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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.
Step-by-step 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
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