Question

Do you dislike waiting in line? Supermarket chain Kroger has used computer simulation and information technology to reduce the average waiting time for customers at 2300 stores. Using a new system called QueVision, which allows Kroger to better predict when shoppers will be checking out, the company was able to decrease average customer waiting time to just 26 seconds (InformationWeek website and The Wall Street Journal website, January 5, 2015) a. Which of the exponential probability distribution is applicable and Show the probability density function of waiting time at Kroger. a. f(z) = e-= e-26 for z 20 b. f(z) =-e e-x for x > 0 c. f(x)--e--=-e-26 for x < 0 d. f(x) =-e--=-e26 for x > 0 μ 26 26 26 26 b. What is the probability that a customer will have to wait between 15 and 30 seconds (to 4 decimals)? 0.0095 c. What is the probability that a customer will have to wait more than 2 minutes (to 4 decimals)? 0.9962

Answer 1

Answer:

a

  $f(x) = \frac{1}{\mu} e^{- \frac{x}{\mu} } = \frac{1}{26} e^{-\frac{x}{26} } \ for \ \ x\ge 0$   option B

b

$P(15 < X < 30) = 0.2462$

c

$P(X > 120 ) = 0.0099$  

Step-by-step explanation:

From the question we are told that

   The number of stores is  n  = 2300

   The mean is  $\mu = 26$

Generally the probability mass function for exponential  distribution is  

       $f(x) = \left \{ {{ \lambda e^{-\lambda x}, \ \ \ \ \ x \ge 0} \atop {0 } \ \ \ \ \ \ \ \ \ \ \ x < 0. } \right.$

Here  $\lambda$ is the rate which is mathematically represented as

        $\lambda = \frac{1}{\mu}$

    =>     $\lambda = \frac{1}{26}$

   =>     $\lambda = 0.03846$

Generally the exponential probability distribution is applicable and Show the probability density function of waiting time at Kroger is  

      $f(x) = \frac{1}{\mu} e^{- \frac{x}{\mu} } = \frac{1}{26} e^{-\frac{x}{26} } \ for \ \ x\ge 0$

Generally the probability that a customer will have to wait between 15 and 30 seconds is mathematically represented as

      $P(15 < X < 30) = \int\limits^{15}_{30} {\frac{1}{26} e^{\frac{x}{26} } } \, dx$

=>   $P(15 < X < 30) =\frac{1}{26}[ { 26 e^{-\frac{x}{26} } } ]|\left \ 15} \atop {30}} \right.$

=>   $P(15 < X < 30) = [ { e^{-\frac{x}{26} } } ]|\left \ 15} \atop {30}} \right.$

=>   $P(15 < X < 30) = [ { e^{-\frac{15}{26} } } - e^{-\frac{30}{26} } } ]$

=>   $P(15 < X < 30) = 0.2462$

Generally 2 minutes =  120  seconds

Generally the probability that a customer will have to wait more than 2 minutes

     $P(X > 120 ) = 1- P(X < 120 )$

Here

     $P(X < 120) = \int\limits^{0}_{120} { \frac{1}{26} e^{-\frac{x}{26} } } \, dx$

=>  $P(X < 120) =\frac{1}{26} [ { 26 * e^{-\frac{x}{26} } } ]|\left \ 0 } \atop {120}} \right.$

=>  $P(X < 120) = [ { e^{-\frac{x}{26} } } ]|\left \ 0 } \atop {120}} \right.$

=>  $P(X < 120) = [ { e^{-\frac{0}{26} } } - e^{-\frac{120}{26} } } ]$

=>  $P(X < 120) = 0.990101$

So

  $P(X > 120 ) = 1- 0.990101$    

  $P(X > 120 ) = 0.0099$