Further Maths The Poisson distribution Practice Questions

Free Further Maths The Poisson distribution practice questions with full step-by-step worked solutions. Covers poisson, probability-function, cumulative-probability, inequalities. Practise exam-style problems and check your method.

poissonprobability-functioncumulative-probabilityinequalitiescomplementmean-and-variance
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.
Show worked solution

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 number of flaws in a randomly chosen 11 metre length of cloth is to be modelled by a Poisson distribution. Which of the following is NOT one of the conditions required for a Poisson model?
Show worked solution

Worked solution

  1. List the conditions for a Poisson model

    singly,  independently,  constant mean rate\text{singly},\;\text{independently},\;\text{constant mean rate}

    Events must occur singly, independently of one another, and at a constant average rate.

  2. Note the consequence for the mean and the variance

    E(X)=Var(X)=λ\text{E}(X)=\text{Var}(X)=\lambda

    Equality of the mean and the variance is a property of the Poisson distribution.

  3. Contrast with the binomial conditions

    B(n,p):  n fixed trials,  p constant\text{B}(n,p):\;n\text{ fixed trials},\;p\text{ constant}

    A fixed number of trials is a BINOMIAL condition, not a Poisson one; a Poisson variable has no upper bound.

  4. Select the option that answers the question

    Events always occur in pairs\text{Events always occur in pairs}

    The other four options are all genuine requirements of a Poisson model.

Answer
Events always occur in pairs\text{Events always occur in pairs}
Question 3
4 marksintermediate
The machines in a factory break down from time to time. When one machine breaks down, the extra load placed on the remaining machines makes a further breakdown more likely. Which one of the conditions required for a Poisson model is violated by this situation?
Show worked solution

Worked solution

  1. List the conditions for a Poisson model

    singly,  independently,  constant mean rate\text{singly},\;\text{independently},\;\text{constant mean rate}

    Events must occur singly, independently of one another, and at a constant average rate.

  2. Note the consequence for the mean and the variance

    E(X)=Var(X)=λ\text{E}(X)=\text{Var}(X)=\lambda

    Equality of the mean and the variance is a property of the Poisson distribution.

  3. Contrast with the binomial conditions

    B(n,p):  n fixed trials,  p constant\text{B}(n,p):\;n\text{ fixed trials},\;p\text{ constant}

    A fixed number of trials is a BINOMIAL condition, not a Poisson one; a Poisson variable has no upper bound.

  4. Test the situation against each condition in turn

    singly?  independently?  constant rate?\text{singly?}\;\text{independently?}\;\text{constant rate?}

    Each condition is checked against the description given in the question.

  5. Identify the condition that fails

    one breakdown changes the chance of another\text{one breakdown changes the chance of another}

    The occurrence of one event alters the probability of the next, so the events are not independent.

  6. State the violated condition

    Events occur independently of one another\text{Events occur independently of one another}

    This is the condition that the situation described breaks.

Answer
Events occur independently of one another\text{Events occur independently of one another}
Question 4
6 markshard
The random variables XX and YY are independent, with X∼Po(1.8)X \sim \text{Po}(1.8) and Y∼Po(2.2)Y \sim \text{Po}(2.2). The random variable TT is defined by T=X+YT = X + Y. Which of the following is the value of P(T≥5)P(T \ge 5), correct to 44 decimal places?
Show worked solution

Worked solution

  1. Quote the additivity property of independent Poisson variables

    X+Y∼Po(λ1+λ2)X+Y\sim\text{Po}\left(\lambda_{1}+\lambda_{2}\right)

    If the variables are independent, the sum of Poisson variables is Poisson with the sum of the means. Independence is stated in the question and is essential.

  2. Add the means

    λ=1.8+2.2=4\lambda=1.8+2.2=4

    The parameter of the combined distribution is the total of the separate means.

  3. Write down the distribution of TT

    T∼Po(4)T\sim\text{Po}(4)

    The total count is Poisson with mean λ=4\lambda=4.

