Question

A California water company has determined that the average customer billing is $1,250 per year and the amounts billed have an exponential distribution. a. What is the mean and lamda? b. Calculate the probability that a random chosen customer would spend more than $5,000 c. Compute the probability that a random chosen customer would spend more than the average amount spent by all customers of this company.

Answer 1

Answer:

a)

Mean $\mu = \dfrac{1}{\lambda }= 1250$

$\lambda = \dfrac{1}{1250}$

b)

$P(X > 5000) = 0.0183$

c)

$P(X > 1250) =0.3679$

Step-by-step explanation:

From the given information:

a.)

Mean $\mu = \dfrac{1}{\lambda }= 1250$

$\lambda = \dfrac{1}{1250}$

Let consider X to be a random variable that follows an exponential distribution; then:

P(X) = 1 - $e^{- \lambda x}$   since $\lambda > 0$

b.)

The required probability that a random chosen customer would spend more than $5,000 can be computed as:

$P(X > 5000) = 1 - \bigg [ 1 - e ^{- \dfrac{5000}{1250}} \bigg]$

$P(X > 5000) =e^ {-4$

$P(X > 5000) = 0.0183$

c.)

$P(X > 1250) = 1 - \bigg [ 1 - e ^{- \dfrac{1250}{1250}} \bigg]$

$P(X > 1250) =e ^{- 1$

$P(X > 1250) =0.3679$