The Poisson distribution Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths The Poisson distribution questions. See exactly how to solve problems on poisson, probability-function, cumulative-probability, inequalities.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The random variable XX has a Poisson distribution, X∼Po(3.5)X \sim \text{Po}(3.5). Find P(X=2)P(X = 2), giving your answer to 44 decimal places.

Worked solution

  1. Write down the distribution and its parameter

    X∼Po(3.5),λ=3.5X\sim\text{Po}(3.5),\quad\lambda=3.5

    The parameter of a Poisson distribution is its mean, so λ=3.5\lambda=3.5.

  2. Evaluate the probability from the Poisson probability function

    P(X=2)=e−3.5(3.5)22!=0.184959P(X = 2)=\frac{e^{-3.5}\left(3.5\right)^{2}}{2!}=0.184959

    Each term is e−λλx/x!e^{-\lambda}\lambda^{x}/x! with λ=3.5\lambda=3.5.

  3. Round to the required accuracy

    P(X=2)=0.1850  (4 d.p.)P(X = 2)=0.1850\;(4\text{ d.p.})

    The unrounded value is 0.184959…0.184959\ldots, which rounds to 0.18500.1850.

  4. State the final answer

    P(X=2)=0.1850P(X = 2)=0.1850

    This is the required probability, correct to the stated accuracy.

Answer
P(X=2)=0.1850P(X = 2)=0.1850
Question 2
2 markseasy
The random variable XX has a Poisson distribution, X∼Po(2)X \sim \text{Po}(2). Find P(X=0)P(X = 0), giving your answer to 44 decimal places.

Worked solution

  1. Write down the distribution and its parameter

    X∼Po(2),λ=2X\sim\text{Po}(2),\quad\lambda=2

    The parameter of a Poisson distribution is its mean, so λ=2\lambda=2.

  2. Evaluate the probability from the Poisson probability function

    P(X=0)=e−2200!=0.135335P(X = 0)=\frac{e^{-2}2^{0}}{0!}=0.135335

    Each term is e−λλx/x!e^{-\lambda}\lambda^{x}/x! with λ=2\lambda=2.

  3. Round to the required accuracy

    P(X=0)=0.1353  (4 d.p.)P(X = 0)=0.1353\;(4\text{ d.p.})

    The unrounded value is 0.135335…0.135335\ldots, which rounds to 0.13530.1353.

  4. State the final answer

    P(X=0)=0.1353P(X = 0)=0.1353

    This is the required probability, correct to the stated accuracy.

Answer
P(X=0)=0.1353P(X = 0)=0.1353
Question 3
2 markseasy
The random variable XX has a Poisson distribution, X∼Po(4.5)X \sim \text{Po}(4.5). Find P(X=3)P(X = 3), giving your answer to 44 decimal places.

Worked solution

  1. Write down the distribution and its parameter

    X∼Po(4.5),λ=4.5X\sim\text{Po}(4.5),\quad\lambda=4.5

    The parameter of a Poisson distribution is its mean, so λ=4.5\lambda=4.5.

  2. Evaluate the probability from the Poisson probability function

    P(X=3)=e−4.5(4.5)33!=0.168718P(X = 3)=\frac{e^{-4.5}\left(4.5\right)^{3}}{3!}=0.168718

    Each term is e−λλx/x!e^{-\lambda}\lambda^{x}/x! with λ=4.5\lambda=4.5.

  3. Round to the required accuracy

    P(X=3)=0.1687  (4 d.p.)P(X = 3)=0.1687\;(4\text{ d.p.})

    The unrounded value is 0.168718…0.168718\ldots, which rounds to 0.16870.1687.

  4. State the final answer

    P(X=3)=0.1687P(X = 3)=0.1687

    This is the required probability, correct to the stated accuracy.

Answer
P(X=3)=0.1687P(X = 3)=0.1687
Question 4
2 markseasy
The random variable XX has a Poisson distribution, X∼Po(1.2)X \sim \text{Po}(1.2). Find P(X=1)P(X = 1), giving your answer to 44 decimal places.

Worked solution

  1. Write down the distribution and its parameter

    X∼Po(1.2),λ=1.2X\sim\text{Po}(1.2),\quad\lambda=1.2

    The parameter of a Poisson distribution is its mean, so λ=1.2\lambda=1.2.

  2. Evaluate the probability from the Poisson probability function

    P(X=1)=e−1.2(1.2)11!=0.361433P(X = 1)=\frac{e^{-1.2}\left(1.2\right)^{1}}{1!}=0.361433

    Each term is e−λλx/x!e^{-\lambda}\lambda^{x}/x! with λ=1.2\lambda=1.2.

  3. Round to the required accuracy

    P(X=1)=0.3614  (4 d.p.)P(X = 1)=0.3614\;(4\text{ d.p.})

    The unrounded value is 0.361433…0.361433\ldots, which rounds to 0.36140.3614.

  4. State the final answer

    P(X=1)=0.3614P(X = 1)=0.3614

    This is the required probability, correct to the stated accuracy.

Answer
P(X=1)=0.3614P(X = 1)=0.3614
Question 5
2 markseasy
The random variable XX has a Poisson distribution, X∼Po(6)X \sim \text{Po}(6). Find P(X=5)P(X = 5), giving your answer to 44 decimal places.

Worked solution

  1. Write down the distribution and its parameter

    X∼Po(6),λ=6X\sim\text{Po}(6),\quad\lambda=6

    The parameter of a Poisson distribution is its mean, so λ=6\lambda=6.

  2. Evaluate the probability from the Poisson probability function

    P(X=5)=e−6655!=0.160623P(X = 5)=\frac{e^{-6}6^{5}}{5!}=0.160623

    Each term is e−λλx/x!e^{-\lambda}\lambda^{x}/x! with λ=6\lambda=6.

  3. Round to the required accuracy

    P(X=5)=0.1606  (4 d.p.)P(X = 5)=0.1606\;(4\text{ d.p.})

    The unrounded value is 0.160623…0.160623\ldots, which rounds to 0.16060.1606.

  4. State the final answer

    P(X=5)=0.1606P(X = 5)=0.1606

    This is the required probability, correct to the stated accuracy.

Answer
P(X=5)=0.1606P(X = 5)=0.1606

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