Chapter 5.4, Problem 1AC

Rockets and Targets

During the latter days of World War II, the Germans developed flying rocket bombs. These bombs were used to attack London. Allied military intelligence didn’t know whether these bombs were fired at random or had a sophisticated aiming device. To determine the answer, they used the Poisson distribution.

To assess the accuracy of these bombs, London was divided into 576 square regions. Each region was $\frac{1}{4}$ square kilometer in area. They then compared the number of actual hits with the theoretical number of hits by using the Poisson distribution. If the values in both distributions were close, then they would conclude that the rockets were fired at random. The actual distribution is as follows:

$\begin{array}{l}\text{Hits}\text{0}\text{1}\text{2 }\text{3}\text{4}\text{5}\\ \text{Regions}\text{229}\text{211}\text{93}\text{35}\text{7}\text{1}\end{array}$

Using the Poisson distribution, find the theoretical values for each number of hits. In this case, the number of bombs was 535, and the number of regions was 576. So

$\lambda =\frac{535}{576}\approx 0.929$

For 3 hits,

$P\left(X\right)=\frac{{\left(2.7183\right)}^{-0.929}{\left(0.929\right)}^3}{3}$

Hence, the number of hits is (0.0528)(576) = 30.4128.

Complete the table for the other number of hits.

$\begin{array}{l}\text{Hints}\text{0}\text{1}\text{2}\text{3}\text{4}\text{5 }\\ \text{Regions}\text{30}\text{.4}\end{array}$

2. Write a brief statement comparing the two distributions.

3. Based on your answer to question 2, can you conclude that the rockets were fired at random?

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.

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.