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on average, a biker inflates their bike tires every 8 days. the interval between each time, namely t1, t2, and so on, is an exponential random variable. Find probability that 40 inflations are required in 1 year.

I know that an exponential distribution is like a poisson distribution but the former is continuouos. Therefore, the modus operandi is pretty the same, and it's straightforward. divide a line into steps, T1 -> T2 -> ... -> Tn, each step is an exponential distribution with 1/8 as parameter k. The expected value of exp(1/8) is equal to 8. and variance is equal to 64.

Now, I have the difficult part. I have to find P(X >= 365). I don't know how to continue.

By watching the solution (because I have the solution), I know that I should've used probability of normal distribution. P(N(0, 1)) and say that this probability distribution is greater than or equal to $$ (365-n*8) / (\sqrt(n*64)) $$

why do I need normal distribution in this exercise? I think my problem is in the logic of the exercise. What do you think about this choice of using normal distribution? why should I use it, and are there other ways to do that? and, why do I have the other part: "$(365-n*8) / (\sqrt(n*64)) $"?

this is the full solution, I've posted it because I think it makes the answer easier to understand, full solution below

EDIT: let's call a generic interval, T. T is an exp(1/8), so by plugging 1/8 in exp formula, and plugging 40 in x, I have $8.4224 * 10^(-4)$. now, I have to multiply it by 8, and I obtain 6.7379 * 10 ^ (-3). I think this is the correct solution.

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    $\begingroup$ You don't need the normal distribution. Please solve the exercise with the exponential distribution and compare it to the solution with the normal distribution. For large parameter of the exponential distribution the results should be close. $\endgroup$
    – Kurt G.
    Commented May 7, 2022 at 14:06
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    $\begingroup$ Take your time and write a nice solution by editing your question. I don't know the answer (yet). I want to see that you understood the priniciples. When I read nonsense like "the expected value of $\exp(1/8)$ is equal to $8$" I have to take a deep breathe . $\endgroup$
    – Kurt G.
    Commented May 7, 2022 at 14:14
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    $\begingroup$ "single interval is $1/8$". Don't get this. The expected value of every $T_i$ is $8$ days. What is the $\lambda$ ? Also be careful in which unit you measure time. days or years ? Your choice. $\endgroup$
    – Kurt G.
    Commented May 7, 2022 at 14:35
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    $\begingroup$ I know ! And what value has it ? $\endgroup$
    – Kurt G.
    Commented May 7, 2022 at 14:45
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    $\begingroup$ That's how the exercise is posed. They want $8$ days on average and we know that that average is $1/\lambda$. Now that we seem to begin to make progress slowly please take your time and write out the solution in a clean edit of the question. $\endgroup$
    – Kurt G.
    Commented May 7, 2022 at 14:49

1 Answer 1

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We know that the expected time between two tyre inflations is $8$ days, or $8/365\approx 0.0219$ years. Therefore the parameter of the exponentially distributed $T_i$ s is $\lambda=365/8=45.625$. The unit of this $\lambda$ is tyre inflations per year. It is known that the number of tyre inflations that are happening before time $t$ is a Poisson process $N_t$ and it is also known that this process has distibution $$\tag{1} \mathbb P(N_t=k)=\frac{(\lambda t)^k}{k!}e^{-\lambda t}\,. $$ The exercise asks for the probability $$\tag{A} \mathbb P(N_t=40)\text{ where }t=1\text{ (year), } $$ or, (there is ambiguity in the exercise) $$\tag{B} \mathbb P(N_t\ge 40)=1-\mathbb P(N_t<40)=1-\sum_{k=0}^{39}P(N_t=k) $$ Using formula (1) can you calculate both ?

I get for (A) a probability of 22.71% and for (B) I get 81.67%.

Why did I use Poisson ?

Because we were asked for the number of tyre inflations per year. The introduction using exponential times $T_i$ is just a test if you know the relationship beteen them and the counting (Poisson) process.

Now to the normal distribution. We know that the expectation and variance of $N_t$ are both equal to $\lambda t$. Then using the standard normal CDF I get the equivalent of (B) as $$ 1-\Phi\Big(\frac{39-\lambda}{\sqrt{\lambda}}\Big)\approx 83.67\%. $$ Close to 81.67% as expected.

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  • $\begingroup$ the second is 0.9562 $\endgroup$ Commented May 7, 2022 at 15:55
  • $\begingroup$ the first probability is 12, it sounds strange $\endgroup$ Commented May 7, 2022 at 15:56
  • $\begingroup$ why did you use poisson? it should be an exponential distribution $\endgroup$ Commented May 7, 2022 at 15:57
  • $\begingroup$ first formula: I'm plugging 40 into k, and t = 1. I think the problem is here, but it's strange because apparently there's nothing difficult. but 12 is an incorrect solution $\endgroup$ Commented May 7, 2022 at 16:09
  • $\begingroup$ I have to do the second again $\endgroup$ Commented May 7, 2022 at 16:12

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