Question

3. The speeds of cars passing a speed camera have approximately a normally distribution with mean 54 km/h and standard deviation 3 km/h. Cars are photographed by the speed camera if they are travelling 57 km/h or faster. a) What percentage of cars would be expected to be travelling between 48 km/h and 60 km/h? b) Almost all the cars are expected to be travelling between km/h and km/h. c) What percentage of cars would NOT be expected to be photographed? d) If 450 cars pass the speed camera in an hour, how many would you expect to be photographed?

Answer 1

Let Xkm/hr denotes the speed of cars.

X~Normal(mean=54,SD=3)

(a) We need to find the percentage of cars that travels between 48 km/h and 60 km/hr. That is we need to find

P(48<X<60)

= P((48-mean)/SD < (X-mean)/SD < (60-mean)/SD)

=P( (48-54)/3 < (X-54)/3 < (60-54)/3)

=P(-2< Z<2)

=P(Z< 2) -P(Z<-2)

=P(Z<2) -1 +P(Z<2)

=2*P(Z<2)-1

=2*0.9772-1 [From normal table]

=0.9544

Ans: 0.9544 or 95.44%.

(b) We know for a normal distribution, 99.7% observations are within ± 3ơ limit. That is 99.7% observations will fall between mean - 3*standard deviation and mean+3*standard deviation.

mean-3*SD =54-3*3=54-9=45

and mean +3*SD=54+3*3=54+9=63

So almost all cars are expected to travel between 45km/h and 63km/h (Ans)

(c) Cars are photographed if they have speed of 57km/hr or more.

So cars are not photographed if they have speed less than 57km/hr.

So we are required to find:

P(X<57)

=P((X-mean)/SD < (57-mean)/SD)

=P(Z<(57-54)/3)

=P(Z<1)

=0.8413 [From normal table)

Hence 0.8413 or 84.13% of the cars are not photographed.

(d) Probability cars are photographed

=P(X≥57)

=1-P(X<57)

=1-0.8413 [From part (c) P(X<57)=0.841]

=0.1587

So among 450 cars, 450*0.1587=71.415=71(approx) cars are expected to be photographed.

Ans : 71 cars.