| Date | May 2016 | Marks available | 2 | Reference code | 16M.2.hl.TZ2.6 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Question number | 6 | Adapted from | N/A |
Question
A company produces rectangular sheets of glass of area 5 square metres. During manufacturing these glass sheets flaws occur at the rate of 0.5 per 5 square metres. It is assumed that the number of flaws per glass sheet follows a Poisson distribution.
Glass sheets with no flaws earn a profit of $5. Glass sheets with at least one flaw incur a loss of $3.
This company also produces larger glass sheets of area 20 square metres. The rate of occurrence of flaws remains at 0.5 per 5 square metres.
A larger glass sheet is chosen at random.
Find the probability that a randomly chosen glass sheet contains at least one flaw.
Find the expected profit, \(P\) dollars, per glass sheet.
Find the probability that it contains no flaws.
Markscheme
\({\text{X}} \sim {\text{Po}}(0.5)\) (A1)
\({\text{P}}(X \geqslant 1) = 0.393{\text{ }}( = 1 - {{\text{e}}^{ - 0.5}})\) (M1)A1
[3 marks]
\({\text{P}}(X = 0) = 0.607 \ldots \) (A1)
\({\text{E}}(P) = (0.607 \ldots \times 5) - (0.393 \ldots \times 3)\) (M1)
the expected profit is $1.85 per glass sheet A1
[3 marks]
\(Y \sim {\text{Po}}(2)\) (M1)
\({\text{P}}(Y = 0) = 0.135{\text{ }}( = {{\text{e}}^{ - 2}})\) A1
[2 marks]
Examiners report
Part (a) was reasonably well done. Some candidates calculated \({\text{P}}(X = 1)\).
Part (b) was not as well done as expected with a surprising number of candidates calculating \(5{\text{P}}(X = 0) + 3{\text{P}}(X \geqslant 1)\) rather than \(5{\text{P}}(X = 0) - 3{\text{P}}(X \geqslant 1)\).
Part (c) was very well done.