  4. Write down the Poisson probability function

    P(T=x)=e−λλxx!,x=0,1,2,…P(T=x)=\frac{e^{-\lambda}\lambda^{x}}{x!},\quad x=0,1,2,\ldots

    This is the probability function of a Poisson distribution with mean λ\lambda.

  5. Convert the upper tail into a cumulative probability

    P(T≥5)=1−P(T≤4)P(T \ge 5)=1-P(T\le 4)

    A Poisson variable has no upper limit, so an upper tail is found as one minus a cumulative probability. Note that P(T≥5)P(T \ge 5) and P(T>5)P(T > 5) are different: here the terms x=0,…,4x=0,\ldots,4 are the ones subtracted.

  6. Evaluate P(T=0)P(T = 0)

    P(T=0)=e−4400!=0.018316P(T=0)=\frac{e^{-4}4^{0}}{0!}=0.018316

    Substitute x=0x=0 and λ=4\lambda=4 into the Poisson probability function.

  7. Evaluate P(T=1)P(T = 1)

    P(T=1)=e−4411!=0.073263P(T=1)=\frac{e^{-4}4^{1}}{1!}=0.073263

    Substitute x=1x=1 and λ=4\lambda=4 into the Poisson probability function.

  8. Evaluate P(T=2)P(T = 2)

    P(T=2)=e−4422!=0.146525P(T=2)=\frac{e^{-4}4^{2}}{2!}=0.146525

    Substitute x=2x=2 and λ=4\lambda=4 into the Poisson probability function.

  9. Evaluate P(T=3)P(T = 3)

    P(T=3)=e−4433!=0.195367P(T=3)=\frac{e^{-4}4^{3}}{3!}=0.195367

    Substitute x=3x=3 and λ=4\lambda=4 into the Poisson probability function.

  10. Evaluate P(T=4)P(T = 4)

    P(T=4)=e−4444!=0.195367P(T=4)=\frac{e^{-4}4^{4}}{4!}=0.195367

    Substitute x=4x=4 and λ=4\lambda=4 into the Poisson probability function.

  11. Add the probabilities

    P(T≤4)=0.018316+0.073263+0.146525+0.195367+0.195367=0.628837P(T\le 4)=0.018316+0.073263+0.146525+0.195367+0.195367=0.628837

    The individual terms are added. The inequality is not strict, so x=5x=5 is included in the tail.

  12. Subtract from 11

    P(T≥5)=1−0.628837=0.371163P(T \ge 5)=1-0.628837=0.371163

    The complement of the cumulative probability gives the upper tail.

  13. Round to the required accuracy

    P(T≥5)=0.3712  (4 d.p.)P(T \ge 5)=0.3712\;(4\text{ d.p.})

    The unrounded value is 0.371163…0.371163\ldots, which rounds to 0.37120.3712.

  14. Select the matching option

    P(T≥5)=0.3712P(T \ge 5)=0.3712

    This is the value of the required probability.

Answer
P(T≥5)=0.3712P(T \ge 5)=0.3712
Question 5
9 markschallenging
The random variable YY has the binomial distribution Y∼B(150,0.02)Y \sim \text{B}(150, 0.02). Which of the following is the exact value of P(Y≤5)P(Y \le 5), using the binomial distribution, correct to 44 decimal places?
Show worked solution

Worked solution

  1. Write down the exact distribution

    Y∼B(150,0.02)Y\sim\text{B}(150,0.02)

    The question asks for the exact binomial probability, not a Poisson approximation.

  2. Write down the binomial probability function

    P(Y=y)=(150y)(0.02)y(0.98)150−yP(Y=y)=\binom{150}{y}\left(0.02\right)^{y}\left(0.98\right)^{150-y}

    This is the exact model; the Poisson distribution would only approximate it.

  3. Evaluate P(Y=0)P(Y = 0)

    P(Y=0)=(1500)(0.02)0(0.98)150=0.048296P(Y=0)=\binom{150}{0}\left(0.02\right)^{0}\left(0.98\right)^{150}=0.048296

    Substitute y=0y=0 into the binomial probability function.

