Probability, expectation and variance of exponential distribution

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Bill Foster 6 years, 1 month ago

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Name	Status	Author	Last Modified
Probability, expectation and variance of exponential distribution	Ready to use	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Probability, expectation and variance of exponential distribution	draft	Xiaodan Leng	10/07/2019 22:36

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Question statement

The time, in {period} between {this} follows an exponential distribution with rate $\var{ra}$ i.e.

$X \sim \operatorname{exp}(\var{ra})$

Give any introductory information the student needs.

			Name	Type	Generated Value

that	string	two customers arriving
this	string	customer arrivals at the RyanJ
ans1	number	2.000
ans2	number	4.000
ans3	number	0.393
period	string	minutes
ra	number	0.5
tol	number	0.001
tans3	number	0.3934693403
thistime	number	1

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
"two customers arriving "

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `string`

two customers arriving

→ Used by:

"Part b)" - prompt

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
Part b) Gap-fill
1. Gap Number entry
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Find $\operatorname{E}[X]$ between {this}:

$\operatorname{E}[X]=\;$ ?Gap 0.{period} (enter as a decimal correct to 3 decimal places).

Find $\operatorname{Var}(X)$ :

$\operatorname{Var}(X)=\;$ ?Gap 1.(enter as a decimal correct to 3 decimal places).

Advice

If $X \sim \operatorname{exp}(\lambda)$ then $\displaystyle \operatorname{E}[X] =\frac{1}{\lambda}$ and $\displaystyle \operatorname{Var}(X)=\frac{1}{\lambda^2}$ .

Also $P(X \lt a)=1-e^{-\lambda a}$ .

If $X \sim \operatorname{exp}(\var{ra})$ then:

$\displaystyle \operatorname{E}[X] =\frac{1}{\lambda}=\frac{1}{\var{ra}}=\var{ans1}$ to 3 decimal places.

$\displaystyle \operatorname{Var}(X) =\frac{1}{\lambda^2}=\frac{1}{\var{ra}^2}=\var{ans2}$ to 3 decimal places.

$P(X \lt \var{thistime}) = 1 -(e ^ {-\var{ ra} \times \var{thistime}}) = 1 -(e ^ { -\var{ra * thistime}}) = \var{ans3}$ to 3 decimal places.

Give a worked solution to the whole question.

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Number entry
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Probability, expectation and variance of exponential distribution

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Bill Foster 6 years, 1 month ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `string`

Parts

Gap-fill

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Error

Probability, expectation and variance of exponential distribution

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 1 month ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: string

Parts

Gap-fill

Generated value: `string`