Normal Distribution

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Question statement

This test assesses your understanding of the normal distribution.

Give any introductory information the student needs.

			Name	Type	Generated Value

Mu1	integer	18
SD1	integer	6
SM1	integer	21
Z1	decimal	dec("0.50")
P1	number	0.6914624613
SM2	integer	25
Z2	decimal	dec("1.17")
P2	number	0.8789995156
SM3	number	20.6
Z3	number	0.43
P3	number	0.6664021794

			Name	Type	Generated Value

Mu2	integer	4
SD2	number	5.4
SM4	number	6.9
Z4	number	0.54
P4	number	0.2945985162
SM5	number	7.9
Z5	number	0.72
P5	number	0.2357624978
SM6	number	8.2
Z6	number	0.78
P6	number	0.2176954376
SM7	number	7.4
Z7	number	0.63
P7	number	0.2643472921

			Name	Type	Generated Value

Mu3	integer	32000
SD3	integer	13500
SM8	integer	30500
Z8	decimal	dec("-0.11")
P8	number	0.4562046875
SM9	integer	43000
Z9	decimal	dec("0.81")
P9	number	0.2089700879
SM10Lower	integer	13500
SM10Upper	integer	44500
Z10Lower	decimal	dec("-1.37")
Z10Upper	decimal	dec("0.93")
P10	number	0.7384710069
h	string

			Name	Type	Generated Value

Mu4	integer	50
SD4	integer	2
SM11	integer	53
Z11	decimal	dec("1.50")
P11	number	0.0668072013
P12	rational	19/20
Z12	number	1.64
SM12	number	46.72
SigLev	integer	95

			Name	Type	Generated Value

Mu5	integer	440
SD5	integer	19
SM13	integer	435
Z13	decimal	dec("-0.26")
P13	number	0.3974318868
n	integer	37
Z14	number	-1.60
P14	number	0.0547992917

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

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Generated value: `integer`

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

(a) Given that $X\sim N(\var{Mu1},\var{SD1}^2)$ . Find,

(i) $P(X<\var{SM1})$

Answer:Gap 0.

(ii) $P(X<\var{SM2})$

Answer:Gap 1.

(iii) $P(X<\var{SM3})$

Answer:Gap 2.

(b) Given that $X\sim N(\var{Mu2},\var{SD2}^2)$ . Find,

(i) $P(X>\var{SM4})$

Answer:Gap 3.

(ii) $P(X>\var{SM5})$

Answer:Gap 4.

(iii) $P(X>\var{SM6})$

Answer:Gap 5.

(iv) $P(X>\var{SM7})$

Answer:Gap 6.

Advice

Here are some tips on how to set out your work, as well as how to complete the questions. None of the answers given below have been rounded to what the question states.

Hint: If your Z-Score is a negative value, or the question is asking for $P(Z>z)$ then ensure you calculate $1 - \phi(z)$. E.g. $P(Z<-0.32) =1-P(Z<0.32) = 1-0.6255 = 0.3745$ or $P(Z>0.32)=1-P(Z<0.32) =1-0.6255 = 0.3745$ or $P(Z>-0.32)=1-(1-P(Z<0.32)) = 1 - (1 - 0.6255) =1 - 0.3745 = 0.6255$.

Question 1

1a: $X\sim N(\var{Mu1}, \var{SD1})$

(i) $P(X<\var{SM1}) =P(Z<\frac{\bar{X} - \mu}{\sigma})= P(Z<\frac{\var{SM1}-\var{Mu1}}{\var{SD1}})= P(Z<\var{Z1}) = \var{P1}$

(ii) $P(X<\var{SM2})=P(Z<\frac{\bar{X} - \mu}{\sigma})= P(Z<\frac{\var{SM2}-\var{Mu1}}{\var{SD1}})= P(Z<\var{Z2}) = \var{P2}$

(iii) $P(X<\var{SM3}) =P(Z<\frac{\bar{X} - \mu}{\sigma})= P(Z<\frac{\var{SM3}-\var{Mu1}}{\var{SD1}})= P(Z<\var{Z3}) = \var{P3}$

1b: $X\sim N(\var{Mu2}, \var{SD2})$

(i) $P(X>\var{SM4}) =P(Z>\frac{\bar{X} - \mu}{\sigma})= P(Z>\frac{\var{SM4}-\var{Mu2}}{\var{SD2}})= P(Z>\var{Z4}) = 1-P(Z<\var{Z4})=\var{P4}$

