Answer to Question #110960 in Statistics and Probability for Li

Question

“FunBakery” is a bakery shop that customers can use the facilities to bake cakes. According to the record, the duration a customer spends in the shop follows a normal distribution with mean 140 minutes and standard deviation 18 minutes.
(a) What is the probability that a customer stays in the shop for more than 3 hours?
(b) There are 15% of all customers would stay in the shop for less than t minutes. Find the value of t. Round
up the answer to the next integer.
The charge in “FunBakery” is $100 for the first hour and then $1.1 per extra minute.
(c) Find the average, median and standard deviation of the amount of money a customer spends in the shop.
(d) For a customer stays in the shop for 3 hours, how much he / she needs to pay?
(e) There are 15% of all customers would pay less than $k. Find the value of k by using your answer in
part (b).
(f) Write a simple summary about the spending of a customer in “FunBakery”.

Answer 1

$a)\xi\text{ is a random variable: the duration a customer spends in the shop}.\\ \xi \in N(140; 18^2).\\ P\{\xi>180\}=1-P\{\xi\leq 180\}=1-F(180)=1-\Phi(\frac{180-140}{18})\approx\\ \approx 1-0.9869=0.0131.\\ F(x)=\frac{1}{\sqrt{2\pi}18}\int_{-\infty}^x e^{-\frac{(z-140)^2}{2\cdot 18^2}}dz\\ \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{z^2}{2}}dz\\ b) P\{\xi\leq t\}=F(t)=\Phi(\frac{t-140}{18})=0.15\\ \frac{t-140}{18}\approx -1.04\\ t\approx 121.28\approx 122.\\ с) \text{Average of the amount of money equals } 100+80(1.1)=\\ =188 (140=60+80).\\ \text{Median equals average because the duration a customer}\\ \text{spends in the shop follows a normal distribution with mean 140 minutes}\\ \text{and standard deviation 18 minutes}.\\ \text{Standard deviation of the amount of money equals } 18(1.1)=19.8.\\ d) 3\text{ hours}=60+2\cdot 60.\\ \text{A customer needs to pay } 100+120(1.1)=232.\\ e)\text{In b) } t\approx 122.\\ k=100+62(1.1)=168.2.\\ f)\text{Spending of a customer in “FunBakery” has normal distribution}\\ \text{with mean 188 and standard deviation 19.8. Median of the amount}\\ \text{of money equals mean}.$