1. HISTOGRAMS Representing Data Module S1

2. Why use a Histogram When there is a lot of data When data is Continuous a mass, height, volume, time etc Presented in a Grouped Frequency Distribution usually in groups or classes that are UNEQUAL

3. Continuous data NO GAPS between Bars

4. Bars are different in width Determined by Grouped Frequency Distribution

5. AREA is proportional to FREQUENCY NOT height, because of UNEQUAL classes! So we use FREQUENCY DENSITY = Frequency Class width

6. Grouped Frequency Distribution Time taken (nearest minute) 5-910-1920-2930-3940-59 Freq1491835 Speed, kph0< v 4040< v 5050< v 6060< v 9090< v 110 Frequency8015259030 Classes No gaps GAPS!Need to adjust to Continuous Ready to graph

7. Adjusting Classes Class Widths Time taken (nearest minute) 5-910-1920-2930-3940-59 Freq1491835 9½4½19½29½39½59½ 105 20

8. Frequency Density Time taken (nearest minute) 5-910-1920-2930-3940-59 Freq 1491835 Class width 510 20 Frequency Density 2.80.91.80.30.25

9. Drawing Sensible Scales Bases correctly aligned Plot the Class Boundaries Heights correct Frequency Density

10. 4.519.59.529.539.549.559.5 3.0 2.0 1.0 Freq Dens Time (Mins)

11. Estimating a Frequency Imagine we want to Estimate the number of people with a time between 12 and 25 mins Because rounded to nearest minute Consider the interval 11.5 to 25.5

12. 4.519.59.529.539.549.559.5 3.0 2.0 1.0 Freq Dens Time (Mins) 11.5 25.5 Frequency = 0.9 x 8 = 7.2 Frequency = 1.8 x 6 = 10.8 Total Frequency = 18

13. …and the other one? Simpler to plot No adjustments required – class widths friendly No ½ values Estimation from the EXACT values given No adjustment required Estimate 15 to 56 would use 15 and 56! Appear LESS OFTEN in the exam Speed, kph0< v 4040< v 5050< v 6060< v 9090< v 110 Frequency8015259030