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.

(a)What is the mean of the average payout x that Joe receives from his 14,000 bets?

(b)What is the standard deviation of the average payout x that Joe receives from his n=14,000 bets?

(c)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?

To answer these questions, we need to calculate the mean and standard deviation of the average payout, and then use the standard normal distribution to find the approximate probability.

(a) The mean of the average payout x can be calculated by multiplying the mean payout for a single bet ($0.60) by the number of bets (14,000):
Mean of x = Mean payout for single bet * Number of bets
Mean of x = $0.60 * 14,000 = $8,400

(b) The standard deviation of the average payout x can be calculated using the formula:
Standard deviation of x = Standard deviation of individual bet / √Number of bets
Standard deviation of x = $18.96 / √14,000 ≈ $0.1602

(c) Now that we have the mean ($8,400) and standard deviation ($0.1602), we can use the standard normal distribution to find the approximate probability that Joe's average payout per bet is between $0.50 and $0.70.

We need to convert the values $0.50 and $0.70 to standardized z-scores. The z-score formula is:
z = (x - mean) / standard deviation

For $0.50:
z1 = ($0.50 - $8,400) / $0.1602

For $0.70:
z2 = ($0.70 - $8,400) / $0.1602

Once we have the z-scores, we can use a standard normal distribution table or a calculator to find the cumulative probabilities associated with each z-score. The probability we need is the difference between the cumulative probabilities for z1 and z2.

Approximate probability = P(z1 ≤ z ≤ z2) = P(z ≤ z2) - P(z ≤ z1)

Note: If you have a standard normal distribution table, you can look up the cumulative probabilities for z1 and z2 and subtract them. If you have a calculator, you can use the cumulative distribution function (CDF) to find the probabilities directly.

ghgh