1-3. Quiz
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The range is the difference between the greatest and least value. The standard deviation is the average amount by which each individual value deviates or differs from the mean.
Female Range: 4
Male Range: 6
Female Standard Deviation: Approximately 1.33
Male Standard Deviation: Approximately 1.94
Comparison: See solution.
We want to find the range, and standard deviation of the two given data sets, as well as compare our results.
Now that we found the ranges, we want to find the standard deviation. To do it, we will need the mean, so we will calculate it first.
The mean of a data set is the sum of the values divided by the total number of values in the set. Let's start by calculating the sum of the given values. Female:&& 6+2+4+3+4= 19 Male:&& 5+3+6+6+9= 29 There are 5 values in each set, so we have to divide their sums by 5. Female Mean:&& 19/5=3.8 Male Mean:&& 29/5=5.8
The standard deviation of a set of data is the average amount by which each individual value deviates or differs from the mean. Standard Deviation sqrt((x_1-x )^2+(x_2-x )^2+... +(x_n-x )^2/n) In the above formula, x_1, ... ,x_n are the values of the set of data, x is the mean, and n is the number of values. For this exercise, we already found the means for both sets. Female $x$:&& 19/5=3.8 Male $x$:&& 29/5=5.8 Let's use these values and apply the formula to each value in the sets. We will do it separately for each data set. Let's consider the female group first.
| x_n | x_n-x | (x_n-x)^2 |
|---|---|---|
| 6 | 6-3.8=2.2 | 2.2^2=4.84 |
| 2 | 2-3.8=- 1.8 | (- 1.8)^2=3.24 |
| 4 | 4-3.8=0.2 | 0.2^2=0.04 |
| 3 | 3-3.8=- 0.8 | (- 0.8)^2=0.64 |
| 4 | 4-3.8=0.2 | 0.2^2=0.04 |
| Sum of Values | = 8.8 | |
Finally, we need to divide by 5 and then calculate the square root. Female Standard Deviation: sqrt(8.8/5)≈ 1.33 Now, let's consider the male group. Remember that we need to consider the other value for x!
| x_n | x_n-x | (x_n-x)^2 |
|---|---|---|
| 5 | 5-5.8=- 0.8 | (- 0.8)^2=0.64 |
| 3 | 3-5.8=- 2.8 | (- 2.8)^2=7.84 |
| 6 | 6-5.8=0.2 | 0.2^2=0.04 |
| 6 | 6-5.8=0.2 | 0.2^2=0.04 |
| 9 | 9-5.8=3.2 | 3.2^2=10.24 |
| Sum of Values | = 18.8 | |
Finally, we need to divide by 5 and then calculate the square root. Male Standard Deviation: sqrt(18.8/5)≈ 1.94
The male range is greater than the female range. This means the values for male absence are more spread out. Similarly, the male standard deviation is greater than that of the female group, which means that girls are more consistent than the boys.