1. pdf and mean problems

1) The length of time, Y, that a customer spends in line at a bank teller's window before being served is described by the exponential pdf:

f(y) = 0.2e-0.2y, y ³ 0.

a) What is the probability that a customer will wait for more than 10 minutes?

I used 1 - P( Y £ 10 ) = 1 - 0.2e-0.2y = 0.9729

b) Suppose the customer will leave if the wait is more than 10 minutes. Assume that the customer goes to the bank 3 times next month. Let the random variable X be the number of times the customer leaves without being served. Find P(X=1).

This I'm not sure how to do. Any hints?

2) A warehouse contains 10 printing machines, 3 of which are defective. A company selects 5 of the machines at random. What is the mean number of defective printers selected?

Since there are 3 defective machines, I found the probability of picking each one.
1: 3/10
2: 2/9
3: 1/8
Then I multiplied the number of the defective machine by its probability of being selected and summed them all up: (3/10) + [2*(2/9)] + [3*(1/8)] = 1.1194

Does this seem right?

2. Originally Posted by engineer22
1) The length of time, Y, that a customer spends in line at a bank teller's window before being served is described by the exponential pdf:

f(y) = 0.2e-0.2y, y ³ 0.

a) What is the probability that a customer will wait for more than 10 minutes?

I used 1 - P( Y £ 10 ) = 1 - 0.2e-0.2y = 0.9729

b) Suppose the customer will leave if the wait is more than 10 minutes. Assume that the customer goes to the bank 3 times next month. Let the random variable X be the number of times the customer leaves without being served. Find P(X=1).

This I'm not sure how to do. Any hints?
You are confusing the pdf with the cumulative distribution function.

$
f(y)=0.2 e^{-0.2y}, \ \ \ 0$

is a pdf, it integrates up to 1, but not a cdf as it goes to zeros as $y \to \infty$.

The cdf corresponding to this pdf is:

$
F(y) = \int_0^y f(\zeta)\ d\zeta = 1 - e^{- 0.2\ x}
$

The second part of the problem is a binomial distribution problem, you will have found the probability that
he leaves without being served on a single visit call this p, then the number of times he leaves without
being served in three visits X has a binomial distribution B(3, p).

RonL

3. for number 2 you have a binomial distribution where the probability of picking a defective part is $\frac{3}{10}$. so

$E[X]=np=5\left(\frac{3}{10}\right)=\frac{3}{2}$