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 \(\mu \) km and standard deviation \(\sigma \) km.

If 10 % of students travel more than 8 km and 5 % of students travel less than 2 km, find the value of \(\mu \) and of \(\sigma \) .

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.

(i)     \({\text{P}}(4.8 < X < 7.5) = {\text{P}}( - 0.8 < Z < 1)\)     (M1)

= 0.629     A1     N2

Note: Accept \({\text{P}}(4.8 \leqslant X \leqslant 7.5) = {\text{P}}( - 0.8 \leqslant Z \leqslant 1)\) .

(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]

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 \(\mu = 7\).