Here Is An Example Of What The Problems Are Like There Is A
Here Is An Example Of What The Problems Are Like There Is A 01755 Pr
Here is an example of what the problems are like: there is a 0.1755 probability that a best-of-seven contest will last four games, a 0.1836 probability that it will last five games, a 0.2073 probability that it will last six games, and a 0.4336 probability that it will last seven games. Verify that this is a probability distribution. Find its mean and standard deviation. Is it unusual for a team to "sweep" by winning in four games? I need to find someone who can help me to complete the question sets which are each 10-20 questions long.
Paper For Above instruction
This problem involves analyzing a probability distribution related to the length of a best-of-seven contest, where the probability of the contest lasting a specific number of games is given. To address this, we will first verify that the probabilities form a valid probability distribution, then calculate the mean and standard deviation, and finally interpret whether winning in four games (a sweep) is unusual.
Verification of the Probability Distribution
A probability distribution must satisfy two main conditions: all probabilities must be between 0 and 1, and the sum of all probabilities must equal 1. Given the probabilities: 0.1755, 0.1836, 0.2073, and 0.4336, we check their sum:
- Sum = 0.1755 + 0.1836 + 0.2073 + 0.4336 = 1.000
Since their sum equals 1, these probabilities satisfy the criteria for a probability distribution.
Calculating the Mean (Expected Value)
The mean or expected value (E[X]) of a discrete probability distribution is computed by multiplying each value of the random variable by its probability and summing all these products. Here, the random variable (X) is the number of games the contest lasts, taking values 4, 5, 6, and 7.
E[X] = (4)(0.1755) + (5)(0.1836) + (6)(0.2073) + (7)(0.4336)
E[X] = 0.702 + 0.918 + 1.244 + 3.0352 = 5.8992
Thus, the expected number of games in a contest is approximately 5.9.
Calculating the Variance and Standard Deviation
The variance (Var[X]) measures how much the values of the distribution vary around the mean. It is computed as:
Var[X] = Σ (x - μ)^2 * P(x)
Alternatively, we can use the shortcut:
Var[X] = E[X^2] - (E[X])^2
Where,
E[X^2] = (4)^2(0.1755) + (5)^2(0.1836) + (6)^2(0.2073) + (7)^2(0.4336)
- E[X^2] = 16 0.1755 + 25 0.1836 + 36 0.2073 + 49 0.4336
- = 2.808 + 4.59 + 7.4628 + 21.240
- = 36.1008
Now, the variance:
Var[X] = 36.1008 - (5.8992)^2 = 36.1008 - 34.784 ≈ 1.3168
The standard deviation (σ) is the square root of variance:
σ ≈ √1.3168 ≈ 1.147
So, the standard deviation of the number of games in such contests is approximately 1.15.
Interpreting the Results: Is a Sweep (Winning in Four Games) Unusual?
Winning in four games corresponds to a probability of 0.1755, or 17.55%. This means that, on average, about 17.55% of these contests result in a sweep. In probability and statistics, a result is often considered unusual if it is less than 5% or greater than 95%. Since 17.55% is well above 5%, winning in four games is not particularly unusual in this context.
Moreover, given the mean of approximately 5.9 games and a standard deviation of roughly 1.15, winning in four games is within one standard deviation below the mean. This suggests that sweeps happen with a reasonable frequency, reflecting realistic variations in competitive outcomes.
In conclusion, the provided probability distribution is valid, and winning in four games does not constitute an unusual event in these contests. Such insights can inform expectations about the typical length and outcomes of best-of-seven series, important in sports analytics and strategic planning.
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