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let us suppose we have following question : A Pizzeria gets an average $4.2$ orders in its closing hour, 11PM to midnight.what the probability that at least $30$ minutes will go by before next order ? exponential distribution is determined by $\lambda$ which is the average number of arrival per unit time, in our case orders are determined in units of hours, which means that per minute we have average

$\frac{60}{4.2}=14.28571429 $

we have formula for exponential distirbution $P(x>k)=1-P(x \leq k)$ or in our case

$p(x>k)=e^{-\lambda*k} $

substitution of $30$ we would have

$p(x>30)=e^{-14.28571429*30} $

did i calculate lambda correctly? it is also the inverse of expected duration ,for instance if we want to calculate average amount of time between orders we have

$\mu=\frac{1}{\lambda}$ or $0.07$

dato datuashvili
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1 Answers1

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I think you managed to reverse $\lambda$ and get an implausibly small probability of no orders in $30$ minutes of about $10^{-186}$

The average number of orders per hour is only $4.2$ so the average number of orders per minute is rather less: $\frac{4.2}{60}=0.07$

Assuming a Poisson process (exponentially distributed times between orders), you then have an expected $\frac{1}{0.07} \approx 14.2857$ minutes between orders. The probability of no orders in $30$ minutes would then be $e^{-0.07 \times 30}=e^{-2.1} \approx 0.122456$

In fact you could have got there another way, since $30$ minutes is half an hour, so the probability can also be written as $e^{-4.2 \times 1/2}$

Henry
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