Question

Problem 4.

a) We define the standard deviation to be the average of all deviations from the mean. To express this definition mathematically, explain why we need to take the square of deviations, divide by n-1, and then take the square root.

b) Use the formula above to find the standard deviation s of the following sample

X= {2, 3, 4, 6, 7 }.

c) Use the built-in statistics function “1-Var Stats” in your TI-84 calculator to verify your

answer above (show the “complete” output).

Answer 1

a) The standard deviation is the average of all deviation from the mean.

But these deviations can be both negative and positive and if we add all these deviations then it will result in 0. So, we won't be able to get an estimate of the actual deviation. That's why we need to take the square of the deviations so that every negative value will turn positive when squared and then we can take an average of them. While taking the average, we don't divide the sum by n rather we divide it by n-1 because only n-1 values can be freely changed. We have known that the sum of all deviation will be 0. So, if we know the deviation of n-1 values from the mean then we can know the deviation of the nth value. Hence, in a sample, the degree of freedom is n-1, that is to say, only n-1 values are required to be considered because the nth value gets fixed automatically.

Since, we took the square of every deviation earlier, hence, in order to now the sample standard deviation, we would again need to take the square root of the average.

b) The given sample is

X = {2,3,4,6,7}

The sample mean = (2+3+4+6+7)/5 = 22/5 = 4.4

Deviation of the first value from the mean = 2-4.4 = -2.4

Square of the deviation of the first value = (-2.4)^2 = 5.76

Deviation of the second value from the mean = 3-4.4 = -1.4

Square of the deviation of the second value = (-1.4)^2 = 1.96

Deviation of the third value from the mean = 4-4.4 = -0.4

Square of the deviation of the second value = (-0.4)^2 = 0.16

Deviation of the fourth value from the mean = 6-4.4 = 1.6

Square of the deviation of the fourth value = (1.6)^2 = 2.56

Deviation of the fifth value from the mean = 7-4.4 = 2.6

Square of the deviation of the fifth value = (2.6)^2 = 6.76

Now, suppose we didn't know the 5th standard deviation. Let it be x

Then the sum of all the deviation must be 0. So

-2.4 - 1.4 - 0.4 + 1.6 + x = 0

x = 4.2 - 1.6 = 2.6 (our fifth deviation value)

Since only 4 values are needed to know the deviation of all the numbers , hence we would divide the sum of square of deviations by n-1.

Sum of the square of the deviations = 5.76 + 1.96 + 0.16+2.56+6.76 = 17.2

Average of the sum of square deviations = 17.2/(5-1) = 17.2/4 = 4.3

Sample standard deviation = square root of 4.3 = 2.074

c) The calculator output is the same = 2.074

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