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", "advice": "\n

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. 

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Question 1

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1a: $X\\sim Po(\\var{Lambda3})$

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(i) $P(X=x)=P(X=\\var{x5})=\\frac{e^{-\\var{Lambda3}}\\times\\var{Lambda3}^{\\var{x5}}}{\\var{x5}!}=\\var{Ans5}$

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(ii) $P(X=x)=P(X=\\var{x6})=\\frac{e^{-\\var{Lambda3}}\\times\\var{Lambda3}^{\\var{x6}}}{\\var{x6}!}=\\var{Ans6}$

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(iii) $P(X=x)=P(X=\\var{x7})=\\frac{e^{-\\var{Lambda3}}\\times\\var{Lambda3}^{\\var{x7}}}{\\var{x7}!}=\\var{Ans7}$

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1b: $X\\sim Po(\\var{Lambda4})$

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(i) $P(X=x)=P(X<\\var{x8})=P(X=0)+P(X=1)+P(X=2)=\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^0} {0!}+\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^1} {1!}+\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^2} {2!}=\\var{Ans8}$

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(ii) $P(X=x)=P(X\\leq\\var{x9})=P(X=0)+P(X=1)+...+P(X={\\var{x9}})=\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^0} {0!}+\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^1} {1!}+...+\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^{\\var{x9}}} {\\var{x9}!}=\\var{Ans9}$

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(iii) $P(X=x)=P(X>\\var{x10})=1-P(X\\leq\\var{x10})=1-(P(X=0)+P(X=1)+...+P(X={\\var{x10}}))=1-(\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^0} {0!}+\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^1} {1!}+...+\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^{\\var{x10}}} {\\var{x10}!})=\\var{Ans10}$

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(iv)$P(X=x)=P(X\\geq\\var{x11})=1-P(X<\\var{x11})=1-(P(X=0)+P(X=1)+...+P(X={5}))=1-(\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^0} {0!}+\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^1} {1!}+...+\\frac{e^{-\\var{Lambda4}}\\times\\var{Lambda4}^{5}} {5!})=\\var{Ans11}$

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Question 2

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2a: $\\lambda=rate  \\times hours = \\var{rate2} \\times 1 = \\var{rate2} \\therefore X\\sim Po(\\var{rate2})$

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(i) $P(X=x) = P(X=\\var{xx1})=\\frac{e^{-\\var{rate2}}\\times\\var{rate2}^\\var{xx1}}{\\var{xx1}!}=\\var{Ansxx}$

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2b: $\\lambda=rate  \\times hours =rate  \\times \\frac{mins}{60} = \\var{rate}\\times\\frac{\\var{mins}}{60}=\\var{Lambda1}  \\therefore X\\sim Po(\\var{Lambda1})$

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(i) $P(X=x) = P(X=\\var{x1})=\\frac{e^{-\\var{Lambda1}}\\times\\var{Lambda1}^\\var{x1}}{\\var{x1}!}=\\var{Ans1}$

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(ii) $P(X=x) = P(X\\geq\\var{x2})=1-P(X<\\var{x2})=1-(P(X=0)+P(X=1)+...+P(X={3}))=1-(\\frac{e^{-\\var{Lambda1}}\\times\\var{Lambda1}^0} {0!}+\\frac{e^{-\\var{Lambda1}}\\times\\var{Lambda1}^1} {1!}+...+\\frac{e^{-\\var{Lambda1}}\\times\\var{Lambda1}^{3}} {3!})=\\var{Ans2}$

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2c: $\\lambda=rate  \\times hours=rate  \\times \\frac{mins}{60} = \\var{rate}\\times\\frac{\\var{Totalmins}}{60}=\\var{Lambda1}  \\therefore X\\sim Po(\\var{Lambda1})$

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(i) This is the time difference in minutes, that the rate of patients arrives within. 

