Question

Suppose that the amount of time T a customer spends in a bank is exponentially distributed with an average of 10 minutes. What is 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

Answer 1

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 T is defined as the amount of time a customer spends in a bank.

The random variable T is exponentially distributed.

The probability density function of a an exponential random variable is:

$f(x)=\lambda e^{-\lambda x};\ x>0$

The average time a customer spends in a bank is β = 10 minutes.

Then the parameter of the distribution is:

$\lambda=\frac{1}{\beta}=\frac{1}{10}=0.10$

An exponential distribution has a memory-less property, i.e the future probabilities are not affected by any past data.

That is, P (X > s + x | X > s) = P (X > x)

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)

$\int\limits^{\infty}_{5} {f(x)} \, dx =\int\limits^{\infty}_{5} {\lambda e^{-\lambda x}} \, dx\\=\int\limits^{\infty}_{5} {0.10 e^{-0.10 x}} \, dx\\=0.10\int\limits^{\infty}_{5} {e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10}|^{\infty}_{5}\\=[-e^{-0.10 \times \infty}+e^{-0.10 \times 5}]\\=0.6065$

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.