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Chapter 4: Q59E (page 175)

Let X= the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with \({\rm{\lambda = I}}\) (which is identical to a standard gamma distribution with \({\rm{\alpha = 1}}\) ), compute the following:

a. The expected time between two successive arrivals

b. The standard deviation of the time between successive arrivals

c. \({\rm{P(X}} \le {\rm{4)}}\)

d. \({\rm{P(2}} \le {\rm{X}} \le {\rm{5)}}\)

Short Answer

Expert verified

(a) The expected time between two successive arrivals is 1.

(b) The standard deviation is 1 .

(c) The value of \(P(X \le 4)\) is\({\rm{0}}{\rm{.982}}\).

(d) The value of \(P(2 \le X \le 5)\)is \({\rm{0}}{\rm{.129}}{\rm{.}}\)

Step by step solution

01

Definition of Concept

Probability: Probability means possibility. It is a branch of mathematics which deals with the occurrence of a random event.

02

Find the expected time between two successive arrivals

(a)

Considering the given information:

Let X be the time between the two successive arrivals at the drive-up window of the local bank which follows exponential distribution with parameter mean 1 .

Here, X follows exponential distribution with parameter $\lambda$ and the parameter \({\rm{\lambda = 1}}\)

The mean of the exponential distribution is given below:

\({\rm{\mu = }}\frac{{\rm{1}}}{{\rm{\lambda }}}\)

The expected time between two successive arrivals is obtained as shown below:

Putting \({\rm{\lambda = 1}}\)

\(\begin{array}{c}{\rm{\mu = }}\frac{{\rm{1}}}{{\rm{1}}}\\{\rm{ = 1}}\end{array}\)

Therefore, the expected time between two successive arrivals is 1.

03

Find the standard deviation of the time between successive arrivals

(b)

Considering the given information:

The variance of exponential distribution is given below:

\({{\rm{\sigma }}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{{\rm{\lambda }}^{\rm{2}}}}}\)

The standard deviation of exponential distribution is given below:

\({\rm{\sigma = }}\sqrt {\frac{{\rm{1}}}{{{{\rm{\lambda }}^{\rm{2}}}}}} \)

The standard deviation of the time between two successive arrivals is obtained as shown below:

Putting\({\rm{\lambda = 1}}\)

\(\begin{array}{c}{\rm{\sigma = }}\sqrt {\frac{{\rm{1}}}{{{{\rm{1}}^{\rm{2}}}}}} \\{\rm{ = }}\sqrt {\rm{1}} \\{\rm{ = 1}}\end{array}\)

Therefore, the standard deviation is 1 .

04

Compute the \({\rm{P(X}} \le {\rm{4)}}\)

(c)

Considering the given information:

The cumulative density function is given by,

The value of\(P(X \le 4)\)is obtained as shown below:

Putting \({\rm{\lambda = 1,x = 4}}\)

\(\begin{array}{c}P(X \le 4) = 1 - {e^{ - 1(4)}}\\ = 1 - {e^{ - 4}}\\ = 1 - 0.0183\\ = 0.982\end{array}\)

Therefore, the value of \(P(X \le 4)\) is\({\rm{0}}{\rm{.982}}\).

05

Compute the \({\rm{P(2}} \le {\rm{X}} \le {\rm{5)}}\)

(d)

Considering the given information:

The\(P(2 \le X \le 5)\)is obtained as shown below:

\(\begin{array}{c}P(2 \le X \le 5) = P(X = 5) - P(X = 2)\\ = \left( {1 - {e^{ - 1(5)}}} \right) - \left( {1 - {e^{ - 1(2)}}} \right)\quad \\ = \left( {1 - {e^{ - (5)}}} \right) - \left( {1 - {e^{ - (2)}}} \right)\\ = (1 - 0.00674) - (1 - 0.135)\\ = 0.99326 - 0.865\\ = 0.129\end{array}\)

Therefore, the value of \(P(2 \le X \le 5)\)is \({\rm{0}}{\rm{.129}}{\rm{.}}\)

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