Find expectation, variance and probability sample mean in range for normal distribution

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Simon Thomas 6 years, 2 months ago

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Question statement

The total {stuff} content of the blood plasma of {nationality} {animals} ($X$, in mg/100ml) is known to follow a $N(\var{m},\var{s^2})$ distribution.

Give any introductory information the student needs.

			Name	Type	Generated Value

upper	number	248.8
zlowsam	number	-2.75
zlow	number	-0.55
m	integer	240
plowsam	number	0.997020
plow	number	0.708840
slow	number	0.55
tol	number	0.0001
sup	number	0.44
prob	number	0.3789
zup	number	0.44
probsam	number	0.9831
zupsam	number	2.2
nationality	string	English
pup	number	0.670031
lower	number	229
animals	string	goats
pupsam	number	0.986097
svb	number	11.11
sva	number	16.00
s	integer	20
stuff	string	fibrinogen
sb	integer	36
sa	integer	25
tol1	integer	0

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Name

Data type

Value

xxxxxxxxxx
 
m+sup*s

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Generated value: `number`

248.8

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Find:

$P(\var{lower} \lt X \lt \var{upper})=\;\;$Unnamed gap

Correct to 4 decimal places.

Advice

a) Converting to $Z$ scores in $N(0,1)$ we have for $X \sim N(\var{m},\var{s^2})$

\[\begin{eqnarray*} P(\var{lower} \lt X \lt \var{upper})&=&P(X \lt \var{upper})-P(X \lt \var{lower})\\ &=&P\left(Z \lt \frac{\var{upper}-\var{m}}{\var{s}}\right)-P\left(Z \lt \frac{\var{lower}-\var{m}}{\var{s}}\right)\\ &=&P(Z \lt \var{zup})-P( Z \lt \var{zlow})\\ &=&\var{pup}-\var{1-plow}\\ &=&\var{prob} \end{eqnarray*} \] to 4 decimal places.

Here the probabilities could have been looked up from normal CDF tables. Alternatively we can simply do the whole calculation in

R by typing $\operatorname{pnorm}(\var{upper},\var{m},\var{s})-\operatorname{pnorm}(\var{lower},\var{m},\var{s})$.

b)
This part reminds you that for the sample mean for samples of size $n$ from a normal distribution $N(\mu,\sigma^2)$ has a normal distribution $\displaystyle N\left(\mu,\frac{\sigma^2}{n}\right)$.

Hence for a sample size $\var{sa}$:

$\displaystyle \operatorname{E}[\overline{X}] = \mu=\var{m}$ and $\displaystyle \operatorname{Var}(\overline{X}) = \frac{\sigma^2}{n}=\frac{\var{s^2}}{\var{sa}}=\var{sva}$

Hence for a sample size $\var{sb}$:

$\operatorname{E}[\overline{X}] = \mu=\var{m}$ and $\displaystyle \operatorname{Var}(\overline{X}) = \frac{\sigma^2}{n}=\frac{\var{s^2}}{\var{sb}}=\var{svb}$

c)
Since the sample size is $\var{sa}$ we are dealing with the normal distribution $N(\var{m},\simplify[std]{{s^2}/{sa}})$.

Converting to $Z$ scores in $N(0,1)$ we have for $\overline{X} \sim N(\var{m},\simplify[std]{{s^2}/{sa}})$

\[\begin{eqnarray*} P(\var{lower} \lt \overline{X} \lt \var{upper})&=&P(\overline{X} \lt \var{upper})-P(\overline{X}\lt \var{lower})\\ &=&P\left(Z \lt \frac{\sqrt{\var{sa}}(\var{upper}-\var{m})}{\var{s}}\right)-P\left(Z \lt \frac{\sqrt{\var{sa}}(\var{lower}-\var{m})}{\var{s}}\right)\\ &=&P(Z \lt \var{zupsam})-P( Z \lt \var{zlowsam})\\ &=&\var{pupsam}-\var{1-plowsam}\\ &=&\var{probsam} \end{eqnarray*} \] to 4 decimal places.

Here the probabilities could have been looked up from normal CDF tables. Alternatively we can simply do the whole calculation in

R by typing $\operatorname{pnorm}(\var{upper},\var{m},\frac{\var{s}}{\sqrt{\var{sa}}})-\operatorname{pnorm}(\var{lower},\var{m},\frac{\var{s}}{\sqrt{\var{sa}}})$ i.e. $\operatorname{pnorm}(\var{upper},\var{m},\var{s/sqrt(sa)})-\operatorname{pnorm}(\var{lower},\var{m},\var{s/sqrt(sa)})$

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Find expectation, variance and probability sample mean in range for normal distribution

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Simon Thomas 6 years, 2 months ago

Error in variable testing condition

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Find expectation, variance and probability sample mean in range for normal distribution

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years, 2 months ago

Error in variable testing condition

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Generated value: number

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Gap-fill

Generated value: `number`