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\n\n

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.

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

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

Answer:[[0]]

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

Answer:[[1]]

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

Answer:[[2]]

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

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

Answer:[[3]]

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

Answer:[[4]]

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

Answer:[[5]]

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

Answer:[[6]]

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

Answer:[[7]]

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

Answer:[[8]]

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(a) The probability of a telesales representative making a sale on a customer call is {p4}. Find the probability that,

(i) no sales are made in {n4} calls.

Answer:[[0]]

(ii) more than {k11} sales are made in {n5} calls.

Answer:[[1]]

(b) Representatives are required to achieve a mean of {Expx} sales each day.

(i) Given that $E[X]=np$, find the least number of calls each day a representative should make to achieve this requirement.

Answer:[[2]]

(ii) Calculate the least number of calls that need to be made by a representative for the probability of at least 1 sale to exceed 0.95.

Answer:[[3]]

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(a) A factory produces components of which 1% are defective. The components are packed in boxes of {n6}. A box is selected at random.

(i) Find the probability that the box contains exactly 1 defective component.

Answer:[[0]]

(ii) Find the probability that there are at least 2 defective components in the box.

Answer:[[1]]

(a) The probability of a treatment for a particular disease alleviating all signs and symptoms is {p6}. Suppose that {n7} patients are treated, determine the probabilities that,

(i) there are no cures.

Answer:[[0]]

(ii) there are exactly 2 cures.

Answer:[[1]]

(iii) there are at least 3 cures.

Answer:[[2]]