In Exercises 9–16, use the Poisson distribution to find the indicated probabilities.

World War II Bombs In Exercise 1 “Notation” we noted that in analyzing hits by V-1 buzz bombs in World War II, South London was partitioned into 576 regions, each with an area of 0.25 km2 . A total of 535 bombs hit the combined area of 576 regions.

a. Find the probability that a randomly selected region had exactly 2 hits.

b. Among the 576 regions, find the expected number of regions with exactly 2 hits.

c. How does the result from part (b) compare to this actual result: There were 93 regions that had exactly 2 hits?

Short Answer

Expert verified

a.The probability of exactly 2 bomb hits per region is equal to 0.170.

b.The expected number of regions corresponding to 2 bomb hits is equal to 97.9.

c.The actual number of regions is approximately equal to the expected number of regions corresponding to 2 bomb hits.

Step by step solution

01

Given information

It is given that there was a total of 535 bomb hits in the area of South London with 576 regions.

02

Mean

The total number of bomb hits is given to be equal to 535.

The total number of regions is equal to 576.

The mean number of bomb hits per region is equal to:

\(\begin{aligned}{c}\mu = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{Bomb}}\;{\rm{Hits}}}}{{{\rm{Number}}\;{\rm{of}}\;{\rm{Regions}}}}\\ = \frac{{535}}{{576}}\\ = 0.929\end{aligned}\)

Thus, the mean number of bomb hits per region is equal to 0.929.

03

Probability

a.

Let X be the number of bomb hits per region.

Here, X follows a Poisson distribution with mean equal to\({\kern 1pt} \mu = 0.929\).

The probability of exactly 2 bomb hits per region is computed below:

\(\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 2 \right) = \frac{{{{\left( {0.929} \right)}^2}{{\left( {2.71828} \right)}^{ - 0.929}}}}{{2!}}\\ = 0.170\end{aligned}\)

Therefore, the probability of exactly 2 bomb hits per region is equal to 0.170.

04

Expected number of regions

b.

The expected number of regions corresponding to 2 bomb hits is computed below:

\(\begin{aligned}{c}{\rm{Expected}}\;{\rm{Frequency}} = 576 \times P\left( 2 \right)\\ = 576 \times 0.170\\ = 97.92\\ \approx 97.9\end{aligned}\)

The expected number of regions corresponding to 2 bomb hits is equal to 97.9.

05

Comparison

c.

The actual number of regions with 2 bomb hits is equal to 93.

The expected number of regions with 2 bomb hits is equal to 97.9.

The actual number of regions is approximately equal to the expected number of regions corresponding to 2 bomb hits.

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