The numbers of printing mistakes found on 100 pages of a book are given below.
| Number of mistakes on a page | 0 | 1 | 2 | 3 |
| Number of pages | 61 | 29 | 9 | 1 |
A Poisson distribution with the same mean is fitted to these data. The expected number of pages with exactly one mistake is approximately (Given \(e^{-0.5} = 0.6065\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 30
The mean of the observed data is the mean of the fitted Poisson distribution:
\( \displaystyle m = \frac{0 \times 61 + 1 \times 29 + 2 \times 9 + 3 \times 1}{100} \)
\( \displaystyle = \frac{50}{100} = 0.5 \)
For a Poisson variate with mean m,
\( \displaystyle P(X = 1) = \frac{e^{-m}\,m^{1}}{1!} = 0.6065 \times 0.5 = 0.30325 \)
The expected number of pages is 100 times this probability:
\(100 \times 0.30325 = 30.3 \approx 30\)
Why the other options are incorrect
Option (1) 61 — Incorrect. This is the expected number of pages with no mistake, \(100 \times 0.6065 \approx 61\).
Option (3) 29 — Incorrect. This is the observed number of pages with one mistake, not the number given by the fitted distribution.
Option (4) 8 — Incorrect. This is the expected number of pages with two mistakes, \(100 \times 0.6065 \times \frac{0.25}{2} \approx 8\).