Question

SUCUL IUR. 2. The time that it takes to complete a marathon is modelled by a normal distribution with a mean of 4.547 hours and a standard deviation of 0.634 hours. a) What is the probability that a randomly selected runner will complete the marathon in less than 3 hours and a half? (4 marks)

Answer 1

2.

X : time takes to complete a marathon

X modelled by a normal distribution with mean : 4.547 and standard deviation 0.634 hours

a)

Probability that a randomly selected runner will complete the marathon in less than 3 hours and a half = P(X<3.5)

Z-score of 3.5 = (3.5-mean)/standard deviation = (3.5-4.547)/0.634 = -1.65

From standard normal tables, P(Z<-1.65) = 0.0495

P(X<3.5) = P(Z<-1.65) = 0.0495

Probability that a randomly selected runner will complete the marathon in less than 3 hours and a half = 0.0495

b)

Median finish time in this marathon.

For a Normal distribution Mean = Median

Median finish time in this marathon = mean finish time = 4.547

Median finish time in this marathon = 4.547

c)

Number of people participate in 2020 Marathon in Ottawa = 20000

Let Longest finish time for the top 50 runners be : X₅₀

Proportion of 50 runners in 20000 = 50/20000=0.0025

i.e

P(X<X₅₀) = 0.0025

Z₅₀ : Z-score for X₅₀ = (X₅₀ - 4.547)/0.634

X₅₀ = 4.547 + 0.634Z₅₀

P(Z<Z₅₀) = P(X<X₅₀) = 0.0025

From standard normal tables,

P(Z<-2.81) = 0.0024

Z₅₀ = -2.81

X₅₀ = 4.547 + 0.634Z₅₀ = 4.547 - 0.634 x (-2.81) = 2.76546

The longest finish time for the top 50 runners = 2.76546 hours.