Use piecewise CDF to find probabilities at given points, and the expectation.

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Bill Foster 6 years, 5 months ago

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Question statement

Suppose that the cumulative distribution function (CDF) of the random variable $X$ is given by

$F_X(b) = \left \{ \begin{array}{l} \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}}\\ \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}} \\ \end{array} \right .$	0,	$b \lt 0,$

	$\var{x1}$	$0 \le b \lt \var{c1},$

	$\var{x2}$	$\var{c1} \le b \lt\var{c2},$

	$\var{x3}$	$\var{c2} \le b \lt \var{c3},$

	1,	$b \ge \var{c3}.$

Answer the following questions:

Give any introductory information the student needs.

			Name	Type	Generated Value

c4	decimal	dec("3")
val	number	0.3
mess	string	$\;$
x2	number	0.45
v2	integer	0
v3	integer	0
v4	integer	0
ex	number	4.8
x3	number	0.65
c3	integer	8
c2	integer	7
c1	integer	4
x1	number	0.3
message	string
c5	integer	8
t	integer	28

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
round(((100-t)/100+t/100*(c3-1)))

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: `decimal`

dec("3")

← Depends on:

→ Used by:

mess
message
v2
v3
v4
val

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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Part b) Gap-fill
1. Gap Number entry
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Part c) Gap-fill
1. Gap Number entry
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Gap-fill

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

Write down the following probabilities: (all as exact decimals )

$P(X=0)=\;\;$ Gap 0. $P(X=\var{c1})=\;\;$ Gap 1.

$P(X=\var{c2})=\;\;$ Gap 2. $P(X=\var{c3})=\;\;$ Gap 3.

Advice

From lectures we know that jumps in the CDF imply a non zero probability at that point.

Thus:

$P(X=0)=\var{x1}$ ,

$P(X=\var{c1})=\var{x2}-\var{x1}=\var{x2-x1}$ ,

$P(X=\var{c2})=\var{x3}-\var{x2}=\var{x3-x2}$ ,

$P(X=\var{c3})=1-\var{x3}=\var{1-x3}$

$\begin{eqnarray*} F_X(\var{c4}) =P(X \le \var{c4}) &=&\simplify[all]{ P(X = 0) + {v2} * P(X = {c1}) + {v3} * P(X = {c2}) + {v4} * P(X = {c3})}\\&=& \simplify[zerofactor,zeroterm,unitfactor]{{x1} + {v2} *({x2-x1}) + {v3} *({x3 -x2}) + {v4} * ({1 -x3}) }\\&=& \var{val}\end{eqnarray*}$

{message}{mess}.

$\displaystyle \operatorname{E}[X]=\sum x P(X=x)=0\times \var{x1}+\var{c1}\times \var{x2-x1}+\var{c2}\times \var{x3-x2}+\var{c3}\times \var{1-x3}=\var{ex}$

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Use piecewise CDF to find probabilities at given points, and the expectation.

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Taxonomy: mathcentre

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Bill Foster 6 years, 5 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `decimal`

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Gap-fill

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Use piecewise CDF to find probabilities at given points, and the expectation.

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 5 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: decimal

Parts

Gap-fill

Generated value: `decimal`