| Frequency distribution: | | | | 10 |
| The table below contains data collected: | | | | Standard deviation: |
| DATE | | | | The standard deviation for grouped data is calculated |
| MINUTES TO DRIVE TO WORK | | | | as follows: |
| MON/5 OCT | | | | Sd = [(FX2/ f) – (FX/F)2]½ |
| 30 MIN | | | | The table below summarizes the midpoints of the |
| TUES/6 OCT | | | | classes and calculations made to determine the |
| 30 MIN | | | | standard deviation:xclass |
| WED/7 OCT | | | | Frequency(FX)mid point |
| 20 MIN | | | | FX |
| THURS/8 OCT | | | | FX2 |
| 20 MIN | | | | 0 to 10 |
| FRI/9 OCT | | | | 6 |
| SICK LEAVE/DID NOT GO TO WORK | | | | 5 |
| MON/12 OCT | | | | 30 |
| COLUMBUS DAY/DID NOT GO TO WORK | | | | 900 |
| TUES/13 OCT | | | | 11 to 21 |
| SICK LEAVE/DID NOT GO TO WORK | | | | 2 |
| WED/14 OCT | | | | 16 |
| SICK LEAVE/DID NOT GO TO WORK | | | | 32 |
| THURS/15 OCT | | | | 1024 |
| SICK LEAVE/DID NOT GO TO WORK | | | | 22 to 32 |
| FRI/16 OCT | | | | 2 |
| SICK LEAVE/DID NOT GO TO WORK | | | | 27 |
| We assume that for the days that the individual did not | | | | 54 |
| report to work the minutes taken to drive to work | | | | 2916total |
| amount to zero, therefore the table is summarized as | | | | 10 |
| follows: | | | | 116 |
| DATE | | | | 4840 |
| MINUTES TO DRIVE TO WORK | | | | Given |
| MON/5 OCT | | | | Sd = [(FX2/ f) – (FX/F)2]½ |
| 30 MIN | | | | Then |
| TUES/6 OCT | | | | Sd = [(4840/ 10) – (116/10)2]½ |
| 30 MIN | | | | Sd = 18.69331 |
| WED/7 OCT | | | | Normal distribution: |
| 20 MIN | | | | The central limit theorem givens the conditions and |
| THURS/8 OCT | | | | properties of a normal distribution, they include: |
| 20 MIN | | | | 68% of data is contained within one standard deviation, |
| FRI/9 OCT | | | | 95% of the data is contained within two standard |
| 0 | | | | deviations, the mean value is determined as follows: |
| MON/12 OCT | | | | Mean = FX/F |
| 0 | | | | Mean = 116/ 10 = 11.6 |
| TUES/13 OCT | | | | 68% of observations: |
| 0 | | | | Standard deviation = 18.69331 |
| WED/14 OCT | | | | Mean = 11.6 |
| 0 | | | | Range of data |
| THURS/15 OCT | | | | (11.6+ 18.69331) and (1.6 – 18.69331) |
| 0 | | | | (30.29331) and (-7.09331) |
| FRI/16 OCT | | | | From our case data that ranged from -7.09331 to |
| 0 | | | | 30.29331 is greater than 68%, therefore the distribution |
| The frequency distribution table will contain three | | | | is not a normal distribution. |
| classes and they include the class 0 to 10, 11 to 21 and | | | | Implications: |
| 22 to 32, the table below summarizes the | | | | Given that this is not a normal distribution this means |
| frequencies:classfrequency | | | | that statistical tests that assume normal distribution |
| 0 to 10 | | | | cannot be applied, also this means that the sample is |
| 6 | | | | not large enough given that the central limit theorem |
| 11 to 21 | | | | states that as the number of random numbers |
| 2 | | | | increase the data assumes a normal distribution. |
| 22 to 32 | | | | Reference: |
| 2total | | | | Mendenhall, W. |