A puppy named Puddles started Douthat 3 pounds and gained 5 pounds every month. At how many months did Puddles weigh 33 pounds?
1 answer:
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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.
