Binomial Distribution

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Question statement

The following questions all require calculations using the binomial distribution. Before you start you may wish to write down any formulae you may need. Also please note that you need to answer every question to 4 decimal places or it will be marked incorrect.

Remember that the numbers will change if you start the test again so you can practice as many times as you like.

Give any introductory information the student needs.

			Name	Type	Generated Value

n1	integer	13
p1	number	0.2
k1	integer	2
Ans1	number	0.2680059593
k2	integer	5
Ans2	number	0.0690952864
k3	integer	6
Ans3	number	0.0230317621
n2	integer	14
p2	number	0.21
k4	integer	3
j1	integer	2
Ans4	number	0.4112650985
k5	integer	4
j2	integer	3
Ans5	number	0.6634141112
k6	integer	6
j3	integer	5
Ans6	number	0.9457333331
n3	integer	12
p3	number	0.8
k7	integer	3
Ans7	number	0.0000621978
k8	integer	4
Ans8	number	0.9994187571
k9	integer	3
j4	integer	2
Ans9	number	0.9999954739

			Name	Type	Generated Value

p4	number	0.15
k10	integer	0
n4	integer	8
Ans10	number	0.272490525
k11	integer	2
n5	integer	22
Ans11	number	0.6618230913
Expx	integer	7
n	number	46.6666666667
Ans12	integer	47
q	number	0.95
r	number	0.05
nn	number	18.4331282689
Ans13	integer	19
t	number	0.85

			Name	Type	Generated Value

p5	rational	1/100
n6	integer	8
k13	integer	1
Ans14	number	0.0745652278
j5	integer	2
k14	integer	1
Ans15	number	0.0001

			Name	Type	Generated Value

p6	number	0.3
n7	integer	9
k15	integer	0
Ans16	number	0.040353607
k16	integer	2
Ans17	number	0.266827932
j6	integer	3
k17	integer	2
Ans18	number	0.537168834

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

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: `integer`

This variable doesn't seem to be used anywhere.

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

(a) Given that $X\sim Bi(\var{n1},\var{p1})$ . Find,

(i) $P(X=\var{k1})$

Answer:Gap 0.

(ii) $P(X=\var{k2})$

Answer:Gap 1.

(iii) $P(X=\var{k3})$

Answer:Gap 2.

(b) Given that $X\sim Bi(\var{n2}, \var{p2})$ . Find,

(i) $P(X<\var{k4})$

Answer:Gap 3.

(ii) $P(X<\var{k5})$

Answer:Gap 4.

(iii) $P(X<\var{k6})$

Answer:Gap 5.

(i) $P(X\leq\var{k7})$

Answer:Gap 6.

(ii) $P(X>\var{k8})$

Answer:Gap 7.

(iii) $P(X\geq\var{k9})$

Answer:Gap 8.

Advice

Here are some tips on how to set out your work, as well as how to complete the questions. As the numbers you saw were randomly generated, these are examples rather than the solutions to the questions you answered. None of the answers given below have been rounded but you need to round yours to 4 d.p. .

Question 1

1a: $X\sim B(\var{n1},\var{p1})$

(i) $P(X=\var{k1})$

$P(X=k)={n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X=\var{k1}) = {\var{n1} \choose \var{k1}} \var{p1}^{\var{k1}} (1-\var{p1})^{\var{n1}-\var{k1}}=\var{Ans1}$

(ii) $P(X=\var{k2})$

$P(X=k)={n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X=\var{k2}) = {\var{n1} \choose \var{k2}} \var{p1}^{\var{k2}} (1-\var{p1})^{\var{n1}-\var{k2}}=\var{Ans2}$

(iii) $P(X=\var{k3})$

$P(X=k)={n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X=\var{k3}) = {\var{n1} \choose \var{k3}} \var{p1}^{\var{k3}} (1-\var{p1})^{\var{n1}-\var{k3}}=\var{Ans3}$

1b: $X\sim B(\var{n2}, \var{p2})$

(i) $P(X<\var{k4})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X<\var{k4})=P(X\leq \var{k4}-1)=P(X\leq{\var{j1}})=\sum_{k=0}^\var{n2} {\var{n2} \choose \var{j1}} \var{p2}^{\var{j1}} (1-\var{p2})^{\var{n2}-\var{j1}}={\var{n2} \choose 0} \var{p2}^{0} (1-\var{p2})^{\var{n2}-0}+{\var{n2} \choose 1} \var{p2}^{1} (1-\var{p2})^{\var{n2}-1} + ... +{\var{n2} \choose \var{j1}} \var{p2}^{\var{j1}} (1-\var{p2})^{\var{n2}-\var{j1}} = \var{Ans4}$

