Question

Customers make purchases at a convenience store, on average, every six minutes. It is fair to assume that the time between customer purchases is exponentially distributed. Jack operates the cash register at this store.

Answer 1

Answer:

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Answer 2

Given is an exponential distribution with mean six minutes, answer the question based on this.

Explanation:

• The PDF- probability density function of an exponential distribution using rate parameter λ is given by $f(x)=\frac{e^{-\frac{x}{\lambda} }}{\lambda} ,\ \ \ x\geq 0$  
• here mean is $\mu=\lambda$ and variance is $\sigma^2=\lambda^2$
• hence, the probability of a random variable 'X' is given by,                                    $P(x=X)=\int\limits^X_0 {f(x)} \, dx$      
• For the given distribution we have , $\mu=6\ min=\lambda$                                           $f(x)=\frac{e^{-\frac{x}{6} }}{6} ,\ \ \ x\geq 0$          
• standard deviation is $\sigma=\lambda= 6\ min$
• variance is $\sigma^2=\lambda^2=36$.          
• the probability that a customer will arrive in less than five minutes is,                                                                                                                          $P(x< 5)=\int\limits^5_0 {\frac{e^{-\frac{x}{6} }}{6}} \, dx \\P(x<5)= 1-e^{\frac{-5}{6} } \\P(x<5)=0.5654(approx.)$  
• the probability that no one shows up for over half hour is,                               $P(x>30)=1-P(x\leq30 )\\P(x>30)=1-\int\limits^{30}_0 {\frac{e^{\frac{-x}{6}} }{6} } \, dx \\P(x>30)=1-(1-e^{\frac{-30}{6} })\\P(x>30)=e^{-5}\\P(x>30)=0.0067(approx.)$