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The number of patients arriving at a dentist’s surgery each afternoon follows
a Poisson distribution, with a mean of four patients per hour.
Calculate the probability that in a particular one-hour period

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The number of patients arriving at a dentist’s surgery each afternoon follows a Poisson distribution, with a mean of $\\var{l}$ patients per hour.

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The Poisson distribution formula: $P(r)=\\frac{\\lambda^re^{-\\lambda}}{r!}$ or $P(r)=e^{-\\lambda}\\left[\\frac{\\lambda^r}{r!}\\right]$

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Part (a)

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Remember that for a Poisson random variable:
\\begin{align}
\\operatorname{P}(X=x)&=\\dfrac{\\lambda^x\\times e^{-\\lambda}}{x!}\\\\
\\end{align}

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1.\$\\begin{eqnarray*}\\operatorname{P}(X = \\var{x}) &=& \\frac{\\var{l} ^ {\\var{x}}e ^ { -\\var{l}}} {\\var{x}!}\\\\& =& \\var{answer1} \\end{eqnarray*} \$ to 3 decimal places.

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Part (b)

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The probability that in a particular one hour period, the number of patients entering the waiting room will be between $\\var{x}$ and $\\var{y}$ inclusive is given by:

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$P(\\var{x} \\leq X\\leq\\var{y}) = P(X=\\var{x}) + P(X=\\var{x+1}) +P(X=\\var{y})$

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where

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$P(X=\\var{x}) =\\frac{\\var{l}^{\\var{x}}e^{-\\var{l}}}{\\var{x}!}=\\var{prx}$

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$P(X=\\var{x+1}) =\\frac{\\var{l}^{\\var{x+1}}e^{-\\var{l}}}{\\var{x+1}!}=\\var{prx1}$

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$P(X=\\var{y}) =\\frac{\\var{l}^{\\var{y}}e^{-\\var{l}}}{\\var{y}!}=\\var{pry}$

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Hence

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$P(\\var{x} \\leq X \\leq \\var{y})=$ $\\var{prx}+\\var{prx1}+\\var{pry}=\\var{answer2}$

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Write down the value of $\\lambda$ for one hour.

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$\\lambda =$ [[0]]

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Calculate the probability that in a particular one-hour period exactly $\\var{x}$ patients will arrive. Please give your answer to at least 3 decimal places.

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$P(r = \\var{x}) =$ [[0]]

Find the probability that in a particular one hour period, the number of patients entering the waiting room will be between $\\var{x}$ and $\\var{y}$ inclusive. Please give your answer to at least 3 decimal places.
$P(\\var{x}\\leq r \\leq\\var{y}) =$ [[0]]