Question

use the fact that the mean of a geometric distribution is \\( \mu = \frac{1}{p} \\) and the variance is \\( \sigma^2 = \frac{q}{p^2} \\).

a daily number lottery chooses three balls numbered 0 to 9. the probability of winning the lottery is \\( \frac{1}{1000} \\). let \\( x \\) be the number of times you play the lottery before winning the first time.
(a) find the mean, variance, and standard deviation. (b) how many times would you expect to have to play the lottery before winning? it costs $1 to play and winners are paid $400. would you expect to make or lose money playing this lottery? explain.

(a) the mean is 1000. (type an integer or a decimal.)
the variance is 999000. (type an integer or a decimal.)
the standard deviation is 999.5. (round to one decimal place as needed.)
(b) you can expect to play the game \\( \square \\) times before winning.

Explanation:

Step1: Recall the mean of geometric distribution

The mean of a geometric distribution is given by \(\mu=\frac{1}{p}\). Here, the probability of winning \(p = \frac{1}{1000}\). So, the expected number of trials before the first success (winning) is the mean of the geometric distribution.

Step2: Calculate the expected number of plays

Using the formula for the mean of geometric distribution \(\mu=\frac{1}{p}\), substitute \(p=\frac{1}{1000}\). Then \(\mu=\frac{1}{\frac{1}{1000}} = 1000\). This means we expect to play 1000 times before winning.

Answer:

1000