Julie's copy of BS2.1

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Julie Crowley 8 years, 5 months ago

Published this.

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Julie Crowley 10 years, 4 months ago

Created this as a copy of BS2.1.

Name	Status	Author	Last Modified
Julie's copy of BS2.1	draft	Julie Crowley	06/03/2024 13:06
Julie's copy of Julie's copy of BS2.1	draft	Julie Crowley	06/02/2018 16:03
Simon's copy of Julie's copy of BS2.1	draft	Simon Thomas	08/05/2019 10:06

There are 7 other versions that do you not have access to.

Question statement

{sc[k]}

{table(data,[' From',' To', ' Loans Made'])}

Give any introductory information the student needs.

			Name	Type	Generated Value

a	list	[ 3000, 4000, 5000 ]
sc	list	List of 1 item
thismany	integer	900
ans1	number	0.07
k	integer	0
ans3	decimal	dec("0.84")
u1	integer	3000
p	integer	0
n	list	List of 4 items
q	integer	2
a0	integer	3000
b0	integer	1000
t	integer	0
v	integer	1
n0	decimal	dec("60")
n1	decimal	dec("300")
n3	decimal	dec("82")
data	list	Nested 4×3 list
ans2	number	0.60
o1	integer	4000

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
[a0,a0+b0,a0+2*b0]

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`

[ 3000, 4000, 5000 ]

← Depends on:

→ Used by:

data
o1
t
u1
v

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
3. Gap Gap 2. Number entry
  Add

Gap-fill

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

One of these loans is sampled randomly for review by the bank. What is the probability that it is :

a) Under € $\var{u1}$ ? Probability = ? Gap 0. (answer to 2 decimal places).

b) Over € $\var{o1-1}$ ? Probability = ? Gap 1. (answer to 2 decimal places).

c) Between € $\var{a[p]}$ and € $\var{a[q]-1}$ ? Probability = ? Gap 2. (answer to 2 decimal places).

Advice

a) The number of loans less than €$\var{u1}$ is $\var{accumdisp(n,t)}$$=\var{sum(n[0..t+1])}$

Since there are $\var{thismany}$ loans the probability of choosing one of these loans is $\displaystyle \frac{\var{sum(n[0..t+1])}}{\var{thismany}}=\var{ans1}$ to 2 decimal places.

b) The number of loans greater than €$\var{o1}$ is $\var{accumdisp(n[v+1..abs(n)],abs(n)-v-2)}=\var{sum(n[v+1..abs(n)])}$.

Since there are $\var{thismany}$ loans the probability of choosing one of these loans is $\displaystyle \frac{\var{sum(n[v+1..abs(n)])}}{\var{thismany}}=\var{ans2}$ to 2 decimal places.

c) There are $\var{accumdisp(n[p+1..q+1],q-p-1)}=\var{sum(n[p+1..q+1])}$ loans between €$\var{a[p]}$ and €$\var{a[q]-1}$.

Hence the probability of selecting one of these loans in this range for review is $\displaystyle \frac{\var{sum(n[p+1..q+1])}}{\var{thismany}}=\var{ans3}$ to 2 decimal places.

Give a worked solution to the whole question.

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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
Number entry
Number entry		Hasn't run yet

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Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Julie Crowley 8 years, 5 months ago

Julie Crowley 10 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: `list`

Parts

Gap-fill

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Julie's copy of BS2.1

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Julie Crowley 8 years, 5 months ago

Julie Crowley 10 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

Generated value: `list`