Normal distribution probability and percentiles,

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Ben Brawn

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Ben Brawn 2 years, 4 months ago

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Question statement

Let $X$ be the {something} which has a normal distribution, mean $\var{m}$ and standard deviation $\var{s}$ i.e. $X \sim \operatorname{N}(\var{m},\var{s^2})$.

Give any introductory information the student needs.

			Name	Type	Generated Value

zl	number	-2
lower1	number	196.8
upper1	number	338.88
this	string	a platelet count
m	number	285.6
athing	string	an individual
l	number	0.0227501319
p	number	0.862
s	number	44.4
percentile	integer	67
u	number	0.8849303298
something	string	blood platelet count (measured
tol	number	0.001
zu	number	1.2
perx	number	305.13

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(-2.0..-0.2#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`

-2

→ Used by:

l
lower1
zu

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Number entry
  Add
Part b) 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.

What is the probability that {athing} has {this} between $\var{lower1}$ and $\var{upper1}$?

Probability = ? Unnamed gap (to 3 decimal places).

Advice

Converting to $\operatorname{N}(0,1)$:

\[\begin{eqnarray*}P(\var{lower1} \lt X \lt \var{upper1})&=&P(X \lt \var{upper1})- P(X \lt \var{lower1})\\&=&P\left(Z \lt \frac{\var{upper1}-\var{m}}{\var{s}}\right)-P\left(Z \lt \frac{\var{lower1}-\var{m}}{\var{s}}\right)\\&=&P(Z \lt \var{zu})-P(Z \lt \var{zl})=\var{u}-\var{l}\\&=&\var{p}\end{eqnarray*}\]

to 3 decimal places.

We need to find $x$ such that:

\[\begin{eqnarray*}P(X \lt x) &\le& \var{percentile/100}\\ \Rightarrow P\left(Z \le \frac{x-\var{m}}{\var{s}}\right) &\le& \var{percentile/100}\\ \Rightarrow \frac{x-\var{m}}{\var{s}} &\le& \Phi^{-1}(\var{percentile/100})=\var{normalinv(percentile/100,0,1)}\\ \Rightarrow x &\le&\var{normalinv(percentile/100,0,1)}\times\var{s}+\var{m}=\var{perX}\end{eqnarray*}\] to 2 decimal places.

Hence $X=\var{perX}$ gives the $\var{percentile}$% percentile.

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Normal distribution probability and percentiles,

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Taxonomy: mathcentre

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Ben Brawn 2 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`

Parts

Gap-fill

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Error

Normal distribution probability and percentiles,

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Ben Brawn 2 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`