Chapter 5.3, Problem 15E

To determine

The probability of the given scenario.

Answer to Problem 15E

Solution:

The value of the required probability is 0.258760

Explanation of Solution

Given Information:

Assume a random variable X follows Poisson distribution with λ = 9.60.

Formula used:

For a Poisson random variable X, the probability of getting successes in a given interval is given by:

$P\left(X=x\right)=\frac{e^{-\lambda }{\lambda }^x}{x!}$

$\begin{array}{rll}P\left(X\ge r\right)& =& 1-P(X<r)\\ & =& 1-(P\left(X=0\right)+P(X& =& 1)+\mathrm{...}+P(X& =& r-1\left)\right)\end{array}$

Where,

X is the number of successes occurred

$\lambda$ is the mean number of successes in a each interval.

Calculation:

Substitute 12 for r and 9.60 for λ in the greater than formula mentioned above,

$\begin{array}{rll}P\left(X\ge 12\right)& =& 1-P\left(X<12\right)\\ & =& 1-P\left(X\le 11\right)\\ & =& P\left(X=0\right)+P(X& =& 1)+P(X& =& 2)+\mathrm{............}+P(X& =& 6)+P(X& =& 7)+\mathrm{..........}+P(X& =& 11)\end{array}$

Substitute 0 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=0\right)$.

$\begin{array}{rll}P(X=0)& =& \frac{e^{-9.60}{9.60}^0}{0!}\\ & =& 0.000067729\end{array}$

Substitute 1 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=1\right)$.

$\begin{array}{rll}P(X=1)& =& \frac{e^{-9.60}9.60}{1}\\ & =& 0.00065019\end{array}$

Substitute 2 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=2\right)$

$\begin{array}{rll}P(X=2)& =& \frac{e^{-9.60}{9.60}^2}{2!}\\ & =& \frac{0.00006772\times 92.16}{2}\\ & =& 0.0031209\end{array}$

Substitute 3 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=3\right)$

$\begin{array}{rll}P(X=3)& =& \frac{e^{-9.60}{9.60}^3}{3!}\\ & =& \frac{0.00006772\times {9.60}^3}{6}\\ & =& 0.009987\end{array}$

Substitute 4 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=4\right)$

$\begin{array}{rll}P(X=4)& =& \frac{e^{-9.60}{9.60}^4}{4!}\\ & =& \frac{0.00006772\times {9.60}^4}{24}\\ & =& 0.0239688\end{array}$

Substitute 5 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=5\right)$

$\begin{array}{rll}P(X=5)& =& \frac{e^{-9.60}{9.60}^5}{5!}\\ & =& \frac{0.00006772\times {9.60}^5}{120}\\ & =& 0.046020\end{array}$

Substitute 6 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=6\right)$

$\begin{array}{rll}P(X=6)& =& \frac{e^{-9.60}{9.60}^6}{6!}\\ & =& \frac{0.00006772\times {9.60}^6}{720}\\ & =& 0.073632\end{array}$

Substitute 7 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=7\right)$

$\begin{array}{rll}P(X=7)& =& \frac{e^{-9.60}{9.60}^7}{7!}\\ & =& \frac{0.00006772\times {9.60}^7}{5040}\\ & =& 0.1009813\end{array}$

Substitute 8 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=8\right)$

$\begin{array}{rll}P(X=8)& =& \frac{e^{-9.60}{9.60}^8}{8!}\\ & =& \frac{0.00006772\times {9.60}^8}{8!}\\ & =& 0.121177\end{array}$

Substitute 9 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=9\right)$

$\begin{array}{rll}P(X=9)& =& \frac{e^{-9.60}{9.60}^9}{9!}\\ & =& \frac{0.00006772\times {9.60}^9}{9!}\\ & =& 0.129256\end{array}$

Substitute 10 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=10\right)$

$\begin{array}{rll}P(X=10)& =& \frac{e^{-9.60}{9.60}^{10}}{10!}\\ & =& \frac{0.00006772\times {9.60}^{10}}{10!}\\ & =& 0.124085\end{array}$

Substitute 11 for $r$, and 9.60 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=11\right)$

$\begin{array}{rll}P(X=11)& =& \frac{e^{-9.60}{9.60}^{11}}{11!}\\ & =& \frac{0.00006772\times {9.60}^{11}}{11!}\\ & =& 0.10829\end{array}$

Add all the values of

$P\left(X=0\right),P\left(X=1\right)\mathrm{....},P\left(X=11\right)$ then,

$\begin{array}{rll}P(X\ge 12)& =& 1-P(X<12)\\ & =& 1-P(X\le 11)\\ & =& P(X=0)+P(X& =& 1)+P(X& =& 2)+\mathrm{............}+P(X& =& 6)+P(X& =& 7)+\mathrm{..........}+P(X& =& 11)\\ & =& 0.2588\end{array}$