Question

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.

Required:
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 14,000 bets?
c. What is the approximate probability that Joe's average payout per bet is between $0.50 and $0.70?

Answer 1

Answer: (a) The mean and standard deviation of the average payout that Joe receives from his 14,000 bets are $0.60 and $0.1602 respectively.

(b) The probability that Joe's average payout per bet is between $0.50 and $0.70 is 0.4647.

Explanation:

Let X = average payout of a bet.

The mean of the random variable X is, μ = $0.60.

The standard deviation of the random variable X is, σ = $18.96.

The number of total bets made is, n = 14000.

Since n = 14000 > 30, the Central limit theorem can be used to approximate the sampling distribution of sample mean.

Thus, the mean and standard deviation of the average payout that Joe receives from his 14,000 bets are $0.60 and $0.1602 respectively.

Compute the probability that Joe's average payout per bet is between $0.50 and $0.70 as follows:

=P (-0.62 < Z < 0.62)

= P (Z < 0.62) - P(Z < -0.62)

= P (Z < 0.62) - [1-P (Z < 0.62)

= 2P (Z < 0.62) - 1

= (2 x 0.73237) - 1

= 0.46474

≈ 0.4647