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How would you do these questions?

How would you do these questions?-example-1
How would you do these questions?-example-1
How would you do these questions?-example-2
User CIAvash
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Part 1

Population = {1,2,4}

Sample 1 = {1,1}, probablity = 1/9

Sample 2 = {1,2}, probablity = 1/9

Sample 3 = {1,4}, probablity = 1/9

Sample 4 = {2,1}, probablity = 1/9

Sample 5 = {2,2}, probablity = 1/9

Sample 6 = {2,4}, probablity = 1/9

Sample 7 = {4,1}, probablity = 1/9

Sample 8 = {4,2}, probablity = 1/9

Sample 9 = {4,4}, probablity = 1/9

Each of the 9 samples has equal probablity of occuring, so that's why each probablity is 1/9

Note: it may be best to write the above in the form of a table

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Part 2

For each sample (1 through 9), compute the mean. So we'll add up the values and divide by 2

xbar represents the sample mean

Sample 1 = {1,1}, xbar = 1

Sample 2 = {1,2}, xbar = 1.5

Sample 3 = {1,4}, xbar = 2.5

Sample 4 = {2,1}, xbar = 1.5

Sample 5 = {2,2}, xbar = 2

Sample 6 = {2,4}, xbar = 3

Sample 7 = {4,1}, xbar = 2.5

Sample 8 = {4,2}, xbar = 3

Sample 9 = {4,4}, xbar = 4

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Now let's list out each possible xbar value and its associated probability

xbar = 1, probability = 1/9

xbar = 1.5, probability = 2/9

xbar = 2, probability = 1/9

xbar = 2.5, probability = 2/9

xbar = 3, probability = 2/9

xbar = 4, probability = 1/9

note how the probabilities listed above add to 9/9 = 1.

Again a table is ideal to help organize the data.

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Part 3

To calculate the value of mu_x, we add up all the xbar values found in part 2, and divide by 9

(1+1.5+2.5+1.5+2+3+2.5+3+4)/9 = 2.333

So mu_x = 2.333 approximately

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Calculating the standard deviation is a bit lengthier, but you subtract each data value from the mean, square each difference, and add up the squares.

Afterward, you divide by n = 9 to get the population variance. Apply the square root to get the population standard deviation.

Luckily a calculator or spreadsheet can make quick work of this. You should get roughly 0.88192 as the standard deviation.

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Part 4

The mean of the population {1,2,4} is (1+2+4)/3 = 2.3333 which matches with mu_xbar

The standard deviation of the population {1,2,4} is roughly 0.88191710368819 which matches with 0.88192 found earlier back in part 3.

How would you do these questions?-example-1
User Jiexi
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