(ii) $P(X>\var{SM5}) =P(Z>\frac{\bar{X} - \mu}{\sigma})= P(Z>\frac{\var{SM5}-\var{Mu2}}{\var{SD2}})= P(Z>\var{Z5}) = 1-P(Z<\var{Z5})=\var{P5}$

(iii) $P(X>\var{SM6}) =P(Z>\frac{\bar{X} - \mu}{\sigma})= P(Z>\frac{\var{SM6}-\var{Mu2}}{\var{SD2}})= P(Z>\var{Z6}) = 1-P(Z<\var{Z6})= \var{P6}$

(iv) $P(X>\var{SM7}) =P(Z>\frac{\bar{X} - \mu}{\sigma})= P(Z>\frac{\var{SM7}-\var{Mu2}}{\var{SD2}})= P(Z>\var{Z7}) = 1-P(Z<\var{Z7})=\var{P7}$

Question 2:

2a: $X\sim N(\var{Mu3}, \var{SD3})$

(i) $P(X<\var{SM8}) =P(Z<\frac{\bar{X} - \mu}{\sigma})= P(Z<\frac{\var{SM8}-\var{Mu3}}{\var{SD3}})= P(Z<\var{Z8}) = \var{P8}$

(ii) $P(X<\var{SM9}) =P(Z<\frac{\bar{X} - \mu}{\sigma})= P(Z<\frac{\var{SM9}-\var{Mu3}}{\var{SD3}})= P(Z<\var{Z9}) = \var{P9}$

(iii) $P(\var{SM10Lower}<X<\var{SM10Upper}) =P(Z<\frac{\bar{X}_{upper} - \mu}{\sigma})-P(Z<\frac{\bar{X}_{lower} - \mu}{\sigma})= P(Z<\frac{\var{SM10Upper}-\var{Mu3}}{\var{SD3}})-P(Z<\frac{\var{SM10Lower}-\var{Mu3}}{\var{SD3}}) = P(Z<\var{Z10Upper}) - P(Z<\var{Z10Lower})=\var{P10}$

2b: $X\sim N(\var{Mu4}, \var{SD4})$

(i) $P(X>\var{SM11}) = 1 - P(X<\var{SM11}) =1 - P(Z<\frac{\bar{X} - \mu}{\sigma})= 1 - P(Z<\frac{\var{SM11}-\var{Mu4}}{\var{SD4}})= 1 - P(Z<\var{Z11}) = 1 - \var{1-P11} = \var{P11}$

(ii) $Z=\frac{X - \mu}{\sigma} \rightarrow X = Z \times \sigma + \mu = \var{-Z12} \times \var{SD4} + \var{Mu4} = \var{SM12}$

[Note: We use the negative version of the Z-score, as we want to know the weight exceeded by 95% of the bags, so the weight should be less than the mean of $\var{Mu4}$kg].

Question 3:

3a: $X\sim N(\var{Mu5}, \var{SD5}^2)$

(i) Since the normal distribution is repersented by $X \sim N(\mu,\sigma^2) \rightarrow X\sim N(\var{Mu5}, \var{SD5}^2)$

(ii) $P(X<\var{SM13}) =P(Z<\frac{\bar{X} - \mu}{\sigma})= P(Z<\frac{\var{SM13}-\var{Mu5}}{\var{SD5}})= P(Z<\var{Z13}) = \var{P13}$

(iii) $P(X=\var{SM13})=P(Z=\frac{\bar{X} - \mu}{\sigma/ \sqrt{n}}) = P(Z= \frac{\var{SM13}-\var{Mu5}}{\var{SD5}/\sqrt{\var{n}}}) = P(Z=\var{Z14}) = \var{P14}$

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This question is used in the following exams:

Week 3 test by Ellen Marshall in Ellen's workspace.
Normal Distribution by Brad Allison in Exam Version of Tests.
Apprentices Year 2 Inferential Statistics Quiz by Aoife O'Brien in Aoife's workspace.

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Brad Allison 4 years, 6 months ago

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Normal Distribution

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Taxonomy: mathcentre

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Brad Allison 4 years, 6 months ago

Brad Allison 4 years, 6 months ago

Brad Allison 5 years, 7 months ago

Brad Allison 5 years, 9 months ago

Brad Allison 5 years, 10 months ago

Brad Allison 5 years, 10 months ago

Error in variable testing condition

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