Question 1192346: The law of large numbers tells us what happens in the long run.
Like many games of chance, the numbers racket has outcomes so variable - one three-digit number wins $600 and all others win nothing - that gamblers never reach the long run.
Even after many bets, their average winnings may not be close to the mean.
For the numbers racket, the mean payout for single bets is $0.60 (60 cents) and the standard deviation of payouts is about $18.96. If Joe plays 350 days a year for 40 years, he makes 14,000 bets.
(Step 1). What is the mean of the average payout x that Joe receives from his 14,000 bets?
Give your answer in dollars to 2 decimal places.
(Step 2). What is the standard deviation of the average payout x that Joe receives from his 14,000 bets?
Give your answer in dollars to 4 decimal places.
(Step 3) The central limit theorem says that his average payout is approximately Normal with the mean and standard deviation you found above. What is the approximate probability that Joe's average payout per bet is between $0.50 and $0.70? (Give your answer to 4 decimal places)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **Step 1: Mean of the average payout**
* The mean of the average payout remains the same as the mean payout per bet, which is $0.60.
**Step 2: Standard deviation of the average payout**
* **Formula:** Standard deviation of the sample mean (average payout) = σ / √n
* where σ is the population standard deviation and n is the sample size (number of bets)
* **Calculation:**
* Standard deviation of the average payout = $18.96 / √14,000
≈ $0.1602
**Step 3: Probability of average payout between $0.50 and $0.70**
* **Central Limit Theorem:** Since the sample size (14,000 bets) is large, we can approximate the distribution of the average payout as a Normal distribution.
* **Standardize the values:**
* Z1 = (0.50 - 0.60) / 0.1602 ≈ -0.62
* Z2 = (0.70 - 0.60) / 0.1602 ≈ 0.62
* **Find the probability:**
* P(0.50 ≤ average payout ≤ 0.70) = P(-0.62 ≤ Z ≤ 0.62)
= P(Z ≤ 0.62) - P(Z ≤ -0.62)
≈ 0.7324 - 0.2676
≈ 0.4648
**Therefore, the approximate probability that Joe's average payout per bet is between $0.50 and $0.70 is 0.4648.**
**Key takeaway:** Even with a large number of bets, the high standard deviation of individual payouts in the numbers racket makes it unlikely that Joe's average winnings will consistently approach the mean payout of $0.60.
|
|
|
