Probability, expectation and variance of uniform distribution

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Bill Foster

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6 years, 1 month ago

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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 uniform distribution	Ready to use	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Probability, expectation and variance of uniform distribution	draft	Xiaodan Leng	10/07/2019 22:35

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Question statement

A new supermarket plans to open somewhere on the outskirts of a town. In fact, $X$ , the distance of a new supermarket from the town centre is Uniformly distributed between $\var{lower}$ metres and $\var{upper}$ metres i.e.

$X \sim \operatorname{U}(\var{lower},\var{upper})$

Give any introductory information the student needs.

			Name	Type	Generated Value

upper	integer	1200
lower	integer	700
ans1	rational	950
ans2	number	20833.333
ans3	decimal	dec("0.200")
thisdis	decimal	dec("0.80")
t	integer	80
tol	number	0.001

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
lower+random(300..500#50)

JME function reference

Description

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

Can an exam override the value of this variable?

Generated value: `integer`

1200

← Depends on:

lower

→ Used by:

ans1
ans2
ans3
thisdis

Statement
Advice

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 $\operatorname{E}[X]$ , the expected distance in metres of the new supermarket from the town centre:

$\operatorname{E}[X]=\;?$ Gap 0.m (to 3 decimal places).

Also find the variance $\operatorname{Var}(X)$ :

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

Advice

a) For a Uniform distribution

$X \sim \operatorname{U}(\var{lower},\var{upper})$ we have:

$\displaystyle \operatorname{E}[X] = \frac{\var{lower}+\var{upper}}{2}=\var{ans1}$ m

$\displaystyle \operatorname{Var}(X) = \frac{(\var{upper}-\var{lower})^2}{12}=\frac{(\var{upper-lower})^2}{12}=\var{ans2}$ to 3 decimal places.

$\displaystyle P(X \le \var{thisdis}\textrm{km})=\frac{\var{thisdis}\times 1000 -\var{lower}}{\var{upper}-\var{lower}}=\var{ans3}$ to 3 decimal places.

Give a worked solution to the whole question.

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There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet

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Probability, expectation and variance of uniform distribution

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Taxonomy: mathcentre

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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: `integer`

Parts

Gap-fill

Loading...

Error

Probability, expectation and variance of uniform 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: integer

Parts

Gap-fill

Generated value: `integer`