Chapter Review
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The range is the difference between the greatest and least value. The standard deviation measures how much the data elements differ from the mean.
Ranges: 56, 86
Standard Deviations: Approximately 15.6 and 30.3
Comparison: See solution.
We want to find the range and standard deviation of each of the given data sets.
| Player A | Player B | ||
|---|---|---|---|
| 205 | 190 | 228 | 205 |
| 185 | 200 | 172 | 181 |
| 210 | 219 | 154 | 240 |
| 174 | 203 | 235 | 235 |
| 194 | 230 | 168 | 192 |
The range is the difference between the greatest and least values in a set of data. Now, we can rewrite each data set as an ordered list of numerical values and find their range.
Now, we can do the same for the scores of the player B. 154, 168, 172, 181, 192, 205, 228, 235, 235, 240 For this data set, the greatest value is 240 and the least value is 154. Range: 240- 154=86
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-μ )^2 +(x_2-μ )^2 + ... +(x_n-μ )^2/n) In the above formula, x_1, ... ,x_n are the values of the set of data, μ is the mean, and n is the number of values.
For this data set, we can calculate the sum of the values. 174+185+190+194+200+203+ 205+210+219+230 = 2010 The mean is the sum of the values 2010 divided by the number of values 10. μ= 2010 10=201 Let's use this value and apply the formula to each value in the set.
| x_n | x_n-μ | (x_n-μ)^2 |
|---|---|---|
| 174 | 174-201=-27 | (-27)^2= 729 |
| 185 | 185-201=-16 | (-16)^2= 256 |
| 190 | 190-201=-11 | (-11)^2=121 |
| 194 | 194-201=-7 | (-7)^2=49 |
| 200 | 200-201=- 1 | (- 1)^2=1 |
| 203 | 203-201=2 | 2^2=4 |
| 205 | 205-201=4 | 4^2=16 |
| 210 | 210-201=9 | 9^2=81 |
| 219 | 219-201=18 | 18^2=324 |
| 230 | 230-201=29 | 29^2=841 |
| Sum of Values | = 2422 | |
Finally, we need to divide by 10 and then calculate the square root. Standard Deviation: sqrt(2422/10)≈ 15.6
For this data set, we can calculate the sum of the values. 154+168+172+181+192+205+ 228+235+235+240 = 2010 The mean is the sum of the values 2010 divided by the number of values 10. μ= 2010 10=201 Let's use this value and apply the formula to each value in the set.
| x_n | x_n-μ | (x_n-μ)^2 |
|---|---|---|
| 154 | 154-201=-47 | (-47)^2= 2209 |
| 168 | 168-201=-33 | (-33)^2= 1089 |
| 172 | 172-201=-29 | (-29)^2=841 |
| 181 | 181-201=- 20 | (- 20)^2=400 |
| 192 | 192-201=-9 | (-9)^2=81 |
| 205 | 205-201=4 | 4^2=16 |
| 228 | 228-201=27 | 27^2=729 |
| 235 | 235-201=34 | 34^2=1156 |
| 235 | 235-201=34 | 34^2=1156 |
| 240 | 240-201=39 | 39^2=1521 |
| Sum of Values | = 9198 | |
Finally, we need to divide by 10 and then calculate the square root. Standard Deviation: sqrt(9198/10)≈ 30.3
The range of the bowling scores of Player A is 56, and the range of the scores of Player B is 86. So, the scores are more spread out for Player B. Both of the data sets have the same mean. The standard deviation for Player A is about 15.6, and the standard deviation for Player B is about 30.3. Both of the data sets have the same mean. Therefore, the typical score of Player A differs by 15.6 from the mean, while the typical score of Player B differs by 30.3.