83, 84, 91, 90, 98,
88, 95, 89, 87, 95
We are asked to find the standard deviation for the given data. To do so, let's first find the mean of the data.
Mean
The mean of a numerical data set is the sum of the data divided by the number of data values. Since we have 10 data values, the sum of the data values will be the sum of the data values divided by 10.
The standard deviation σ of a numerical data set is given by the following formula.
In this formula n is the number of data values in the data set, x_1, x_2,..., x_n are the data values, and x_1-x, x_2-x,... x_n-x are the deviations of each data value. The deviation is given by the difference of the data value and the mean of the data set. We already found out the mean of the data set.
x=90
With this value we can now calculate the deviation and the square
of each deviation. Let's do this in a table.
x
x
x-x
(x-x)^2
83
90
-7
49
84
90
-6
36
91
90
1
1
90
90
0
0
98
90
8
64
88
90
-2
4
95
90
5
25
89
90
-1
1
87
90
-3
9
95
90
5
25
Now, let's find the mean of the squared deviations. This is called the variance. Let's do it!
Finally, we will take the square root of the variance to get the standard deviation.
σ=sqrt(21.4) ⇒ σ≈ 4.6
Therefore, the standard deviation is about 4.6. This means that Golfer A's typical score differs from the mean by about 4.6 strokes.
b Consider the data for Golfer B.
89, 93, 92, 88, 89,
87, 95, 94, 91, 92
Again, we are asked to calculate the standard deviation for the given data. Proceeding in a similar way as we did in Part A, let's first calculate the mean and then the standard deviation.
Mean
In this case we also have 10 data values. Therefore, the mean will be the sum of the data divided by 10.
Finally, we will take the square root of the variance to get the standard deviation.
σ=sqrt(6.4) ⇒ σ≈ 2.5
Therefore, the standard deviation is about 2.5. This means that Golfer B's typical score differs from the mean by about 2.5 strokes.
c The standard deviation for Golfer A is about 4.6, and for Golfer B it is about 2.5 strokes. Since Golfer A's standard deviation is greater, Golfer A's scores are more spread out and Golfer B is more consistent. Please note that there are many possible interpretations. Here we are only showing one possibility.
