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Answer

Prerequisites​

  • Poisson Probability Calculation
  • Expected Value of Frequencies

Step-by-Step Derivation​

  1. From part (c), we have the ML estimate λ^≈0.9288\hat{\lambda} \approx 0.9288 and the total number of cells N=576N = 576.

  2. To find the expected number of cells with exactly kk hits, we use the formula: Expected Cells=N×p(x=k∣λ^)=576×1k!e−λ^λ^k\text{Expected Cells} = N \times p(x = k | \hat{\lambda}) = 576 \times \frac{1}{k!}e^{-\hat{\lambda}}\hat{\lambda}^k

  3. Calculate the expected probability and expected count for each kk:

    • For k=0k = 0: p(x=0∣0.9288)=10!e−0.9288(0.9288)0=e−0.9288≈0.3950p(x = 0 | 0.9288) = \frac{1}{0!}e^{-0.9288}(0.9288)^0 = e^{-0.9288} \approx 0.3950 Expected Cells=576×0.3950≈227.5\text{Expected Cells} = 576 \times 0.3950 \approx 227.5

    • For k=1k = 1: p(x=1∣0.9288)=11!e−0.9288(0.9288)1=0.9288×e−0.9288≈0.3669p(x = 1 | 0.9288) = \frac{1}{1!}e^{-0.9288}(0.9288)^1 = 0.9288 \times e^{-0.9288} \approx 0.3669 Expected Cells=576×0.3669≈211.3\text{Expected Cells} = 576 \times 0.3669 \approx 211.3

    • For k=2k = 2: p(x=2∣0.9288)=12!e−0.9288(0.9288)2=0.86272×e−0.9288≈0.1704p(x = 2 | 0.9288) = \frac{1}{2!}e^{-0.9288}(0.9288)^2 = \frac{0.8627}{2} \times e^{-0.9288} \approx 0.1704 Expected Cells=576×0.1704≈98.1\text{Expected Cells} = 576 \times 0.1704 \approx 98.1

    • For k=3k = 3: p(x=3∣0.9288)=13!e−0.9288(0.9288)3=0.80136×e−0.9288≈0.0528p(x = 3 | 0.9288) = \frac{1}{3!}e^{-0.9288}(0.9288)^3 = \frac{0.8013}{6} \times e^{-0.9288} \approx 0.0528 Expected Cells=576×0.0528≈30.4\text{Expected Cells} = 576 \times 0.0528 \approx 30.4

    • For k=4k = 4: p(x=4∣0.9288)=14!e−0.9288(0.9288)4=0.744224×e−0.9288≈0.0123p(x = 4 | 0.9288) = \frac{1}{4!}e^{-0.9288}(0.9288)^4 = \frac{0.7442}{24} \times e^{-0.9288} \approx 0.0123 Expected Cells=576×0.0123≈7.1\text{Expected Cells} = 576 \times 0.0123 \approx 7.1

    • For k≥5k \ge 5 (5 and over): Since the probabilities must sum to 1, we can compute this as 1−∑k=04p(x=k∣λ^)1 - \sum_{k=0}^4 p(x=k|\hat{\lambda}): p(x≥5∣0.9288)=1−(0.3950+0.3669+0.1704+0.0528+0.0123)=1−0.9974=0.0026p(x \ge 5 | 0.9288) = 1 - (0.3950 + 0.3669 + 0.1704 + 0.0528 + 0.0123) = 1 - 0.9974 = 0.0026 Expected Cells=576×0.0026≈1.5\text{Expected Cells} = 576 \times 0.0026 \approx 1.5

  4. Comparison Table:

number of hits (kk)012345 and over
Observed Data229211933571
Expected Counts227.5211.398.130.47.11.5

Conclusion​

The theoretically derived expected counts under the Poisson model fit the actual observed hit data remarkably well. Because exactly this Poisson distribution explicitly models random, independent events happening over area/time, we can conclude that the bomb hits were essentially random and uniform.

The Germans were highly unlikely to be systematically or precisely targeting specific London neighborhoods. The occurrences of "clustered" hits on particular cells are indistinguishable from what one would expect from pure chance.