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Date May 2008 Marks available 8 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.

[7]
a.

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 .

[6]
b.

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.

[8]
c.

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]

a.

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]

b.

(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]

c.

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.

a.

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.

b.

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.

c.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.6 » Poisson distribution, its mean and variance.
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