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Bill Foster 6 years, 4 months ago

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Question statement

In a laboratory experiment $\var{n}$ yeast cultures are grown and the number that are still alive after one week is counted. Each culture independently survives longer than a week with probability $p = \var{p1}$ . However, there is a probability $\var{q}$ that the laboratory equipment is contaminated. If that is the case, then all the cultures are affected, and the probability of survival after one week drops to $p = \var{p2}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

p2	number	0.1
p1	number	0.5
psurvive	number	0.1580914272
m	integer	3
n	integer	9
q	number	0.05
pcont	number	0.0141187428
tol	number	0.001
anspsurvive	number	0.158
anspcont	number	0.014

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(0.1..p1-0.1)

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: `number`

0.1

← Depends on:

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
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Part b) Number entry
Part c) Number entry

Gap-fill

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

Which probability distribution is best suited to represent the number of cells that die (assuming you know whether or not the equipment is contaminated)?Unnamed gap

Advice

a) The binomial distribution.

b) Using the law of total probability:

\[\begin{eqnarray*}\operatorname{Pr}(\var{m}\;\text{ survive})&=&\operatorname{Pr}(\var{m}\; \text{survive|uncontaminated})\operatorname{Pr}(\text{uncontaminated})+\operatorname{Pr}(\var{m} \;\text{survive|contaminated})\operatorname{Pr}(\text{contaminated})\\&=&{\var{n} \choose\var{m}}\var{p1}^{\var{m}}(1-\var{p1})^{\var{n-m}} \times (1-\var{q})+{\var{n} \choose\var{m}}\var{p2}^{\var{m}}(1-\var{p2})^{\var{n-m}} \times \var{q}\\&=&\var{psurvive}\\&=&\var{anspsurvive}\end{eqnarray*}\] to 3 decimal places.

c) By Baye's theorem:

\[\begin{eqnarray*}\operatorname{Pr}\text{contaminated|}\var{m}\;\text{survive})&=&\frac{\operatorname{Pr}(\var{m} \;\text{survive|contaminated})\operatorname{Pr}(\text{contaminated})}{\operatorname{Pr}(\var{m}\;\text{ survive})}\\&=&\frac{{\var{n} \choose\var{m}}\var{p2}^{\var{m}}(1-\var{p2})^{\var{n-m}} \times \var{q}}{\var{psurvive}}\\&=&\var{anspcont}\end{eqnarray*}\] to 3 decimal places.

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Bill Foster 6 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `number`

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Tom 1341cba2b extra

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`