# S1 Averages and Measures of Dispersion

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S1 Averages and Measures of Dispersion

S1 Measures of Dispersion
Objectives: To be able to find the median and quartiles for discrete data To be able to find the median and quartiles for continuous data using interpolation

Can you work out the rule for finding median and quartiles from discrete data?
LQ Median UQ Can you spot any rules for n amount of numbers in a list?

LQ  n/4 If n/4 is a whole number find the mid point of corresponding term and the term above If n is not a whole number, round the number up and find the corresponding term UQ  3n/4 If 3n/4 is a whole number find the mid point of corresponding term and the term above If n is not a whole number, round the number up and find the corresponding term

Median  n/2 If n/2 is a whole number find the midpoint of the corresponding term and the term above
If n/2 is not a whole number, round up and find the corresponding term

Calculate the mean, median and inter quartile range from a table of discrete data
Number of CDs(x) Number of students (f) 35 3 36 17 37 29 38 34 39 12 Mean = Σfx Σf

Calculate the mean, median and inter quartile range from a table of discrete data
Median = n/2 Number of CDs(x) Number of students (f) 35 3 36 17 37 29 38 34 39 12 Cumulative frequency Median = 95/2 = 47.5 = 48th value 3 20 49 83 95 Median = 37 CDs LQ = 95/4 = 23.75 LQ = 24th value LQ (Q1) = 37 CDs UQ (Q3) = 95/4 x 3 = 71.25 UQ = 72nd value UQ (Q3) = 38 CDs IQR = Q3-Q1 = 38-37=1

Length of flower stem (mm)
Calculate the mean, median and inter quartile range from a table of continuous data Median = n/2 We do not need to do any rounding because we are dealing with continuous data Length of flower stem (mm) Number of flowers (f) 30-31 2 32-33 25 34-36 30 37-39 13 Cumulative frequency 2 27 57 70 Median = 70/2 = 35th value This lies in the class but we don’t know the exact value of the term

Using interpolation to find an estimate for the median
33.5mm m 36.5mm 27 35 57 m – = m – = 8 36.5 – = m – = 0.26 x 3 m = = 34.3

Using interpolation to find an estimate for the lower quartile
LQ = 70/4 = 17.5 (in the group) 31.5mm Q1 33.5mm 2 17.5 27 Q1 – = Q1 – = 15.5 33.5 – = Q1 – = 0.62 x 2 Q1 = = 32.74

Using interpolation to find an estimate for the upper quartile
UQ = 70/4x3 = 52.5 (in the group) 33.5mm Q3 36.5mm 27 52.5 57 Q3 – = Q3 – = 25.5 36.5 – = Q3 – = 0.85 x 3 Q1 = = 36.05

Summary of rules n = total frequency w = class width
fB = cumulative frequency below median/lq/uq fU = cumulative frequency above median/lq/uq Median = LB + ½n – fB x w fU - fB LQ = LB + ¼n – fB x w fU - fB UQ = LB + ¾n – fB x w fU - fB

The lengths of a batch of 2000 rods were measured to the nearest cm
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 60-64 11 65-69 49 70-74 190 75-79 488 80-84 632 85-89 470 90-94 137 95-99 23 Cumulative frequency Q1= x 5 Q1=77.06 11 60 250 738 1370 1840 1977 2000 Q2= x 5 Q2=81.57 Q3= x 5 Q3=85.88 By altering the formula slightly can you work out how to find the 3rd decile (D3) and the 67th percentile (P67)?

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