:

\(P(X \le 5) = p(1) + p(2) + p(3) + p(4) + p(5)\)

\(\begin{array}{c} = 0.254 + 0.294 + 0.227 + 0.131 + 0.061\\ = 0.967\\ = 96.7\% \end{array}\)

Hence, the required value is\(96.7\% \).

(c)

Given:

\(p(x) = k\frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}\)

\(x = 1,2,3, \ldots \)

\(\theta = \mu = 2.313035\)

Result part (a):

\(k = \frac{1}{{1 - {e^{ - \theta }}}}\)

With a mean of \(\mu \), we have a Poisson distribution.

\(P(X = x) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\)

\(x = 0,1,2,3, \ldots \)

We know that the poisson distribution is a valid probability distribution and that the variance of the poisson distribution is \(\mu .\)The variance is the expected value of \({(X - \mu )^2}\)

\(\sum\limits_{x = 0}^{ + \infty } {{{(x - \mu )}^2}} \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}} = \mu \)

Replace \(\mu \)by \(\theta :\)

\(\sum\limits_{x = 0}^{ + \infty } {{{(x - \theta )}^2}} \frac{{{\theta ^x}{e^{ - \theta }}}}{{x!}} = \theta \)

Remove the term with \(x = 0\)from the equation:

\({(0 - \theta )^2}\frac{{{e^{ - \theta }}{\theta ^0}}}{{0!}} + \sum\limits_{x = 1}^{ + \infty } {{{(x - \theta )}^2}} \frac{{{\theta ^x}{e^{ - \theta }}}}{{x!}} = 1\)

Simplify:

\({\theta ^2}{e^{ - \theta }} + \sum\limits_{x = 1}^{ + \infty } {{{(x - \theta )}^2}} \frac{{{\theta ^x}{e^{ - \theta }}}}{{x!}} = 1\)

Subtract \({\theta ^2}{e^{ - \theta }}\)from each side:

\(\sum\limits_{x = 1}^{ + \infty } {{{(x - \theta )}^2}} \frac{{{\theta ^x}{e^{ - \theta }}}}{{x!}} = 1 - {\theta ^2}{e^{ - \theta }}\)

Next, calculate the expected value of \({(X - \mu )^2}\$ :\), which is the variance of the given distribution:

\({\sigma ^2} = {\mathop{\rm Var}\nolimits} (X)\)

\( = \sum\limits_{x = 1}^{ + \infty } {{{(x - \theta )}^2}} \frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}\)

\( = \frac{1}{{1 - {e^{ - \theta }}}}\sum\limits_{x = 1}^{ + \infty } {\frac{{{e^{ - \theta }}{\theta ^x}}}{{x!}}} \)

\( = \frac{1}{{1 - {e^{ - \theta }}}}\left( {1 - {\theta ^2}{e^{ - \theta }}} \right)\)

Replace \(\theta \) by \(2.313035\)and evaluate:

\({\sigma ^2} = \frac{1}{{1 - {e^{ - 2.313035}}}}\left( {1 - {{2.313035}^2}{e^{ - 2.313035}}} \right) \approx 0.522229\)

The square root of the variance is the standard deviation:

\(\sigma = \sqrt {0.522229} \approx 0.7227\)