1. S1 Measures of Dispersion The mean, variance and standard deviation

2. S1 Measures of Dispersion Objectives: To be able to find the variance and standard deviation for discrete data To be able to find the variance and standard deviation for continuous data

3. A factory worker counted the numbers of nuts in a packet. The results are shown in the table Number of nuts Freq CF 688 71220 83656 91874 101589 111099 Calculate P 80 and P 40 P 80 = 99/100x80 = 79.2 P 80 = 80th term P 80 = 10 nuts P 40 = 99/100x40 = 39.6 P 40 = 40th term P 40 = 8 nuts Calculate the 40 th to 80 th interpercentile range 40 th to 80 th interpercentile range = 10 - 8 = 2 nuts

4. The lengths of a batch of 2000 rods were measured to the nearest cm. The measurements are summarised below. Length (nearest cm) Number of rods 60-6411 65-6949 70-74190 75-79488 80-84632 85-89470 90-94137 95-9923 Cumulative frequency 11 60 250 738 1370 1840 1977 2000 Q1=74.5 + 500-250 x 5 738-250 Q1=77.06 Q2=79.5+1000-738 x 5 1370-738 Q2=81.57 Q3=84.5+1500-1370 x 5 1840-1370 Q3=85.88 By altering the formula slightly can you work out how to find the 3 rd decile (D3) and the 67 th percentile (P67)?

5. Answers D3=74.5 + 600-250 x 5 738-250 D3=78.09 P67=79.5 + 1340-738 x 5 1370-738 P67=84.26

6. Variance of discrete data The deviation (difference) of the data (x) from the mean (x) is one way of measuring the dispersion (spread) of a set of data. _ Variance = Σ(x – x)² n _ Variance = Σx² – Σx ² n n _

7. As variance is measured in units ² you usually take the square root. This gives you the standard deviation. Standard deviation. Standard deviation = variance Standard deviation symbol is σ

8. The marks scored in a test by seven students are 3,4,6,2,8,8,5. Calculate the variance and standard deviation. x 3 4 6 2 8 8 5 x² 9 16 36 4 64 25 Σx = 36Σx² = 218 Variance = Σx² – Σx ² n n _ σ ² = 218 – 36 ² = 4.69 7 7 σ = 4.69 = 2.17

9. The variance and standard deviation from a frequency table fx 105 612 1073 1292 1014 Number of rods (x) Freq of rods (f) 353 3617 3729 3834 3926 fx² 3675 22032 39701 49096 39546 Σf=109 Σfx=4096Σfx²=154050 Variance = Σfx² – Σfx ² Σf Σf Variance = 154050 – 4096 ² 109 109 σ ²=1.19805 σ = 1.109

10. Variance and standard deviation from a grouped frequency distribution Height of plant (cm) Freq (f) 0 < h 54 5 < h 1015 10 < h 155 15 < h 202 20 < h 250 25 < h 301 Total σ ² = 6487.5 – 285 ² = 128.85802 27 27 σ = 128.85802 σ = 11.35