Hint
The rate and time use matching units. Use the zero-count Poisson probability.
Worked solution
(a)
\(\displaystyle e^{-\frac{5}{2}}=0.0820849986238988\)
Answer: \(0.0820849986238988\).
Answer: \(\approx 0.0820849986238988\)
(b)
\(\displaystyle 1- \sum_{j= 0}^{3} \frac{e^{- \lambda }\lambda^{j}}{j!}= 0.242423866866934\)
Answer: \(0.242423866866934\).
Answer: \(\approx 0.242423866866934\)
(c)
\(\displaystyle 3\lambda=\frac{15}{2}\)
Answer: \(\displaystyle \frac{15}{2}\).
Answer: \(\displaystyle \frac{15}{2}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Use the zero-count Poisson probability.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not leave the mean unchanged when the time interval changes.
(b) · 2 marks
- 1 mark: Subtract all excluded lower counts.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: At least r is \(\displaystyle 1-P(X\le r-1)\).
(c) · 2 marks
- 1 mark: Scale the rate by three hours.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not leave the mean unchanged when the time interval changes.
Total: 6 marks.