(ii) $P(X<\var{k5})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X<\var{k5})=P(X\leq \var{k5}-1)=P(X\leq{\var{j2}})=\sum_{k=0}^\var{n2} {\var{n2} \choose \var{j2}} \var{p2}^{\var{j2}} (1-\var{p2})^{\var{n2}-\var{j2}}={\var{n2} \choose 0} \var{p2}^{0} (1-\var{p2})^{\var{n2}-0}+{\var{n2} \choose 1} \var{p2}^{1} (1-\var{p2})^{\var{n2}-1} + ... +{\var{n2} \choose \var{j2}} \var{p2}^{\var{j2}} (1-\var{p2})^{\var{n2}-\var{j2}} = \var{Ans5}$

(iii) $P(X<\var{k6})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X<\var{k6})=P(X\leq \var{k6}-1)=P(X\leq{\var{j3}})=\sum_{k=0}^\var{n2} {\var{n2} \choose \var{j3}} \var{p2}^{\var{j3}} (1-\var{p2})^{\var{n2}-\var{j3}}={\var{n2} \choose 0} \var{p2}^{0} (1-\var{p2})^{\var{n2}-0}+{\var{n2} \choose 1} \var{p2}^{1} (1-\var{p2})^{\var{n2}-1} + ... +{\var{n2} \choose \var{j3}} \var{p2}^{\var{j3}} (1-\var{p2})^{\var{n2}-\var{j3}} = \var{Ans6}$

1c: $X\sim B(\var{n3}, \var{p3})$

(i) $P(X\leq\var{k7})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X\leq \var{k7})=\sum_{k=0}^\var{n3} {\var{n3} \choose \var{k7}} \var{p3}^{\var{k7}} (1-\var{p3})^{\var{n3}-\var{k7}}={\var{n3} \choose 0} \var{p3}^{0} (1-\var{p3})^{\var{n3}-0}+{\var{n3} \choose 1} \var{p3}^{1} (1-\var{p3})^{\var{n3}-1} + ... +{\var{n3} \choose \var{k7}} \var{p3}^{\var{k7}} (1-\var{p3})^{\var{n3}-\var{k7}} = \var{Ans7}$

(ii) $P(X>\var{k8})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k}\rightarrow P(X>\var{k8})=1-P(X\leq \var{k8})=1 - (\sum_{k=0}^\var{n3} {\var{n3} \choose \var{k8}} \var{p3}^{\var{k8}} (1-\var{p3})^{\var{n3}-\var{k8}})=1-({\var{n3} \choose 0} \var{p3}^{0} (1-\var{p3})^{\var{n3}-0}+{\var{n3} \choose 1} \var{p3}^{1} (1-\var{p3})^{\var{n3}-1} + ... +{\var{n3} \choose \var{k8}} \var{p3}^{\var{k8}} (1-\var{p3})^{\var{n3}-\var{k8}}) = \var{Ans8}$

(iii) $P(X\geq\var{k9})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k}\rightarrow P(X\geq\var{k9})=1-P(X\leq \var{k9} - 1)=1-P(X\leq \var{j4} )=1 - (\sum_{k=0}^\var{n3} {\var{n3} \choose \var{j4}} \var{p3}^{\var{j4}} (1-\var{p3})^{\var{n3}-\var{j4}})=1-({\var{n3} \choose 0} \var{p3}^{0} (1-\var{p3})^{\var{n3}-0}+{\var{n3} \choose 1} \var{p3}^{1} (1-\var{p3})^{\var{n3}-1} + ... +{\var{n3} \choose \var{j4}} \var{p3}^{\var{j4}} (1-\var{p3})^{\var{n3}-\var{j4}}) = \var{Ans9}$

Question 2

2a: $X\sim B(\var{n4},\var{p4})$

(i) $P(X=\var{k10})$

$P(X=k)={n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X=\var{k10}) = {\var{n4} \choose \var{k10}} \var{p4}^{\var{k10}} (1-\var{p4})^{\var{n4}-\var{k10}}=\var{Ans10}$

(ii) $P(X>\var{k11})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k}\rightarrow P(X>\var{k11})=1-P(X\leq \var{k11})=1 - (\sum_{k=0}^\var{n4} {\var{n4} \choose \var{k11}} \var{p4}^{\var{k11}} (1-\var{p4})^{\var{n4}-\var{k11}})=1-({\var{n4} \choose 0} \var{p4}^{0} (1-\var{p4})^{\var{n4}-0}+{\var{n4} \choose 1} \var{p4}^{1} (1-\var{p4})^{\var{n4}-1} + ... +{\var{n4} \choose \var{k11}} \var{p4}^{\var{k11}} (1-\var{p4})^{\var{n4}-\var{k11}}) = \var{Ans11}$

2b: $X\sim B(n,\var{p4})$

(i) Find n.

