Link Poisson to exponential

10 The number of hits per minute on a particular website has a Poisson distribution with mean 0.8. The time between successive hits is denoted by \(T\) minutes. Show that \(\mathrm { P } ( T > t ) = \mathrm { e } ^ { - 0.8 t }\) and hence show that \(T\) has a negative exponential distribution. Using a suitable approximation, which should be justified, find the probability that the time interval between the 1st hit and the 51st hit exceeds one hour.

9 Cotton cloth is sold from long rolls of cloth. The number of flaws on a randomly chosen piece of cloth of length \(a\) metres has a Poisson distribution with mean \(0.8 a\). The random variable \(X\) is the length of cloth, in metres, between two successive flaws.

1. Explain why, for \(x \geqslant 0 , \mathrm { P } ( X > x ) = \mathrm { e } ^ { - 0.8 x }\).
2. Find the probability that there is at least one flaw in a 4 metre length of cloth.
3. Find
   1. the distribution function of \(X\),
   2. the probability density function of \(X\),
   3. the interquartile range of \(X\).

7 The number of goals scored by a hockey team in an interval of time of length \(t\) minutes follows a Poisson distribution with mean \(\frac { 1 } { 24 } t\). The random variable \(T\) is defined as the length of time, in minutes, between successive goals.

1. (a) Show that \(\mathrm { P } ( T < t ) = 1 - \mathrm { e } ^ { - \frac { 1 } { 24 } t }\) for \(t \geqslant 0\).
   (b) Hence find the probability density function of \(T\).
2. Find the exact value of the interquartile range of \(T\).

The number of flaws in a randomly chosen 100 metre length of ribbon is modelled by a Poisson distribution with mean 1.6. The random variable \(X\) metres is the distance between two successive flaws. Show that the distribution function of \(X\) is given by $$\text{F}(x) = \begin{cases} 1 - e^{-0.016x} & x \geq 0, \\ 0 & x < 0, \end{cases}$$ and deduce that \(X\) has a negative exponential distribution, stating its mean. [4] Find

1. the median distance between two successive flaws, [3]
2. the probability that there is a distance of at least 50 metres between two successive flaws. [2]

The number of flaws in a randomly chosen 100 metre length of ribbon is modelled by a Poisson distribution with mean 1.6. The random variable \(X\) metres is the distance between two successive flaws. Show that the distribution function of \(X\) is given by $$\text{F}(x) = \begin{cases} 1 - e^{-0.016x} & x \geqslant 0, \\ 0 & x < 0, \end{cases}$$ and deduce that \(X\) has a negative exponential distribution, stating its mean. [4] Find

1. the median distance between two successive flaws, [3]
2. the probability that there is a distance of at least 50 metres between two successive flaws. [2]