Date | November 2012 | Marks available | 2 | Reference code | 12N.2.hl.TZ0.11 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 11 | Adapted from | N/A |
Question
The number of visitors that arrive at a museum every minute can be modelled by a Poisson distribution with mean 2.2.
If the museum is open 6 hours daily, find the expected number of visitors in 1 day.
Find the probability that the number of visitors arriving during an hour exceeds 100.
Find the probability that the number of visitors in each of the 6 hours the museum is open exceeds 100.
The ages of the visitors to the museum can be modelled by a normal distribution with mean \(\mu \) and variance \({\sigma ^2}\) . The records show that 29 % of the visitors are under 35 years of age and 23 % are at least 55 years of age.
Find the values of \(\mu \) and \(\sigma \) .
The ages of the visitors to the museum can be modelled by a normal distribution with mean \(\mu \) and variance \({\sigma ^2}\) . The records show that 29 % of the visitors are under 35 years of age and 23 % are at least 55 years of age.
One day, 100 visitors under 35 years of age come to the museum. Estimate the number of visitors under 50 years of age that were at the museum on that day.
Markscheme
\(2.2 \times 6 \times 60 = 792\) (M1)A1
[2 marks]
\(V \sim {\text{Po}}(2.2 \times 60)\) (M1)
\({\text{P}}(V > 100) = 0.998\) (M1)A1
[3 marks]
\({(0.997801...)^6} = 0.987\) (M1)A1
[2 marks]
\(A \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})\)
\({\text{P}}(A < 35) = 0.29{\text{ and P}}(A > 55) = 0.23 \Rightarrow {\text{P}}(A < 55) = 0.77\)
\({\text{P}}\left( {Z < \frac{{35 - \mu }}{\sigma }} \right) = 0.29{\text{ and P}}\left( {Z < \frac{{55 - \mu }}{\sigma }} \right) = 0.77\) (M1)
use of inverse normal (M1)
\(\frac{{35 - \mu }}{\sigma } = - 0.55338...{\text{ and }}\frac{{55 - \mu }}{\sigma } = 0.738846...\) (A1)
solving simultaneously (M1)
\(\mu = 43.564...{\text{ and }}\sigma = 15.477...\) A1A1
\(\mu = 43.6{\text{ and }}\sigma = 15.5{\text{ (3sf)}}\)
[6 marks]
\(0.29n = 100 \Rightarrow n = 344.82...\) (M1)(A1)
\({\text{P}}(A < 50) = 0.66121...\) (A1)
expected number of visitors under 50 = 228 (M1)A1
[5 marks]
Examiners report
This question was generally well done by most candidates. It was evident that candidates had been well prepared in Poisson and normal distribution. In parts (a)-(d) candidates were usually successful and appropriate methods were shown although many candidates used labored algebraic approaches to solving simultaneous equations and wasted time answering part (d). Part (e) was very well answered by a smaller number of candidates but it was obviously more demanding in its level of abstraction.
This question was generally well done by most candidates. It was evident that candidates had been well prepared in Poisson and normal distribution. In parts (a)-(d) candidates were usually successful and appropriate methods were shown although many candidates used labored algebraic approaches to solving simultaneous equations and wasted time answering part (d). Part (e) was very well answered by a smaller number of candidates but it was obviously more demanding in its level of abstraction.
This question was generally well done by most candidates. It was evident that candidates had been well prepared in Poisson and normal distribution. In parts (a)-(d) candidates were usually successful and appropriate methods were shown although many candidates used labored algebraic approaches to solving simultaneous equations and wasted time answering part (d). Part (e) was very well answered by a smaller number of candidates but it was obviously more demanding in its level of abstraction.
This question was generally well done by most candidates. It was evident that candidates had been well prepared in Poisson and normal distribution. In parts (a)-(d) candidates were usually successful and appropriate methods were shown although many candidates used labored algebraic approaches to solving simultaneous equations and wasted time answering part (d). Part (e) was very well answered by a smaller number of candidates but it was obviously more demanding in its level of abstraction.
This question was generally well done by most candidates. It was evident that candidates had been well prepared in Poisson and normal distribution. In parts (a)-(d) candidates were usually successful and appropriate methods were shown although many candidates used labored algebraic approaches to solving simultaneous equations and wasted time answering part (d). Part (e) was very well answered by a smaller number of candidates but it was obviously more demanding in its level of abstraction.