$E(x) =np \rightarrow n=\frac{E(x)}{p} \therefore n= \frac{\var{Expx}}{\var{p4}} \therefore n=\var{n} \therefore n=\var{Ans12}$

(ii) Find n.

$P(X\geq 1) > \var{q} \therefore 1-P(X=0)>\var{q} \therefore P(X=0)< \var{r} \rightarrow{n \choose k} p^{k} (1-p)^{n-k} \therefore {n \choose 0} \var{p4}^{0} (1-\var{p4})^{n-0} < \var{r} \therefore \var{t}^n < \var{r} \therefore nln(\var{t})<ln(\var{r}) \therefore n>\frac{ln(\var{r})}{ln(\var{t})} \therefore n > \var{nn} \therefore n>\var{Ans13} $

Question 3

3a: $X\sim B(\var{n6}, \var{p5})$

(i) $P(X=\var{k13})$

$P(X=k)={n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X=\var{k13}) = {\var{n6} \choose \var{k13}} \var{p5}^{\var{k13}} (1-\var{p5})^{\var{n6}-\var{k13}}=\var{Ans14}$

(ii) $P(X\geq\var{j5})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k}\rightarrow P(X\geq\var{j5})=1-P(X\leq \var{j5} - 1)=1-P(X\leq \var{k14} )=1 - (\sum_{k=0}^\var{n6} {\var{n6} \choose \var{k14}} \var{p5}^{\var{k14}} (1-\var{p5})^{\var{n6}-\var{k14}})=1-({\var{n6} \choose 0} \var{p5}^{0} (1-\var{p5})^{\var{n6}-0}+{\var{n6} \choose 1} \var{p5}^{1} (1-\var{p5})^{\var{n6}-1})= \var{Ans15}$

Question 4

4a: $X\sim B(\var{n7},\var{p6})$

(i) $P(X=\var{k15})$

$P(X=k)={n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X=\var{k15}) = {\var{n7} \choose \var{k15}} \var{p6}^{\var{k15}} (1-\var{p6})^{\var{n7}-\var{k15}}=\var{Ans16}$

(ii) $P(X=\var{k16})$

$P(X=k)={n \choose k} p^{k} (1-p)^{n-k} \rightarrow P(X=\var{k16}) = {\var{n7} \choose \var{k16}} \var{p6}^{\var{k16}} (1-\var{p6})^{\var{n7}-\var{k16}}=\var{Ans17}$

(iii) $P(X\geq\var{j6})$

$ P(X\leq k) =\sum_{k=0}^n {n \choose k} p^{k} (1-p)^{n-k}\rightarrow P(X\geq\var{j6})=1-P(X\leq \var{j6} - 1)=1-P(X\leq \var{k17} )=1 - (\sum_{k=0}^\var{n7} {\var{n7} \choose \var{k17}} \var{p6}^{\var{k17}} (1-\var{p6})^{\var{n7}-\var{k17}})=1-({\var{n7} \choose 0} \var{p6}^{0} (1-\var{p6})^{\var{n7}-0}+{\var{n7} \choose 1} \var{p6}^{1} (1-\var{p6})^{\var{n7}-1}+...+{\var{n7} \choose \var{k17}} \var{p6}^{\var{k17}} (1-\var{p6})^{\var{n7}-\var{k17}})= \var{Ans18} $

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Brad Allison 5 years, 7 months ago

Brad Allison 5 years, 9 months ago

Brad Allison 5 years, 9 months ago

Brad Allison 5 years, 9 months ago

Error in variable testing condition

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Generated value: `integer`

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Binomial Distribution

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

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History

Comment

Checkpoint description

Brad Allison 5 years, 7 months ago

Brad Allison 5 years, 9 months ago

Brad Allison 5 years, 9 months ago

Brad Allison 5 years, 9 months ago

Error in variable testing condition

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Generated value: integer

Parts

Gap-fill

Generated value: `integer`