Proving Poisson properties from first principles

4

A discrete random variable $X$ has a probability function defined by $$\mathrm { P } ( X = x ) = \frac { \mathrm { e } ^ { - \lambda } \lambda ^ { x } } { x ! } \quad \text { for } x = 0,1,2,3,4 , \ldots \ldots$$
1. State the name of the distribution of $X$.
2. Write down, in terms of $\lambda$, expressions for $\mathrm { E } ( 3 X - 1 )$ and $\operatorname { Var } ( 3 X - 1 )$.
3. Write down an expression for $\mathrm { P } ( X = x + 1 )$, and hence show that $$\mathrm { P } ( X = x + 1 ) = \frac { \lambda } { x + 1 } \mathrm { P } ( X = x )$$
The number of cars and the number of coaches passing a certain road junction may be modelled by independent Poisson distributions.
1. On a winter morning, an average of 500 cars per hour and an average of 10 coaches per hour pass this junction. Determine the probability that a total of at least 10 such vehicles pass this junction during a particular 1 -minute interval on a winter morning.
2. On a summer morning, an average of 836 cars per hour and an average of 22 coaches per hour pass this junction. Calculate the probability that a total of at most 3 such vehicles pass this junction during a particular 1 -minute interval on a summer morning. Give your answer to two significant figures.
  (3 marks)

7

The random variable $X$ has a Poisson distribution with $\mathrm { E } ( X ) = \lambda$.
1. Prove, from first principles, that $\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }$.
2. Hence deduce that $\operatorname { Var } ( X ) = \lambda$.
The independent Poisson random variables $X _ { 1 }$ and $X _ { 2 }$ are such that $\mathrm { E } \left( X _ { 1 } \right) = 5$ and $\mathrm { E } \left( X _ { 2 } \right) = 2$. The random variables $D$ and $F$ are defined by $$D = X _ { 1 } - X _ { 2 } \quad \text { and } \quad F = 2 X _ { 1 } + 10$$
1. Determine the mean and the variance of $D$.
2. Determine the mean and the variance of $F$.
3. For each of the variables $D$ and $F$, give a reason why the distribution is not Poisson.
The daily number of black printer cartridges sold by a shop may be modelled by a Poisson distribution with a mean of 5 . Independently, the daily number of colour printer cartridges sold by the same shop may be modelled by a Poisson distribution with a mean of 2. Use a distributional approximation to estimate the probability that the total number of black and colour printer cartridges sold by the shop during a 4 -week period ( 24 days) exceeds 175.

7 The random variable $X$ has a Poisson distribution with parameter $\lambda$.

1. Prove, from first principles, that $\mathrm { E } ( X ) = \lambda$.
2. Hence, given that $\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }$, find, in terms of $\lambda$, an expression for $\operatorname { Var } ( X )$.
The mode, $m$, of $X$ is such that $$\mathrm { P } ( X = m ) \geqslant \mathrm { P } ( X = m - 1 ) \quad \text { and } \quad \mathrm { P } ( X = m ) \geqslant \mathrm { P } ( X = m + 1 )$$
1. Show that $\lambda - 1 \leqslant m \leqslant \lambda$.
2. Given that $\lambda = 4.9$, determine $\mathrm { P } ( X = m )$.
The random variable $Y$ has a Poisson distribution with mode $d$ and standard deviation 15.5. Use a distributional approximation to estimate $\mathrm { P } ( Y \geqslant d )$.
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7

The random variable $X$ has a Poisson distribution with $\mathrm { E } ( X ) = \lambda$.
1. Prove, from first principles, that $\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }$.
2. Hence deduce that $\operatorname { Var } ( X ) = \mathrm { E } ( X )$.
The random variable $Y$ has a Poisson distribution with $\mathrm { E } ( Y ) = 2.5$. Given that $Z = 4 Y + 30$ :
1. show that $\operatorname { Var } ( Z ) = \mathrm { E } ( Z )$;
2. give a reason why the distribution of $Z$ is not Poisson.
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6 marks

6

The discrete random variable $X$ has probability distribution given by $$\mathrm { P } ( X = x ) = \begin{cases} \frac { \mathrm { e } ^ { - \lambda } \lambda ^ { x } } { x ! } & x = 0,1,2 , \ldots
0 & \text { otherwise } \end{cases}$$ Show that $\mathrm { E } ( X ) = \lambda$ and that $\operatorname { Var } ( X ) = \lambda$.
In light-weight chain, faults occur randomly and independently, and at a constant average rate of 0.075 per metre.
1. Calculate the probability that there are no faults in a 10 -metre length of this chain.
2. Use a distributional approximation to estimate the probability that, in a 500 -metre reel of light-weight chain, there are:
  (A) fewer than 30 faults;
  (B) at least 35 faults but at most 45 faults.
As part of an investigation into the quality of a new design of medium-weight chain, a sample of fifty 10 -metre lengths was selected. Subsequent analysis revealed a total of 49 faults.
Assuming that faults occur randomly and independently, and at a constant average rate, construct an approximate $98 \%$ confidence interval for the average number of faults per metre.
[0pt] [6 marks] \section*{DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED}

6 The random variable $X$ has a Poisson distribution with parameter $\lambda$.

Prove that $\mathrm { E } ( X ) = \lambda$.
By first proving that $\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }$, or otherwise, prove that $\operatorname { Var } ( X ) = \lambda$.

1 The random variable $X$ has a Poisson distribution with mean 16 Find the standard deviation of $X$ Circle your answer.
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