According to the Daily Mail, there was an average of one complaint every 12 seconds against Britain’s biggest banks in 2011. It is reasonable to assume that the time between complaints is exponentially distributed. a. What is the mean time between complaints?b. What is the probability that the next complaint will take less than the mean time?c. What is the probability that the next complaint will take between 5 and 10 seconds?

Given Information:

Exponential distribution

Mean = μ = 12 seconds

Required Information:

Mean time between complaints = ?

P(X < mean) = ?

P(5 < X < 10)  = ?

Explanation:

X = time of complaint against Britain’s biggest banks in 2011 has an exponential distribution with the average complaint time of 12 seconds.

Mean = μ = 12 seconds

Decay rate = λ = 1 /μ = 1/12 = 0.083  

The standard deviation is equal to the mean in exponential distribution = σ = μ = 12

a. What is the mean time between complaints?

The mean time between complaints is 12 seconds

The probability of that happening is

P(X = 12) = 0.083e^–0.083x(12) = 0.030

b. What is the probability that the next complaint will take less than the mean time?

P(X < mean) = 1 – e^–0.083x

P(X < 12) = 1 – e^–0.083(12)

P(X < 12) = 1 – 0.369

P(X < 12) = 0.63 = 63%

There is 63% probability that the next complaint will take less than the mean time of 12 seconds.

c. What is the probability that the next complaint will take between 5 and 10 seconds?

P(5 < X < 10) = P(X < 10) – P(X < 5)

P(X < 10) = 1 – e^(–0.083)(10) = 0.563

P(X < 5) = 1 – e^(–0.083)(5) = 0.339

P(5 < X < 10) = 0.563 − 0.339 = 0.224

P(5 < X < 10)  = 22.4 %

There is 22.4% probability that the next complaint will take between 5 and 10 seconds.

answer: it's a

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answer: i don't understand. what are the options

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