Chapter 5.3, Problem 16E

To determine

(a)

The specified probability for the given scenario.

Answer to Problem 16E

Solution:

The value of the required probability is 0.099261.

Explanation of Solution

Formula Used:

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

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

$P\left(X<r\right)=P\left(X=0\right)+P\left(X=1\right)+\mathrm{...}+P\left(X=r-1\right)$

Where,

X is the number of successes occurred

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

Calculation:

The shop averages 4 sales per hour, the total sales in 2 hours that is from 1:00 to 3:00 will be,

$\begin{array}{rll}\lambda & =& \frac{4\text{sales}}{1\text{hour}}\times 2\text{hours}\\ & \text{=}& \text{8}\text{sales}\end{array}$

Probability that Betty rings up exactly ten customers will be,

Substitute 10 in x and 8 in λ in the formula of probability distribution function mentioned above,

$\begin{array}{rll}P(X=10)& =& \frac{e^{-8}\times 8^{10}}{10!}\\ & =& 0.099261\end{array}$

Conclusion:

Thus, the probability that Betty rings up exactly ten customers is 0.099261.

To determine

(b)

The specified probability for the given scenario.

Answer to Problem 16E

Solution:

The value of the required probability is 0.184114.

Explanation of Solution

Formula Used:

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

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

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

Where,

X is the number of successes occurred

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

Calculations:

The shop averages 4 sales per hour, the total sales in 2 hours that is from 1:00 to 3:00 will be,

$\begin{array}{rll}\lambda & =& \frac{4\text{sales}}{1\text{hour}}\times 2\text{hours}\\ & \text{=}& \text{8}\text{sales}\end{array}$

Probability that no more than five major trauma patients will be admitted on any given night will be,

Substitute 10 for r and 8 for λ in the greater than formula mentioned above,

Probability that Betty rings up more than ten customers will be,

$\begin{array}{rll}P(X>10)& =& 1-P(X\le 10)\\ & =& 1-(P(X=0)+P(X& =& 1)+P(X& =& 2)+\mathrm{.............}+\text{P}(X& =& 9)+P(X& =& 10))\end{array}$

Substitute 0 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=0\right)$.

$\begin{array}{rll}P(X=0)& =& \frac{e^{-8}{8}^0}{0!}\\ & =& 0.0003354\end{array}$

Substitute 1 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=1\right)$.

$\begin{array}{rll}P\left(X=1\right)& =& \frac{e^{-8}8}{1}\\ & =& 0.0026837\end{array}$

Substitute 2 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=2\right)$

$\begin{array}{rll}P\left(X=2\right)& =& \frac{e^{-8}{8}^2}{2!}\\ & =& 0.010734\end{array}$

Substitute 3 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=3\right)$

$\begin{array}{rll}P\left(X=3\right)& =& \frac{e^{-8}{8}^3}{3!}\\ & =& \frac{0.0003354\times 64}{6}\\ & =& 0.02862\end{array}$

Substitute 4 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=4\right)$

$\begin{array}{rll}P\left(X=4\right)& =& \frac{e^{-8}{8}^4}{4!}\\ & =& \frac{0.0003354\times 256}{24}\\ & =& 0.05725\end{array}$

Substitute 5 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=5\right)$

$\begin{array}{rll}P\left(X=5\right)& =& \frac{e^{-8}{8}^5}{5!}\\ & =& \frac{0.0003354\times 8^5}{120}\\ & =& 0.091603\end{array}$

Substitute 6 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=6\right)$

$\begin{array}{rll}P\left(X=6\right)& =& \frac{e^{-8}{8}^6}{6!}\\ & =& 0.12213\end{array}$

Substitute 7 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=7\right)$

$\begin{array}{rll}P\left(X=7\right)& =& \frac{e^{-8}{8}^7}{7!}\\ & =& 0.13958\end{array}$

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

$\begin{array}{rll}P\left(X=8\right)& =& \frac{e^{-8}{8}^8}{8!}\\ & =& 0.13958\end{array}$

Substitute 9 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=9\right)$

$\begin{array}{rll}P\left(X=9\right)& =& \frac{e^{-8}{8}^9}{9!}\\ & =& 0.12407\end{array}$

Substitute 10 for $r$, and 8 for λ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=10\right)$

$\begin{array}{rll}P\left(X=10\right)& =& \frac{e^{-8}{8}^{10}}{10!}\\ & =& 0.0992615\end{array}$

Adding all these values $P\left(X=0\right),P\left(X=1\right)\mathrm{....}P\left(X=10\right)$, then

$\begin{array}{rll}P(X>10)& =& 1-P(X\le 10)\\ & =& 1-(P(X=0)+P(X=1)+P(X=2)+\mathrm{.............}+\text{P}(X=9)+P(X=10))\\ & =& 1-\left(0.0003354+0.0026837+0.10734+0.02862+0.05725+0091603+0.12213+0.13958+0.13958+0.13958+0.12407+0.0992615\right)\\ & =& 1-0.815886\\ & =& 0.184114\end{array}$

Conclusion:

Thus, the probability that Betty rings up more than ten customers is 0.184114.