Question

Peter is running a bike rental shop in Science Park. The rental fee of the bike must be paid by credit card and
the fee is calculated as $0.5 per minute with an additional service charge of $10. According to his observation,
customers would rent the bike on the average of 75 minutes with standard deviation of 15 minutes. Assuming
that the number of minutes a customer rents the bike follows a normal distribution.
(a) What proportion of customers would rent the bike for more than 2 hours?
(b) There are 9% customers would rent the bike for less than K minutes. What is the value of K?
(c) What are the (i) mean, (ii) median, (iii) variance, and (iv) standard deviation of the rental fee paid by
a customer?
(d) Suppose (L1, L2) indicates the 90% symmetric range around the mean rental fee paid by a customer.
What are the values of L1 and L2 respectively?

Answer 1

a) P (x >120) =$1-p(z<\tfrac {120-75} {15})=1-P(z<3)=1-0.9987 =0.0013$

b)  P(x<k) = $p(z< \frac {k−75} {15} ​ )=0.09,therefore,z=ϕ(0.09)=−1.341$

$\frac {k−75}{15} =−1.341,therefore, k=75+15∗(−1.341)= 54.885 minutes$

c) mean = median = $0.5*75 +10 =47.5$

standard deviation =0.5*15 = 7.5

Variance =  $7.5^2 =56.25$

d) Margin of error = z* SD = 1.645*7.5= 12.34

L1 = $47.50-12.34=35$.16

L2 = $47.50 + 12.34 = 59.84$