Answer:
The probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes is 0.6065.
Step-by-step explanation:
The random variable <em>T</em> is defined as the amount of time a customer spends in a bank.
The random variable <em>T</em> is exponentially distributed.
The probability density function of a an exponential random variable is:
![f(x)=\lambda e^{-\lambda x};\ x>0](https://tex.z-dn.net/?f=f%28x%29%3D%5Clambda%20e%5E%7B-%5Clambda%20x%7D%3B%5C%20x%3E0)
The average time a customer spends in a bank is <em>β</em> = 10 minutes.
Then the parameter of the distribution is:
![\lambda=\frac{1}{\beta}=\frac{1}{10}=0.10](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1%7D%7B%5Cbeta%7D%3D%5Cfrac%7B1%7D%7B10%7D%3D0.10)
An exponential distribution has a memory-less property, i.e the future probabilities are not affected by any past data.
That is, <em>P</em> (<em>X</em> > <em>s</em> + <em>x</em> | <em>X</em> ><em> s</em>) = <em>P</em> (<em>X</em> > <em>x</em>)
So the probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes is:
P (X > 15 | X > 10) = P (X > 5)
![\int\limits^{\infty}_{5} {f(x)} \, dx =\int\limits^{\infty}_{5} {\lambda e^{-\lambda x}} \, dx\\=\int\limits^{\infty}_{5} {0.10 e^{-0.10 x}} \, dx\\=0.10\int\limits^{\infty}_{5} {e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10}|^{\infty}_{5}\\=[-e^{-0.10 \times \infty}+e^{-0.10 \times 5}]\\=0.6065](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B5%7D%20%7Bf%28x%29%7D%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B5%7D%20%20%7B%5Clambda%20e%5E%7B-%5Clambda%20x%7D%7D%20%5C%2C%20dx%5C%5C%3D%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B5%7D%20%20%7B0.10%20e%5E%7B-0.10%20x%7D%7D%20%5C%2C%20dx%5C%5C%3D0.10%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B5%7D%20%20%7Be%5E%7B-0.10%20x%7D%7D%20%5C%2C%20dx%5C%5C%3D0.10%7C%5Cfrac%7Be%5E%7B-0.10%20x%7D%7D%7B-0.10%7D%7C%5E%7B%5Cinfty%7D_%7B5%7D%5C%5C%3D%5B-e%5E%7B-0.10%20%5Ctimes%20%5Cinfty%7D%2Be%5E%7B-0.10%20%5Ctimes%205%7D%5D%5C%5C%3D0.6065)
Thus, the probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes is 0.6065.