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cupoosta [38]
3 years ago
5

Use the fact that the mean of a geometric distribution is μ= 1 p and the variance is σ2= q p2. A daily number lottery chooses th

ree balls numbered 0 to 9. The probability of winning the lottery is 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 ​$500. Would you expect to make or lose money playing this​ lottery? Explain.
Mathematics
1 answer:
butalik [34]3 years ago
4 0

Answer:

a). The mean = 1000

     The variance = 999,000

     The standard deviation = 999.4999

b). 1000 times , loss

Step-by-step explanation:

The mean of geometric distribution is given as , $\mu = \frac{1}{p}$

And the variance is given by, $\sigma ^2=\frac{q}{p^2}$

Given : $p=\frac{1}{1000}$

             = 0.001

The formulae of mean and variance are :

$\mu = \frac{1}{p}$

$\sigma ^2=\frac{q}{p^2}$

$\sigma ^2=\frac{1-p}{p^2}$

a). Mean =   $\mu = \frac{1}{p}$

              = $\mu = \frac{1}{0.001}$

              = 1000

  Variance =   $\sigma ^2=\frac{1-p}{p^2}$

                  = $\sigma ^2=\frac{1-0.001}{0.001^2}$

                           = 999,000

   The standard deviation is determined by the root of the variance.

    $\sigma = \sqrt{\sigma^2}$

        = $\sqrt{999,000}$ = 999.4999

b). We expect to have play lottery 1000  times to win, because the mean in part (a) is 1000.

When we win the profit is 500 - 1 = 499

When we lose, the profit is -1

Expected value of the mean μ is the summation of a product of each of the possibility x with the probability P(x).

$\mu=\Sigma\ x\ P(x)= 499 \times 0.001+(-1) \times (1-0.001)$

  = $ 0.50

Since the answer is negative, we are expected to make a loss.

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