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 two balls numbered 0 to 9. the probability of winning the lottery is $\frac{1}{100}$. 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 $\square$. (type an integer or a decimal.)

Explanation:

Step1: Identify the value of \(p\)

Given that the probability of winning \(p=\frac{1}{100} = 0.01\)

Step2: Calculate the mean \(\mu\)

Using the formula \(\mu=\frac{1}{p}\), substitute \(p = 0.01\)
\(\mu=\frac{1}{0.01}=100\)

Step3: Calculate \(q\)

Since \(q = 1 - p\), substitute \(p=0.01\)
\(q=1 - 0.01=0.99\)

Step4: Calculate the variance \(\sigma^{2}\)

Using the formula \(\sigma^{2}=\frac{q}{p^{2}}\), substitute \(p = 0.01\) and \(q = 0.99\)
\(\sigma^{2}=\frac{0.99}{(0.01)^{2}}=\frac{0.99}{0.0001}=9900\)

Step5: Calculate the standard deviation \(\sigma\)

Using the formula \(\sigma=\sqrt{\sigma^{2}}\), substitute \(\sigma^{2}=9900\)
\(\sigma=\sqrt{9900}\approx99.5\)

Answer:

The mean is \(100\), the variance is \(9900\), and the standard deviation is approximately \(99.5\)