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 calculate the mean and standard deviation, we can use the formulas for the sample mean and sample standard deviation.

(a) The mean of the average payout x that Joe receives from his 14,000 bets can be calculated using the formula:

Mean (μ) = (Sum of payouts) / (Number of bets)

In this case, the sum of payouts is (14,000 * $0.60) = $8,400, and the number of bets is 14,000. So,

Mean (μ) = $8,400 / 14,000 ≈ $0.60

Therefore, the mean of the average payout x is approximately $0.60.

(b) The standard deviation of the average payout x that Joe receives from his 14,000 bets can be calculated using the formula:

Standard Deviation (σ) = sqrt[(Sum of (payout - mean)^2) / (Number of bets - 1)]

In this case, the sum of (payout - mean)^2 can be calculated by multiplying the standard deviation of individual payouts ($18.96) by the number of bets (√14,000) and then squaring the result. So,

Sum of (payout - mean)^2 = ($18.96 * √14,000)^2

The number of bets is 14,000. So,

Standard Deviation (σ) = sqrt[(($18.96 * √14,000)^2) / (14,000 - 1)]

Therefore, the standard deviation of the average payout x is approximately the square root of that expression.

(c) To find the approximate probability that Joe's average payout per bet is between $0.50 and $0.70, we can use the properties of the normal distribution. We know that the mean is $0.60, and the standard deviation is the value calculated in part (b).

Using a standard normal distribution table or a calculator, we can calculate the z-scores for $0.50 and $0.70. Let's represent the z-score for $0.50 as z1 and the z-score for $0.70 as z2. Then, we can calculate:

z1 = ($0.50 - $0.60) / (σ / √n)
z2 = ($0.70 - $0.60) / (σ / √n)

Here, n represents the number of bets (14,000). Plug in the values of z1, z2, and σ to calculate the z-scores. Once you have the z-scores, you can use the standard normal distribution table or a calculator to find the probability associated with the range between these z-scores.

Keep in mind that this probability is an approximation based on the central limit theorem and assumes that the sample size (number of bets) is large enough.