Date | May 2008 | Marks available | 7 | Reference code | 08M.2.hl.TZ2.11 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 11 | Adapted from | N/A |
Question
The distance travelled by students to attend Gauss College is modelled by a normal distribution with mean 6 km and standard deviation 1.5 km.
(i) Find the probability that the distance travelled to Gauss College by a randomly selected student is between 4.8 km and 7.5 km.
(ii) 15 % of students travel less than d km to attend Gauss College. Find the value of d.
At Euler College, the distance travelled by students to attend their school is modelled by a normal distribution with mean μ km and standard deviation σ km.
If 10 % of students travel more than 8 km and 5 % of students travel less than 2 km, find the value of μ and of σ .
The number of telephone calls, T, received by Euler College each minute can be modelled by a Poisson distribution with a mean of 3.5.
(i) Find the probability that at least three telephone calls are received by Euler College in each of two successive one-minute intervals.
(ii) Find the probability that Euler College receives 15 telephone calls during a randomly selected five-minute interval.
Markscheme
(i) P(4.8<X<7.5)=P(−0.8<Z<1) (M1)
= 0.629 A1 N2
Note: Accept P(4.8⩽ .
(ii) Stating {\text{P}}(X < d) = 0.15 or sketching an appropriately labelled diagram. A1
\frac{{d - 6}}{{1.5}} = - 1.0364… (M1)(A1)
d = (−1.0364...)(1.5) + 6 (M1)
= 4.45 (km) A1 N4
[7 marks]
Stating both {\text{P}}(X > 8) = 0.1 and {\text{P}}(X < 2) = 0.05 or sketching an appropriately labelled diagram. R1
Setting up two equations in \mu and \sigma (M1)
8 = \mu + (1.281…)\sigma and 2 = \mu − (1.644…)\sigma A1
Attempting to solve for \mu and \sigma (including by graphical means) (M1)
\sigma = 2.05 (km) and \mu = 5.37 (km) A1A1 N4
Note: Accept \mu = 5.36, 5.38 .
[6 marks]
(i) Use of the Poisson distribution in an inequality. M1
{\text{P}}(T \geqslant 3) = 1 - {\text{P}}(T \leqslant 2) (A1)
= 0.679... A1
Required probability is {(0.679…)^2} = 0.461 M1A1 N3
Note: Allow FT for their value of {\text{P}}(T \geqslant 3) .
(ii) \tau \sim {\text{Po(17.5)}} A1
{\text{P}}(\tau = 15) = \frac{{{{\text{e}}^{ - 17.5}}{{(17.5)}^{15}}}}{{15!}} (M1)
= 0.0849 A1 N2
[8 marks]
Examiners report
This question was generally well done despite a large proportion of candidates being awarded an accuracy penalty. Candidates found part (a) (i) to be quite straightforward and was generally done very well. In part (a) (ii), a number of candidates used \frac{{d - 6}}{{1.5}} = 1.0364… instead of \frac{{d - 6}}{{1.5}} = - 1.0364… . In part (b), a pleasingly high number of candidates were able to set up and solve a pair of simultaneous linear equations to correctly find the values of \mu and \sigma . Some candidates prematurely rounded intermediate results. In part (c), a number of candidates were unable to express a correct Poisson inequality. Common errors included stating {\text{P}}(T \geqslant 3) = 1 - {\text{P}}(T \leqslant 3) and using \mu = 7.
This question was generally well done despite a large proportion of candidates being awarded an accuracy penalty. Candidates found part (a) (i) to be quite straightforward and was generally done very well. In part (a) (ii), a number of candidates used \frac{{d - 6}}{{1.5}} = 1.0364… instead of \frac{{d - 6}}{{1.5}} = - 1.0364… . In part (b), a pleasingly high number of candidates were able to set up and solve a pair of simultaneous linear equations to correctly find the values of \mu and \sigma . Some candidates prematurely rounded intermediate results. In part (c), a number of candidates were unable to express a correct Poisson inequality. Common errors included stating {\text{P}}(T \geqslant 3) = 1 - {\text{P}}(T \leqslant 3) and using \mu = 7.
This question was generally well done despite a large proportion of candidates being awarded an accuracy penalty. Candidates found part (a) (i) to be quite straightforward and was generally done very well. In part (a) (ii), a number of candidates used \frac{{d - 6}}{{1.5}} = 1.0364… instead of \frac{{d - 6}}{{1.5}} = - 1.0364… . In part (b), a pleasingly high number of candidates were able to set up and solve a pair of simultaneous linear equations to correctly find the values of \mu and \sigma . Some candidates prematurely rounded intermediate results. In part (c), a number of candidates were unable to express a correct Poisson inequality. Common errors included stating {\text{P}}(T \geqslant 3) = 1 - {\text{P}}(T \leqslant 3) and using \mu = 7.