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Find the standard deviation for the data in [link] .
| Class | Frequency, f | Midpoint, m | m 2 | 2 | fm 2 | Standard Deviation |
|---|---|---|---|---|---|---|
| 0–2 | 1 | 1 | 1 | 7.58 | 1 | 3.5 |
| 3–5 | 6 | 4 | 16 | 7.58 | 96 | 3.5 |
| 6–8 | 10 | 7 | 49 | 7.58 | 490 | 3.5 |
| 9–11 | 7 | 10 | 100 | 7.58 | 700 | 3.5 |
| 12–14 | 0 | 13 | 169 | 7.58 | 0 | 3.5 |
| 15–17 | 2 | 16 | 256 | 7.58 | 512 | 3.5 |
For this data set, we have the mean, = 7.58 and the standard deviation, s x = 3.5. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since 7.58 – 3.5 – 3.5 = 0.58. While the formula for calculating the standard deviation is not complicated, where s x = sample standard deviation, = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations.
Find the standard deviation for the data from the previous example
| Class | Frequency, f |
|---|---|
| 0–2 | 1 |
| 3–5 | 6 |
| 6–8 | 10 |
| 9–11 | 7 |
| 12–14 | 0 |
| 15–17 | 2 |
Use your calculator to find the standard deviation.
The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading.
#ofSTDEVs is often called a " z -score"; we can use the symbol z . In symbols, the formulas become:
| Sample | = + zs | |
| Population | = + zσ |
Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school?
| Student | GPA | School Mean GPA | School Standard Deviation |
|---|---|---|---|
| John | 2.85 | 3.0 | 0.7 |
| Ali | 77 | 80 | 10 |
For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.
For John,
For Ali,
John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean.
John's z -score of –0.21 is higher than Ali's z -score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.
Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team?
| Swimmer | Time (seconds) | Team Mean Time | Team Standard Deviation |
|---|---|---|---|
| Angie | 26.2 | 27.2 | 0.8 |
| Beth | 27.3 | 30.1 | 1.4 |
For Angie: z = = –1.25
For Beth: z = = –2
The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data.
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