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

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.

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.

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.

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.