IP3.3

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Bill Foster

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Christian Lawson-Perfect

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Ready to use

10 years, 1 month ago

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Christian Lawson-Perfect 10 years, 1 month ago

Gave some feedback: Ready to use

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Bill Foster 12 years, 4 months ago

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Question statement

Two of the integers $\var{d[0]},\;\var{d[1]}, \;\var{d[2]}, \;\var{d[3]}, \;\var{d[4]} $ are chosen at random, without replacement.

Let $X$ denote the sum of the chosen two values.

Answer the following two parts:

Give any introductory information the student needs.

			Name	Type	Generated Value

a	list	Nested 3×5 list
b	list	Nested 3×9 list
d	list	[ -1, 0, 1, 3, 5 ]
p	integer	1
pmf	list	[ -1, 0, 1, 2, 3, 4, 5, 6, 8 ]
q	integer	-1
twice	number	4
x8	number	0.1
expx	number	3.2
t	integer	1
m2	number	0.1
x2	number	0.1
x3	number	0.1
x0	number	0.1
x1	number	0.1
x6	number	0.1
x7	number	0.1
x4	number	0.1
x5	number	0.2

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
[[0,1,2,3,7],[0,1,2,4,6],[0,1,3,5,7]]

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: `list`

[ [ 0, 1, 2, 3, 7 ], [ 0, 1, 2, 4, 6 ], [ 0, 1, 3, 5, 7 ] ]

→ Used by:

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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Part b) 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.

Input the possible values of $X$ in the following table.

You have to input these in increasing order.

Also find the probability mass function $P(X=x)$ values and input them into the table.

Input all values as exact values.

$X=x$	Gap 0.	Gap 1.	Gap 2.	Gap 3.	Gap 4.	Gap 5.	Gap 6.	Gap 7.	Gap 8.
$P(X=x)$	Gap 9.	Gap 10.	Gap 11.	Gap 12.	Gap 13.	Gap 14.	Gap 15.	Gap 16.	Gap 17.

Advice

a)

First we find the sums that can occur by simply adding two different numbers together from the given.

Note that these are pairs of different numbers as we choose without replacement.

We get:

$X=x$

$\var{pmf[0]}$

$\var{pmf[1]}$

$\var{pmf[2]}$

$\var{pmf[3]}$

$\var{pmf[4]}$

$\var{pmf[5]}$

$\var{pmf[6]}$

$\var{pmf[7]}$

$\var{pmf[8]}$

We have to find the probabilities that each of these sums occur.

There are $10$ ways of selecting $2$ numbers from the $5$ given, but it may be the case that two different pairs produce the same sum.

In this case we find that there are two ways of producing the sum $\var{twice}$. All other sums have only one way.

So since each selection of a pair of numbers has probability $0.1$.

The probability of producing the sum $\var{twice}$ is $0.2$ , and the other sums have probability $0.1$ .

b)

The expectation is given by:

\[ \begin{eqnarray*} E[X]&=& \sum xP(X=x)\\ &=&\simplify[]{ {pmf[0]}*{x0}+{pmf[1]}*{x1}+{pmf[2]}*{x2}+{pmf[3]}*{x3}+{pmf[4]}*{x4}+{pmf[5]}*{x5}+{pmf[6]}*{x6}+{pmf[7]}*{x7}+{pmf[8]}*{x8}}\\ &=&\var{expx} \end{eqnarray*} \]

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

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

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This question is used in the following exams:

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Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

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Comment

Checkpoint description

Christian Lawson-Perfect 10 years, 1 month ago

Bill Foster 12 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: `list`

Parts

Gap-fill

a)

b)

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Error

IP3.3

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 10 years, 1 month ago

Bill Foster 12 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

a)

b)

Generated value: `list`