Respuesta :
Answer:
(a) The mean and standard deviation of the average payout [tex]\bar x[/tex] 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:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the distribution of sample mean is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
(a)
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.
The mean of the distribution of sample mean is given by,
[tex]\mu_{\bar x}=\mu=\$0.60[/tex]
And the standard deviation of the distribution of sample mean is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{18.96}{\sqrt{14000}}=0.1602[/tex]
[tex]\bar X\sim N(0.60,\ 0.1602^{2})[/tex]
Thus, the mean and standard deviation of the average payout [tex]\bar x[/tex] that Joe receives from his 14,000 bets are $0.60 and $0.1602 respectively.
(b)
Compute the probability that Joe's average payout per bet is between $0.50 and $0.70 as follows:
[tex]P(0.50<\bar X<0.70)=P(\frac{0.50-0.60}{0.1602}<\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}<\frac{0.70-0.60}{0.1602})[/tex]
[tex]=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\times 0.73237)-1\\=0.46474\\\approx 0.4647[/tex]
Thus, the probability that Joe's average payout per bet is between $0.50 and $0.70 is 0.4647.