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(ii) $P(X=x) = P(X\\geq1)=1-P(X<1)=1-P(X=0)=1-\\frac{e^{-\\var{Lambda1}}\\times\\var{Lambda1}^0} {0!}=\\var{Ans4}$

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"definition": "random(15 .. 45#15)", "description": "", "templateType": "randrange"}, "Ans9": {"name": "Ans9", "group": "Question 1b", "definition": "if(x9=2,((e^(-Lambda4) * Lambda4^0)/0!)+((e^(-Lambda4) * Lambda4^1)/1!)+((e^(-Lambda4) * Lambda4^2)/2!), if(x9=3,((e^(-Lambda4) * Lambda4^0)/0!)+((e^(-Lambda4) * Lambda4^1)/1!)+((e^(-Lambda4) * Lambda4^2)/2!)+((e^(-Lambda4) * Lambda4^3)/3!),if(x9=4,((e^(-Lambda4) * Lambda4^0)/0!)+((e^(-Lambda4) * Lambda4^1)/1!)+((e^(-Lambda4) * Lambda4^2)/2!)+((e^(-Lambda4) * Lambda4^3)/3!)+((e^(-Lambda4) * Lambda4^4)/4!),if(x9=5,((e^(-Lambda4) * Lambda4^0)/0!)+((e^(-Lambda4) * Lambda4^1)/1!)+((e^(-Lambda4) * Lambda4^2)/2!)+((e^(-Lambda4) * Lambda4^3)/3!)+((e^(-Lambda4) * Lambda4^4)/4!)+((e^(-Lambda4) * Lambda4^5)/5!),0))))", "description": "", "templateType": "anything"}, "mins": {"name": "mins", "group": "Question 2a(i)", "definition": "random(30 .. 120#30)", "description": "", "templateType": "randrange"}, "Hours": {"name": "Hours", "group": "Question 2b", "definition": "Totaltime-11", "description": "", "templateType": "anything"}, "x7": {"name": "x7", "group": "Question 1a", "definition": "random(5 .. 5#1)", "description": "", "templateType": "randrange"}, "Lambda1": {"name": "Lambda1", "group": "Question 2a(i)", "definition": "hrs*rate", "description": "", "templateType": "anything"}, "x3": {"name": "x3", "group": "Question 2b", "definition": "0", "description": "", "templateType": "anything"}, "x8": {"name": "x8", "group": "Question 1b", "definition": "3", "description": "", "templateType": "number"}, "Ans2": {"name": "Ans2", "group": "Question 2a(ii)", "definition": "1-(if(x2=2,((e^(-Lambda1) * Lambda1^0)/0!)+((e^(-Lambda1) * Lambda1^1)/1!), if(x2=3,((e^(-Lambda1) * Lambda1^0)/0!)+((e^(-Lambda1) * Lambda1^1)/1!)+((e^(-Lambda1) * Lambda1^2)/2!),if(x2=4,((e^(-Lambda1) * Lambda1^0)/0!)+((e^(-Lambda1) * Lambda1^1)/1!)+((e^(-Lambda1) * Lambda1^2)/2!)+((e^(-Lambda1) * Lambda1^3)/3!),if(x2=5,((e^(-Lambda1) * 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And

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(a) Given that  $X\\sim Po\\left({\\var{Lambda3}}\\right)$. Find,

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(i) $P\\left(X=\\var{x5}\\right)$

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Answer:[[0]]

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(ii) $P\\left(X=\\var{x6}\\right)$

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Answer:[[1]]

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(iii) $P\\left(X=\\var{x7}\\right)$

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Answer:[[2]]

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(b) Given that  $X\\sim Po\\left({\\var{Lambda4}}\\right)$. Find,

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(i) $P(X<\\var{x8})$ 

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Answer:[[3]]

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(ii) $P(X\\le\\var{x9})$

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Answer:[[4]]

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(iii) $P(X>\\var{x10})$

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Answer:[[5]]

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(iv) $P(X\\ge\\var{x11})$

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Answer:[[6]]

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(a) Patients arrive at a hospital A & E department at random at a rate of {rate2} per hour.  

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Let x be the r.v of the number of patients arriving per hour. 

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Find the probability that, during any hour period, the number of patients arriving at the hospital accident and emergency department is

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(i) exactly {xx1}

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Answer: [[4]]

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(b) Patients arrive at a hospital A & E department at random at a rate of {rate} per hour.  

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Let x be the r.v of the number of patients arriving per {mins} mins. 

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Find the probability that, during any {mins} minute period, the number of patients arriving at the hospital accident and emergency department is

\n

(i) exactly {x1} 

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Answer: [[0]]

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(ii) at least {x2}

\n

Answer: [[1]]

\n

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(c) A patient arrives at 11.00am. To find the probability that the next patient arrives before {Timehrs}.{Timemins}{Daypart}, x needs to be redefined.

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(i) Redefine the random variable x for part b by completing the below statement. 

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Let x be the r.v of the number of patients arriving per [[2]] mins. 

\n

(ii) Now find the probability that the next patient arrives before {Timehrs}.{Timemins}{Daypart}.

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Answer: [[3]]

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