Ricardo's copy of IS2.2

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Ricardo Monge

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Christian Lawson-Perfect

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Ready to use

8 years, 9 months ago

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Christian Lawson-Perfect 8 years, 9 months ago

Gave some feedback: Ready to use

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Ricardo Monge 9 years, 6 months ago

Created this as a copy of IS2.2.

Name	Status	Author	Last Modified
IS2.2	Ready to use	Bill Foster	01/06/2016 09:47
Ricardo's copy of IS2.2	Ready to use	Ricardo Monge	01/06/2016 09:47

Question statement

A random variable $X$ has a probability density function (PDF) given by:

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	0
ux	decimal	dec("5.5")
xl	integer	0
i1	decimal	dec("4")
lo	number	0.12500
j1	decimal	dec("11")
up	number	0.80469
exans	number	0.67969
p	integer	8
u	integer	69
t	integer	58
tol	number	0.01
ans	number	0.68
cval	number	0.0625
lx	decimal	dec("2")
xu	integer	8

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
0

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`

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Mathematical expression
  Add
Part b) Gap-fill
Part c) 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.

$f_X(x) = \left \{ \begin{array}{l} \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}} \\\phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}}\end{array} \right .$	$0$	$ x \leq \var{xl},$

	$cx$	$\var{xl} \lt x \leq \simplify[std]{{xu+xl}/2},$

	$c(\var{xu}-x)$	$\simplify[std]{{xu+xl}/2} \lt x \leq \var{xu},$

	$0$	$x \gt \var{xu}.$

What value of $c$ makes $f_X(x)$ into the pdf of a distribution?

Input your answer here as a fraction and not as a decimal.

$c=\;\;$Unnamed gap

Advice

a)

Note that in order for $f_X(x)$ to be a pdf it must satisfy two important conditions:

1. $f_X(x) \ge 0$ in the range $\var{xl} \le x \le \var{xu}$

2. The area under the curve given by $f_X(x)$ is $1$ and this implies that:
\[\int_{\var{xl}}^{\var{xu}}f_X(x)\;dx = 1\] as the value of the function is $0$ outside this range.

We first check condition 2. and then check that condition 1. is satisfied.

Hence \[\begin{eqnarray*} \int_{-\infty}^{\infty}f_X(x)\;dx=\int_{\var{xl}}^{\var{xu}}f_X(x)\;dx&=&\int_{\var{xl}}^{\simplify[std]{{xu+xl}/2}}cx\;dx+\int_{\simplify[std]{{xu+xl}/2}}^{\var{xu}}c(\var{xu}-x)\;dx\\ &=&\simplify[std]{{(xu-xl)^2}/8}c+\simplify[std]{{(xu-xl)^2}/8}c=\simplify[std]{{(xu-xl)^2}/{4}}c \end{eqnarray*} \]
But this has to equal $1$ and so \[c=\simplify[std]{4/{p^2}}\]

Hence with this value of $c$ we see that condition 2. is satisfied i.e.

\[\int_{-\infty}^{\infty}f_X(x)\;dx=1\]

Condition 1. is clearly satisfied as $c \gt 0$.

b)

We must have $F_X(x)=0,\;\;\;x \le \var{xl},\;\;\;\textrm{and}\;\;\;F_X(x)=1,\;\;\;x \ge \var{xu}$

Apart from that we find an expression for $F_X(x)$ in each of the ranges $[\var{xl},\simplify[std]{{xl+xu}/2}],\;\;\;[\simplify[std]{{xl+xu}/2},\var{xu}]$

1. $x \in [\var{xl},\simplify[std]{{xl+xu}/2}]$

We have:\[\begin{eqnarray*} f_X(x) &=& \simplify[std]{{4}/{p^2}} x \\ \Rightarrow F_X(x)&=& \int_{-\infty}^xf_X(x)\;dx \\ &=& \simplify[std]{{4}/{p^2}}\int_{\var{xl}}^x x\;dx\\ &=&\simplify[std]{{2}/{p^2}}x^2 \end{eqnarray*} \]

2. $x \in [\simplify[std]{{xl+xu}/2},\var{xu}]$

We have:\[\begin{eqnarray*} f_X(x) &=& \simplify[std]{{4}/{p^2}}(\var{xu}-x) \\ \Rightarrow F_X(x)&=& \int_{-\infty}^xf_X(x)\;dx = \int_{\var{xl}}^{\simplify[std]{{xl+xu}/2}}f_X(x)\;dx+\int_{\simplify[std]{{xl+xu}/2}}^x f_X(x)\;dx\\ &=& \simplify[std]{{4}/{p^2}}\int_{\var{xl}}^{\simplify[std]{{xl+xu}/2}} x\;dx + \simplify[std]{{4}/{p^2}}\int_{\simplify[std]{{xl+xu}/2}}^x(\var{xu}-x)\;dx\\ &=&1-\simplify[std]{{2}/{p^2}}(\var{xu}-x)^2 \end{eqnarray*} \]

c)

Using the distribution function we have just found we have that:

$P(\var{lx} \lt X \lt \var{ux}) = F_X(\var{ux})-F_X(\var{lx})$

But $\var{ux}$ is in the range between $\var{(xl+xu)/2}$ and $\var{xu}$ and the distribution function is given by:

\[F_X(x)=1-\simplify[std]{{2}/{p^2}}(\var{xu}-x)^2\]

Hence $F_X(\var{ux})=1-\simplify[std]{{2}/{p^2}}(\var{xu}-\var{ux})^2 = \var{up}$ to 5 decimal places.

Similarly, $\var{lx}$ is in the range between $\var{xl}$ and $\var{(xl+xu)/2}$ and the distribution function is given by:

\[F_X(x)=\simplify[std]{{2}/{p^2}}x^2\]

Hence $F_X(\var{lx})=\simplify[std]{{2}/{p^2}}(\var{lx})^2 = \var{lo}$ to 5 decimal places.

Hence $P(\var{lx} \lt X \lt \var{ux}) = F_X(\var{ux})-F_X(\var{lx})=\var{up}-\var{lo}=\var{up-lo}=\var{ans}$ to 2 decimal places.

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Checkpoint description

Christian Lawson-Perfect 8 years, 9 months ago

Ricardo Monge 9 years, 6 months ago

Error in variable testing condition

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Warning

Generated value: `integer`

Parts

Gap-fill

a)

b)

c)

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Ricardo's copy of IS2.2

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 8 years, 9 months ago

Ricardo Monge 9 years, 6 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

Generated value: `integer`