Answer to Question #98061 in Statistics and Probability for amy

Question

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

Let $X=$ the number of minutes a customer rents the bike in minutes: $X\sim N(\mu,\sigma^2).$

Then

$$Z={X-\mu \over \sigma}\sim N(0,1)$$

Given that $\mu=75\ minutes, \sigma=15\ minutes.$

(a) What proportion of customers would rent the bike for more than 2 hours? 

$$P(X>120)=1-P(X\leq120)=$$ $$=1-P(Z\leq{120-75 \over15})=1-P(Z\leq3)\approx$$ $$\approx1-0.99865010\approx0.0013,\ ( 0.13\%)$$

(b) There are 9% customers would rent the bike for less than K minutes. What is the value of K? 

$$P(X<K)=P(Z<{K-75 \over15})=0.09$$ $${K-75 \over15}\approx-1.340755$$ $$K=55$$

(c) What are the (i) mean, (ii) median, (iii) variance, and (iv) standard deviation of the rental fee paid by a customer? 

$(i) \ \ mean=\$10+\$0.5\cdot75=\$47.50,$

$(ii) \ \ median=mean=\$47.50,$

$(iii)Variance=(0.5)^2(15)^2=56.25,$

$(iv) Standard\ deviation=\sqrt{Variance}=\sqrt{56.25}=\$7.50$

(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?

$$P(\mu-\delta<X<\mu+\delta)=0.9$$ $$P(X\leq\mu-\delta)={1-0.9 \over 2}=0.05$$ $$P(Z\leq{\mu-\delta-\mu \over \sigma})=0.05$$ $$-{\delta\over \sigma}\approx-1.644853$$

$$\delta\approx1.644852\cdot7.5\approx12.3364$$

$$L_1=\mu-\delta\approx47.50-12.34=35.16$$

$$L_2=\mu+\delta\approx47.50+12.34=59.84$$

$$L_1=\$35.16, L_2=\$59.84$$