Revision as of 16:12, 13 May 2023 by Admin (Created page with "'''Key: B''' Let n be the number of observations. For full credibility, <math display = "block"> \begin{aligned} n = \left(\frac{1.96}{0.01} \right)^2 \frac{mq(1-q)}{(mq)^2}...")
Exercise
ABy Admin
May 13'23
Answer
Key: B
Let n be the number of observations. For full credibility,
[[math]]
\begin{aligned}
n = \left(\frac{1.96}{0.01} \right)^2 \frac{mq(1-q)}{(mq)^2} = 38,416 \frac{1-q}{mq}.
\end{aligned}
[[/math]]
The required expected number of claims is
[[math]]
nmq = 34,574 = 38, 416 \frac{1-q}{mq} mq = 38,416(1-q).
[[/math]]
Then q = 1 – 34,574/38,416 = 0.1.