A speed camera on Peterson Road records the speed of each passing vehicle. The speeds are found to be normally distributed with a mean of $67\,{\text{km}}\,{{\text{h}}^{ - 1}}$ and a standard deviation of $3.4\,{\text{km}}\,{{\text{h}}^{ - 1}}$.

Sketch a diagram of this normal distribution and shade the region representing the probability that the speed of a vehicle is between $60$ and $70\,{\text{km}}\,{{\text{h}}^{ - 1}}$.

A vehicle on Peterson Road is chosen at random.

Find the probability that the speed of this vehicle is

(i)      more than $60\,{\text{km}}\,{{\text{h}}^{ - 1}}$;

(ii)     less than $70\,{\text{km}}\,{{\text{h}}^{ - 1}}$;

(iii)    between $60$ and $70\,{\text{km}}\,{{\text{h}}^{ - 1}}$.

It is found that $19\,\%$ of the vehicles are exceeding the speed limit of $s\,{\text{km}}\,{{\text{h}}^{ - 1}}$.

Find the value of $s$ , correct to the nearest integer.

There is a fine of ${\text{US}}\$ 65$ for exceeding the speed limit on Peterson Road. On a particular day the total value of fines issued was ${\text{US}}\$ 14\,820$.

(i)     Calculate the number of fines that were issued on this day.

(ii)    Estimate the total number of vehicles that passed the speed camera on Peterson Road on this day.

(A1)(A1)

Note: Award (A1) for normal curve with mean of $67$ indicated or two vertical lines drawn approximately in correct place. Award (A1) for correct shaded region (between the vertical lines.).

(i)      $0.980\,\,\,(0.980244...,\,\,98.0\,\% )$         (G1)

(ii)     $0.811\,\,\,(0.811207...,\,\,81.1\,\% )$         (G1)

(iii)    $0.791\,\,\,(0.791451...,\,\,79.1\,\% )$         (G1)

${\text{P}}\,\left( {S > s} \right) = 19\,\% \,\,(0.19)$ OR ${\text{P}}\,\left( {S > s} \right) = 81\,\% \,\,(0.81)$          (M1)

OR

(M1)

Note: Award (M1) for the correct probability equation OR for a correct region indicated on labelled diagram.

$(s = )\,\,70.0\,\,\,(69.9848...)$           (A1)(G2)

Note: Award (M1) for any correct method.

(i)     $\frac{{14\,820}}{{65}}$         (M1)

Note: Award (M1) for dividing $14\,820$ by $65$.

$= 228$         (A1)(G2)

(ii)    $\frac{{{\text{their 228}}}}{{0.19}}$ (or equivalent)          (M1)

$= 1200$ (vehicles)         (A1)(ft)(G2)

Note: Award (M1) for correct method. Follow through from their part (d)(i).

Question 3: The normal distribution
Candidates showed comprehensive understanding of the normal distribution. The graphic display calculator was used efficiently by most of the candidates. There was much variability in the ability to sketch the curve in part (a). Instead of drawing the straight-forward sketch with the mean line and two vertical lines as required at 60 and 70, many linked it to standard deviations. It was very rare to see any method in part (c). Most candidates managed part (d)(i) but few went on to complete part (d)(ii).

Question 3: The normal distribution
Candidates showed comprehensive understanding of the normal distribution. The graphic display calculator was used efficiently by most of the candidates. There was much variability in the ability to sketch the curve in part (a). Instead of drawing the straight-forward sketch with the mean line and two vertical lines as required at 60 and 70, many linked it to standard deviations. It was very rare to see any method in part (c). Most candidates managed part (d)(i) but few went on to complete part (d)(ii).

Question 3: The normal distribution
Candidates showed comprehensive understanding of the normal distribution. The graphic display calculator was used efficiently by most of the candidates. There was much variability in the ability to sketch the curve in part (a). Instead of drawing the straight-forward sketch with the mean line and two vertical lines as required at 60 and 70, many linked it to standard deviations. It was very rare to see any method in part (c). Most candidates managed part (d)(i) but few went on to complete part (d)(ii).

Question 3: The normal distribution
Candidates showed comprehensive understanding of the normal distribution. The graphic display calculator was used efficiently by most of the candidates. There was much variability in the ability to sketch the curve in part (a). Instead of drawing the straight-forward sketch with the mean line and two vertical lines as required at 60 and 70, many linked it to standard deviations. It was very rare to see any method in part (c). Most candidates managed part (d)(i) but few went on to complete part (d)(ii).