A student was asked to compute the mean and standard deviation for the following sample of n _ 5 scores: 81, 87, 89, 86, and 87. To simplify the arithmetic, the student first subtracted 80 points from each score to obtain a new sample consisting of 1, 7, 9, 6, and 7. The mean and standard deviation for the new sample were then calculated to be M _ 6 and s _ 3. What are the values of the mean and standard deviation for the original sample?
Mean would be 6+80= 86
while standard deviation would be just the same i.e., 3
The equal sign (=) is the button just to the right of the underscore.(_)
To calculate the mean and standard deviation for the original sample, we need to reverse the transformation done by subtracting 80 points from each score.
Mean:
To find the mean, we add the 80 points back to each score in the new sample and then calculate the mean.
Original sample: 81, 87, 89, 86, and 87
New sample: 1, 7, 9, 6, and 7
Adding 80 to each score in the new sample: 1+80, 7+80, 9+80, 6+80, and 7+80
The new scores are: 81, 87, 89, 86, and 87
To calculate the mean, we sum up all the scores and divide by the number of scores:
Mean (original sample) = (81 + 87 + 89 + 86 + 87) / 5 = 430 / 5 = 86
Therefore, the mean for the original sample is 86.
Standard Deviation:
To find the standard deviation, we need to use the formula:
s (original sample) = √[∑[(x - M)^2] / (n - 1)]
Here, x is each score, M is the mean, and n is the number of scores.
Using the new sample's values:
New sample: 1, 7, 9, 6, and 7
Mean (new sample) = 6
Now, calculating the sum of squares:
Sum of squares (original sample) = [(81 - 86)^2 + (87 - 86)^2 + (89 - 86)^2 + (86 - 86)^2 + (87 - 86)^2] = (25 + 1 + 9 + 0 + 1) = 36
Substituting the values into the standard deviation formula:
s (original sample) = √[36 / 4] = √9 = 3
Therefore, the standard deviation for the original sample is 3.
To find the values of the mean and standard deviation for the original sample, we need to reverse the modifications that were made to obtain the new sample.
1. Reversing the subtraction of 80 points:
To obtain the original sample, we need to add 80 points to each score of the new sample: 1 + 80 = 81, 7 + 80 = 87, 9 + 80 = 89, 6 + 80 = 86, 7 + 80 = 87
So the original sample is: 81, 87, 89, 86, 87
2. Reversing the calculation of the mean:
The mean is calculated by summing up all the scores and dividing by the number of scores.
For the new sample, the sum of the scores is 1 + 7 + 9 + 6 + 7 = 30, and since there are 5 scores, the mean is 30/5 = 6.
To obtain the mean for the original sample, we need to reverse this calculation. We know that the sum of the scores in the original sample is 81 + 87 + 89 + 86 + 87 = 430. Since there are 5 scores in the original sample, the mean is 430/5 = 86.
So the mean for the original sample is 86.
3. Reversing the calculation of the standard deviation:
The standard deviation is a measure of the dispersion of the scores in a sample. To obtain the standard deviation for the original sample, we need to reverse the calculation that was done for the new sample.
For the new sample, the standard deviation is given as s = 3.
To obtain the standard deviation for the original sample, we use the formula:
original_s = new_s * (original_n / new_n)^0.5
where original_n is the number of scores in the original sample and new_n is the number of scores in the new sample.
For the original sample, original_n = 5 and new_n = 5.
original_s = 3 * (5 / 5)^0.5 = 3 * 1^0.5 = 3
So the standard deviation for the original sample is 3.
In conclusion, the values of the mean and standard deviation for the original sample are:
Mean = 86
Standard Deviation = 3