can anyone please check if i've done it correctly?

The waiting time in minutes for a bus in a certain station(from getting to the station until the bus arrival) is a random exponential variable with a mean of $10$. But if the the traffic is busy, then its mean is $20$. The probability that the traffic will be busy when waiting for the bus is $0.18$.

1)what is the probability that a man that arrives in random time will wait longer than $15$ minutes?

2)if a man waits longer than $15$ minutes,what is the probability that the traffic is busy?

3)if a man comes in a random day and after $8$ minutes still waiting, what is the probability he'll have to wait for not longer than $15$ minutes?

what i did:

1)i calculated the expected mean and variance in order to understand the deviation from the mean. so using the formulas of expected value and expected mean i obtained that: expected mean $= 11.8$, expected variance $= 14.76$, then took the root of the expected variance to find the normal distribution $(\sim3.84)$ and then i had to find the $z$ score, so i computed it and obtained $0.2$

2)i defined the events a - being late, $b - x>15$, and then used bayes decomposed formula $p(a|b)=\frac{p(b|a)p(a)}{p(b)}$ where $p(a) = 0.18, p(b) = 0.2$, and calculated $p(b|a)$ by using the mean 20 so that $\frac{0.18*0.9}{0.2}=0.81$

3)i subtracted the expected mean from 1) by $8$, i.e $11.8-8=3.8$, but i don't know how to continue in order to use the $z$-score here.

would appreciate your help and assistance with this. thank you very much for your help!

  • $\begingroup$ For what reason you used the normal distribution? $\endgroup$ – callculus May 31 '18 at 13:08
  • $\begingroup$ i used the normal distribution in order to understand the deviation from the mean $\endgroup$ – BeginningMath May 31 '18 at 15:54
  • For the first question, I don't see a justification to use the normal distribution.

Using the notation in your second part (for your second part, being late should be being busy), \begin{align}P(B)&=P(B|A)P(A)+P(B|A^c)P(A^c) \\ &=\exp\left(-\frac{15}{20}\right)\cdot 0.18+\exp\left(-\frac{15}{10} \right)\cdot 0.82\end{align}

  • For the second part, Bayes rule is indeed the right tool. You might like to change the numbers for $P(B|A)$ and $P(B)$.

  • Guide for the third part, let $X$ denotes the waiting time.

\begin{align} P(X \le 15|X>8) &= \frac{P(8<X \le 15)}{P(X>8)}\\ &=\frac{P(8<X \le 15|A)P(A)+P(8<X \le 15|A^C)P(A^C)}{P(8<X |A)P(A)+P(8<X |A^C)P(A^C)} \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.