There is a 0.0416 probability that a best-of-seven contest will last four games, a 0.0804 probability that it will last five games, a 0.2132 probability that it will last six games, and a 0.6648 probability that it will last seven games.
Find the mean and standard deviation.
Is it unusual for a team to win in four games? Choose the correct answer below.
A - No, because the probability that a team wins in four games is less than or equal to 0.05
B -
Yes because the probability that a team wins in four games is greater than 0.05
C -
Yes, because the probability that a team wins in four games is less than or equal to 0.05
D -
No because the probability that a team wins in four games is greater than 0.05
To find the mean and standard deviation, we can use the following formulas:
Mean (μ) = Σ(xi * Pi)
Standard Deviation (σ) = sqrt(Σ(xi - μ)^2 * Pi)
where xi represents the number of games (4, 5, 6, or 7) and Pi represents their respective probabilities.
Given the probabilities:
P(4 games) = 0.0416
P(5 games) = 0.0804
P(6 games) = 0.2132
P(7 games) = 0.6648
Calculating the mean and standard deviation:
Mean (μ) = (4 * 0.0416) + (5 * 0.0804) + (6 * 0.2132) + (7 * 0.6648) ≈ 6.96
Standard Deviation (σ) = sqrt((4-6.96)^2 * 0.0416 + (5-6.96)^2 * 0.0804 + (6-6.96)^2 * 0.2132 + (7-6.96)^2 * 0.6648) ≈ 1.571
Therefore, the mean is approximately 6.96 and the standard deviation is approximately 1.571.
To determine if it is unusual for a team to win in four games, we compare the probability of winning in four games (0.0416) to a significance level of 0.05 or 5%.
Since the probability of winning in four games (0.0416) is less than 0.05, we can conclude that it is unusual for a team to win in four games. The correct answer is option C - Yes, because the probability that a team wins in four games is less than or equal to 0.05.
To find the mean and standard deviation, we can use the formula for expected value and variance.
The expected value (mean) of a random variable X in this case is given by the formula:
E(X) = Σ(xi * Pi)
where xi are the possible values of X and Pi are their corresponding probabilities.
In this case, the possible values of X are the number of games in the contest: 4, 5, 6, and 7. The corresponding probabilities are: 0.0416, 0.0804, 0.2132, and 0.6648.
Let's calculate the mean:
E(X) = (4 * 0.0416) + (5 * 0.0804) + (6 * 0.2132) + (7 * 0.6648)
E(X) = 0.1664 + 0.402 + 1.2792 + 4.6536
E(X) = 6.5012
So, the mean is 6.5012 games.
To calculate the standard deviation, we need to find the variance first. The variance of a random variable X can be calculated using the formula:
Var(X) = Σ((xi - μ)^2 * Pi)
where μ is the mean, xi are the possible values of X and Pi are their corresponding probabilities.
Let's calculate the variance:
Var(X) = ((4 - 6.5012)^2 * 0.0416) + ((5 - 6.5012)^2 * 0.0804) + ((6 - 6.5012)^2 * 0.2132) + ((7 - 6.5012)^2 * 0.6648)
Var(X) = (2.5012^2 * 0.0416) + (1.5012^2 * 0.0804) + (0.5012^2 * 0.2132) + (0.4988^2 * 0.6648)
Var(X) = 0.313709184 + 0.181306814 + 0.0533513552 + 0.1659901584
Var(X) = 0.7143575116
Therefore, the variance is 0.7143575116.
Finally, we can calculate the standard deviation by taking the square root of the variance:
Standard Deviation = √0.7143575116
Standard Deviation ≈ 0.844437
So, the mean is 6.5012 games and the standard deviation is approximately 0.844437 games.
Now, to answer the second part of the question about whether it is unusual for a team to win in four games, we need to compare the probability (0.0416) to a threshold.
If we consider a threshold of 0.05 (5%), then the answer is option C - Yes, because the probability that a team wins in four games is less than or equal to 0.05. Since 0.0416 is less than 0.05, winning in four games would be considered unusual according to this threshold.
Note that the choice of threshold may vary depending on the context. For statistical significance, a threshold of 0.05 is commonly used, but it can change depending on specific requirements or conventions.