All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Firstly, the average number of errors per page is constant.
Secondly, the number of errors in a page is independent of the number of errors in another page.

(ii)
Let X denote the number of errors per page.
X \sim \mathrm{Po}(1.3)

X_1 \ldots X_6 \sim \mathrm{Po}(7.8)

\mathrm{P}(X_1 \ldots X_6 > 10)

= 1 - \mathrm{P}(X_1 \ldots X_6 \le 10)

\approx 0.165 (3 SF)

(iii)
X_1 \ldots X_n \sim \mathrm{Po}(1.3n)

\mathrm{P}(X_1 \ldots X_n ~ \textless ~ 2) ~ \textless ~ 0.05

e^{-1.3n} + e^{-1.3n}(1.3n) ~ \textless ~ 0.05

e^{-1.3n} (1  + 1.3n) ~ \textless ~ 0.05

Using GC, n > 3.6491

Least n = 4

Leave a Reply