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

a)

Mean
`\mu = (1)/(\lambda )= 1250`

`\lambda = (1)/(1250)`

b)

`P(X > 5000) = 0.0183`

c)

`P(X > 1250) =0.3679`

Explanation:

From the given information:

a.)

Mean
`\mu = (1)/(\lambda )= 1250`

`\lambda = (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 ^{- (5000)/(1250)} \bigg]`

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

`P(X > 5000) = 0.0183`

c.)

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

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

`P(X > 1250) =0.3679`