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The time,  in minutes between customer arrivals at the RyanJet check-in desk at Newcastle Airport follows an exponential distribution with rate 0.6 i.e. 

X∼Exp(0.6)

a)

Find E[X] between customer arrivals at the RyanJet check-in desk at Newcastle Airport :

E[X]=? Expected answer: minutes (enter as a decimal correct to 3 decimal places).

Find Var(X):

Var(X)=? Expected answer: (enter as a decimal correct to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Find the probability that the time between two customers arriving is less than 1.3 minutes:

 P(X<1.3)=? Expected answer: (enter as a decimal correct to 3 decimal places)

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

If X∼Exp(λ) then E[X]=1λ and  Var(X)=1λ2.

Also P(X<a)=1−e−λa.

a)

If X∼Exp(0.6) then:

E[X]=1λ=10.6=1.667 to 3 decimal places.

Var(X)=1λ2=10.62=2.778 to 3 decimal places.

b)

P(X<1.3)=1−(e−0.6×1.3)=1−(e−0.78)=0.542 to 3 decimal places.
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A new supermarket plans to open somewhere on the outskirts of a town. In fact, $X$, the distance of a new supermarket from the town centre is Uniformly distributed between $\var{lower}$ metres and $\var{upper}$ metres i.e. 

\[X \sim \operatorname{U}(\var{lower},\var{upper})\]

a)

Find $\operatorname{E}[X]$, the expected distance in metres of the new supermarket from the town centre:

$\operatorname{E}[X]=\;?$ Expected answer: m  (to 3 decimal places).

Also find the variance $\operatorname{Var}(X)$:

 $\operatorname{Var}(X)=\;$? Expected answer: (to 3 decimal places).

 

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Find the probability that the supermarket opens within $\var{thisdis}$ kilometres of the town centre.

$P(X \le \var{thisdis}\textrm{km})=\;$? Expected answer:  (to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

a) For a Uniform distribution \[X \sim \operatorname{U}(\var{lower},\var{upper})\] we have:

$\displaystyle \operatorname{E}[X] = \frac{\var{lower}+\var{upper}}{2}=\var{ans1}$m

$\displaystyle \operatorname{Var}[X] = \frac{(\var{upper}-\var{lower})^2}{12}=\frac{(\var{upper-lower})^2}{12}=\var{ans2}$ to 3 decimal places.

b)

$\displaystyle P(X \le \var{thisdis}\textrm{km})=\frac{\var{thisdis}\times 1000 -\var{lower}}{\var{upper}-\var{lower}}=\var{ans3}$ to 3 decimal places.
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