Consider an infinite population with 25% of the elements having the value 1, 25% the value 2, 25% the value 3, and 25% the value 4. If X is the value of a randomly selected item, then X is a discrete random variable whose possible values are 1, 2, 3, and 4.
(a) Find the population mean and population variance for the random variable X.
(b) List all 16 possible distinguishable samples of size 2, for each calculate the sample mean.
(c) Calculate the mean and variance of the distriubtion of X(bar) and show that, as expected, they are equal to the population mean and the (population variance)/n, respectively.
ok, (b) is fine, and I got the means to match from (a) and (c), but I can't figure out the variances.
Oh, variance can be quite tricky sometimes, can't it? It's like trying to hold a slippery banana peel in your hand. But don't worry, I'll help you peel away this problem!
(a) To find the population mean, we just need to calculate the weighted average of the possible values of X. Given that each value occurs with equal probability of 25%, the mean would be:
(1 * 0.25) + (2 * 0.25) + (3 * 0.25) + (4 * 0.25) = 2.5
So the population mean is 2.5.
To find the population variance, we need to calculate the weighted average of the squared deviations from the population mean. Since the values 1, 2, 3, and 4 are equally probable, the variance would be:
((1 - 2.5)^2 * 0.25) + ((2 - 2.5)^2 * 0.25) + ((3 - 2.5)^2 * 0.25) + ((4 - 2.5)^2 * 0.25) = 1.25
Therefore, the population variance is 1.25.
(b) Glad to hear you've got the means sorted! Now, to calculate the sample means for all 16 possible distinguishable samples of size 2, let's have some fun. Remember, in each sample, we randomly select two values (with replacement) from the population. Here they are:
1. 1, 1
2. 1, 2
3. 1, 3
4. 1, 4
5. 2, 1
6. 2, 2
7. 2, 3
8. 2, 4
9. 3, 1
10. 3, 2
11. 3, 3
12. 3, 4
13. 4, 1
14. 4, 2
15. 4, 3
16. 4, 4
Now, all you need to do is calculate the mean for each sample by taking the average of the two values in each sample.
(c) For the distribution of X bar, which represents the mean of random samples, the mean would be expected to remain the same as the population mean, which is 2.5. This is because the mean of sample means should be equal to the population mean.
The variance of X bar, on the other hand, would be equal to (population variance)/n, where n is the sample size. In this case, since we're dealing with samples of size 2, the variance of X bar would be (1.25)/2 = 0.625.
So, in summary:
Population mean: 2.5
Population variance: 1.25
Distribution of X bar mean: 2.5
Distribution of X bar variance: 0.625
I hope that helps you slice through the variance problem like a clown juggling bowling pins!
To calculate the population mean (μ) and population variance (σ^2) for the random variable X, we can use the formulas:
μ = Σ(xi * pi)
σ^2 = Σ((xi - μ)^2 * pi)
where xi represents the possible values of X, and pi represents their respective probabilities.
In this case, we have four possible values for X (1, 2, 3, 4), and each value has a probability of 0.25.
(a) Calculating the population mean (μ):
μ = (1 * 0.25) + (2 * 0.25) + (3 * 0.25) + (4 * 0.25)
= 0.25 + 0.5 + 0.75 + 1
= 2.5
Therefore, the population mean (μ) for X is 2.5.
(b) To list all 16 possible distinguishable samples of size 2, we can simply combine the values 1, 2, 3, and 4:
{1, 1}, {1, 2}, {1, 3}, {1, 4},
{2, 1}, {2, 2}, {2, 3}, {2, 4},
{3, 1}, {3, 2}, {3, 3}, {3, 4},
{4, 1}, {4, 2}, {4, 3}, {4, 4}
For each sample, we can calculate the respective sample mean.
(c) Calculating the mean and variance of the distribution of Xbar:
The mean of the distribution of sample means (Xbar) is the same as the population mean (μ), which is 2.5.
To calculate the variance of the distribution of Xbar, we divide the population variance (σ^2) by the sample size (n):
σ^2(Xbar) = σ^2 / n
In this case, since the sample size is 2 (n = 2), we can calculate the variance as:
σ^2(Xbar) = σ^2 / 2
From part (a), we need to calculate the population variance (σ^2) of X.
σ^2 = Σ((xi - μ)^2 * pi)
= ((1 - 2.5)^2 * 0.25) + ((2 - 2.5)^2 * 0.25) + ((3 - 2.5)^2 * 0.25) + ((4 - 2.5)^2 * 0.25)
= 2.25
Therefore, the population variance (σ^2) for X is 2.25.
Now, calculating the variance of the distribution of Xbar:
σ^2(Xbar) = σ^2 / 2
= 2.25 / 2
= 1.125
Therefore, the variance of the distribution of Xbar is 1.125.
In summary:
(a) Population mean (μ) for X: 2.5
Population variance (σ^2) for X: 2.25
(c) Mean of the distribution of Xbar: 2.5
Variance of the distribution of Xbar: 1.125
To find the population mean and variance for the random variable X, we need to use the formulas for mean (μ) and variance (σ^2).
(a) Population Mean (μ):
The population mean is the average value of the random variable X when considering the entire population.
Given that 25% of the elements have the value 1, 25% have the value 2, 25% have the value 3, and 25% have the value 4, we can calculate the population mean as follows:
μ = (1 * 0.25) + (2 * 0.25) + (3 * 0.25) + (4 * 0.25)
= 0.25 + 0.5 + 0.75 + 1
= 2.5
So the population mean for X is 2.5.
(b) Population Variance (σ^2):
The population variance measures the spread or variability of the random variable X values around the mean.
To calculate the population variance, we need to find the variance of each distinct value of X (1, 2, 3, 4) and weight them according to their probability.
The formula for population variance is:
σ^2 = (Σ(X - μ)^2 * P(X)) / n
Where Σ denotes summation, X represents the distinct values, μ is the population mean, P(X) is the probability of each distinct value, and n is the number of distinct values.
Let's calculate the population variance step by step:
For X = 1:
Variance of X = (1 - 2.5)^2 * 0.25
= 2.25 * 0.25
= 0.5625
For X = 2:
Variance of X = (2 - 2.5)^2 * 0.25
= 0.25 * 0.25
= 0.0625
For X = 3:
Variance of X = (3 - 2.5)^2 * 0.25
= 0.25 * 0.25
= 0.0625
For X = 4:
Variance of X = (4 - 2.5)^2 * 0.25
= 2.25 * 0.25
= 0.5625
Next, we sum up the variances of each value, weighted by their probability:
σ^2 = (0.5625 * 0.25) + (0.0625 * 0.25) + (0.0625 * 0.25) + (0.5625 * 0.25)
= 0.5625
So the population variance for X is 0.5625.
(c) Distribution of X(bar):
To calculate the mean and variance of the distribution of X(bar), we use the same formulas as in (a) and (b).
The mean of the distribution of X(bar) is equal to the population mean (μ), which we have already calculated as 2.5.
The variance of the distribution of X(bar) is equal to the population variance (σ^2) divided by the sample size (n). Since in this case, n is 2, we have:
Variance of X(bar) = σ^2 / n
= 0.5625 / 2
= 0.28125
So the variance of the distribution of X(bar) is 0.28125, which matches the (population variance)/n as expected.
I hope this helps! If you have any further questions, please let me know.