Below are the numbers of games played in the World Series from 1903 to 2012. Find the mean, variance and standard deviation.
Games played Probability
4 0.186
5 0.231
6 0.203
7 0.341
8 0.039
To find the mean, variance, and standard deviation for the number of games played in the World Series from 1903 to 2012, we will use the following formulas:
Mean:
The mean is calculated by multiplying each value by its corresponding probability, and then summing all the products.
Variance:
The variance is calculated by finding the square of the difference between each value and the mean, multiplying it by its corresponding probability, and then summing all the products.
Standard Deviation:
The standard deviation is the square root of the variance.
Let's calculate each of these values step by step:
1. Calculate the mean:
Mean = (4 * 0.186) + (5 * 0.231) + (6 * 0.203) + (7 * 0.341) + (8 * 0.039)
Mean = 0.744 + 1.155 + 1.218 + 2.387 + 0.312
Mean = 5.816
So, the mean number of games played in the World Series is 5.816.
2. Calculate the variance:
Variance = [(4 - Mean)^2 * 0.186] + [(5 - Mean)^2 * 0.231] + [(6 - Mean)^2 * 0.203] + [(7 - Mean)^2 * 0.341] + [(8 - Mean)^2 * 0.039]
Variance = [(4 - 5.816)^2 * 0.186] + [(5 - 5.816)^2 * 0.231] + [(6 - 5.816)^2 * 0.203] + [(7 - 5.816)^2 * 0.341] + [(8 - 5.816)^2 * 0.039]
Variance = [(-1.816)^2 * 0.186] + [(-0.816)^2 * 0.231] + [(0.184)^2 * 0.203] + [(1.184)^2 * 0.341] + [(2.184)^2 * 0.039]
Variance = 0.606 + 0.148 + 0.071 + 1.457 + 0.198
Variance = 2.480
So, the variance of the number of games played in the World Series is 2.480.
3. Calculate the standard deviation:
Standard Deviation = √Variance
Standard Deviation = √2.480
Standard Deviation ≈ 1.57
So, the standard deviation of the number of games played in the World Series is approximately 1.57.
To find the mean (average) of the games played, we multiply each value by its probability and then add them up:
Mean (μ) = (4 * 0.186) + (5 * 0.231) + (6 * 0.203) + (7 * 0.341) + (8 * 0.039)
μ = 0.744 + 1.155 + 1.218 + 2.387 + 0.312
μ = 5.816
To find the variance, we need to calculate the squared difference between each value and the mean, multiply each squared difference by its probability, and then sum them up:
Variance (σ^2) = ((4 - 5.816)^2 * 0.186) + ((5 - 5.816)^2 * 0.231) + ((6 - 5.816)^2 * 0.203) + ((7 - 5.816)^2 * 0.341) + ((8 - 5.816)^2 * 0.039)
σ^2 = (2.816^2 * 0.186) + (0.184^2 * 0.231) + (0.184^2 * 0.203) + (1.184^2 * 0.341) + (2.184^2 * 0.039)
σ^2 = 1.574464 + 0.039714816 + 0.034262688 + 0.446637984 + 0.214681856
σ^2 = 2.309761344
Finally, to find the standard deviation, we take the square root of the variance:
Standard Deviation (σ) = √(2.309761344)
σ ≈ 1.519
Therefore, the mean (average) number of games played in the World Series from 1903 to 2012 is approximately 5.816 games, the variance is approximately 2.310, and the standard deviation is approximately 1.519.
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
I'll let you do the calculations.