Chapter 5.4, Problem 1AC

1.

To determine

The theoretical values for the number of hits using Poisson distribution and to complete the given table.

Answer to Problem 1AC

The theoretical value for 0 hit is 227.5 regions.

The theoretical value for 1 hit is 211 regions.

The theoretical value for 2 hits is 97.9 regions.

The theoretical value for 3 hits is 30.4 regions.

The theoretical value for 4 hits is 7.1 regions.

The theoretical value for 5 hits is 1.3 regions.

The completed table is given below:

Hits	0	1	2	3	4	5
Regions	227.5	211	97.9	30.4	7.1	1.3

Explanation of Solution

Given info:

The table shows the number of bombs for 576 regions.

Hits	0	1	2	3	4	5
Regions	229	211	93	35	7	1

The theoretical values for the number of hits are:

Hits	0	1	2	3	4	5
Regions				30.4

The mean number of hits is given to be 0.929.

Calculation:

For 0 hits, the theoretical value is,

$\begin{array}{c}P\left(0\right)=\frac{e^{-\lambda }{\lambda }^X}{X!}\\ =\frac{{\left(2.7183\right)}^{-0.929}{\left(0.929\right)}^0}{0!}\\ =0.394\end{array}$

Thus, $227.5\left(=0.395\times 576\right)$ regions had no hits.

For 1 hit, the theoretical value is,

$\begin{array}{c}P\left(1\right)=\frac{e^{-\lambda }{\lambda }^X}{X!}\\ =\frac{{\left(2.7183\right)}^{-0.929}{\left(0.929\right)}^1}{1!}\\ =0.394\left(0.929\right)\\ =0.366\end{array}$

Thus, $211\left(=0.366\times 576\right)$ regions had 1 hit.

For 2 hits, the theoretical value is,

$\begin{array}{c}P\left(2\right)=\frac{e^{-\lambda }{\lambda }^X}{X!}\\ =\frac{{\left(2.7183\right)}^{-0.929}{\left(0.929\right)}^2}{2!}\\ =\frac{0.394\left(0.863\right)}{2}\\ =\frac{0.34}{2}\end{array}$

         =0.17

Thus, $97.9\left(=0.17\times 576\right)$ regions had 2 hits.

For 3 hits, the theoretical value is,

$\begin{array}{c}P\left(3\right)=\frac{e^{-\lambda }{\lambda }^X}{X!}\\ =\frac{{\left(2.7183\right)}^{-0.929}{\left(0.929\right)}^3}{3!}\\ =\frac{0.394\left(0.802\right)}{6}\\ =\frac{0.316}{6}\end{array}$

         =0.053

Thus, $30.4\left(=0.0527\times 576\right)$ regions had 3 hits.

For 4 hits, the theoretical value is,

$\begin{array}{c}P\left(4\right)=\frac{e^{-\lambda }{\lambda }^X}{X!}\\ =\frac{{\left(2.7183\right)}^{-0.929}{\left(0.929\right)}^4}{4!}\\ =\frac{0.394\left(0.745\right)}{24}\\ =\frac{0.294}{24}\end{array}$

         =0.0123

Thus, $7.1\left(=0.0123\times 576\right)$ regions had 4 hits.

For 5 hits, the theoretical value is,

$\begin{array}{c}P\left(5\right)=\frac{e^{-\lambda }{\lambda }^X}{X!}\\ =\frac{{\left(2.7183\right)}^{-0.929}{\left(0.929\right)}^5}{5!}\\ =\frac{0.394\left(0.692\right)}{120}\\ =\frac{0.273}{120}\end{array}$

         =0.0023

Thus, $1.3\left(=0.0023\times 576\right)$ regions had 5 hits.

2.

To determine

To write: A brief statement comparing the two distributions.

Explanation of Solution

Statement:

The actual values are very close to the theoretical values. Hence Poisson distribution fits the data well.

3.

To conclude: Whether the rockets are fired at random.

Conclusion:

Yes, the rockets were fired at random because the theoretical values and the actual values are closer.