  4. Evaluate P(Y=1)P(Y = 1)

    P(Y=1)=(1501)(0.02)1(0.98)149=0.147845P(Y=1)=\binom{150}{1}\left(0.02\right)^{1}\left(0.98\right)^{149}=0.147845

    Substitute y=1y=1 into the binomial probability function.

  5. Evaluate P(Y=2)P(Y = 2)

    P(Y=2)=(1502)(0.02)2(0.98)148=0.224785P(Y=2)=\binom{150}{2}\left(0.02\right)^{2}\left(0.98\right)^{148}=0.224785

    Substitute y=2y=2 into the binomial probability function.

  6. Evaluate P(Y=3)P(Y = 3)

    P(Y=3)=(1503)(0.02)3(0.98)147=0.226314P(Y=3)=\binom{150}{3}\left(0.02\right)^{3}\left(0.98\right)^{147}=0.226314

    Substitute y=3y=3 into the binomial probability function.

  7. Evaluate P(Y=4)P(Y = 4)

    P(Y=4)=(1504)(0.02)4(0.98)146=0.169735P(Y=4)=\binom{150}{4}\left(0.02\right)^{4}\left(0.98\right)^{146}=0.169735

    Substitute y=4y=4 into the binomial probability function.

  8. Evaluate P(Y=5)P(Y = 5)

    P(Y=5)=(1505)(0.02)5(0.98)145=0.101148P(Y=5)=\binom{150}{5}\left(0.02\right)^{5}\left(0.98\right)^{145}=0.101148

    Substitute y=5y=5 into the binomial probability function.

  9. Add the terms

    P(Y≤5)=0.048296+0.147845+0.224785+0.226314+0.169735+0.101148=0.918123P(Y\le 5)=0.048296+0.147845+0.224785+0.226314+0.169735+0.101148=0.918123

    These are all the outcomes required.

  10. Round to the required accuracy

    P(Y≤5)=0.9181  (4 d.p.)P(Y \le 5)=0.9181\;(4\text{ d.p.})

    The unrounded exact value is 0.918123…0.918123\ldots.

  11. Compare with the Poisson approximation

    λ=np=3,P(Y≤5)≈0.9161  (Poisson)versus0.9181  (exact binomial)\lambda=np=3,\quad P(Y \le 5)\approx0.9161\;\text{(Poisson)}\quad\text{versus}\quad 0.9181\;\text{(exact binomial)}

    The Poisson approximation with λ=np=3\lambda=np=3 gives 0.91610.9161, compared with the exact binomial value 0.91810.9181, both to the stated accuracy. The Poisson value is only an estimate of the exact binomial probability.

  12. Recall the conditions for a Poisson model

    singly, independently, at a constant mean rate\text{singly, independently, at a constant mean rate}

    A Poisson model requires events to occur singly, independently and at a constant mean rate.

  13. Recall the mean and variance of a Poisson distribution

    E(X)=Var(X)=λ\text{E}(X)=\text{Var}(X)=\lambda

    For a Poisson distribution the mean and the variance are both equal to λ\lambda.

  14. Note that the probabilities of a Poisson distribution sum to 11

    ∑x=0∞e−λλxx!=1\sum_{x=0}^{\infty}\frac{e^{-\lambda}\lambda^{x}}{x!}=1

    The Poisson probabilities form a complete probability distribution.

  15. Select the matching option

    P(Y≤5)=0.9181P(Y \le 5)=0.9181

    This is the exact binomial probability.

Answer
P(Y≤5)=0.9181P(Y \le 5)=0.9181

Unlock 65 more The Poisson distribution questions

Create a free account to work through every Further Maths The Poisson distribution question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More The Poisson distribution practice

Hard QuestionsWorked Solutions

Related Statistics topics

Central limit theoremChi-squared testsCombinations of random variablesContinuous random variablesThe continuous uniform distributionDiscrete random variables