Just a reminder: We use the cumulative function to find probabilities
Hey, I need some help with the below questions, any help would be great. Thanks.
The length of time in minutes, Y, that a customer spends in line at a bank teller’s window before being served is described by the exponential pdf f(y) = 0.2exp(-0.2y) y>0.
a) What is the median (m) waiting time in minutes and seconds?
Is this 0.2 = 1/m thus m = 5?
b) What is the probability that a customer will wait more than 10 minutes? Answer correct to 3 decimal places.
Do I just replace y with 10 here? Thus -0.4
c) Suppose the customer will leave if the wait is more than 10 minutes. Assume that the customer goes to the bank twice next month. Let the random variable X be the number of times in the next month that the customer leaves without being served. Find P(X=1) correct to 3 decimal places.
I'm not sure about this one at all.
Just a reminder: We use the cumulative function to find probabilities
Which one is that?
You have to integrate you pdf with respect to x,i.e you get for an exponential Pr(X<=x)=1-exp(-.2x) . Its derivative is the pdf.
Try search the forum and you'll get this problem solved,it's a really common one.I think i replyed for the median time.
So for:
b) P( y > 10) = ∫ 0.2exp-0.2y dy
= [ -e-0.2y ] Between 10 and infinity.
= e-2
= 0.135 (3 d.p.)
Yes?
But what about a) and c) ?
Would a) be:
a) Median = ln 2 / β
= ln 2 / 0.2
= 3.4657
= 3 minutes 28 seconds
You gor it.
c is a common textbook problem.Use a distribution to find a probability (b) and then use it as the probabality of a binomial problem.Two trials with Pr(X>10)
So c) would be the double integral between 10 and infinity of 0.2exp-0.2y?
No,the probability is an input to the binomial distribution (the common black-or-white,success-failure,0-1 distribution)
Pr(X=1)=Pr(Y>10)*[1-Pr(Y>10)],where Y is exponentially distributed as in (a),(b)
> http://en.wikipedia.org/wiki/Binomial_distribution
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