Question

Approximate the measures of center for following GFDT.

Data Frequency

50 - 54 2

55 - 59 1

60 - 64 3

65 - 69 6

70 - 74 7

75 - 79 11

80 - 84 10

85 - 89 23

90 - 94 14

mode =

median =

mean =

Report mode and median accurate to one decimal place. Report the mean accurate to two decimal places (or enter as a fraction).

Answer 1

Solution

First we need to make class intervals continous so we subtract 0.5 from lower limit and add 0.5 in upper bound so we get the following table

Solution: Class (1) Frequency () Mid value (x) (2) (3) x - 72 he 5 4 = 72, h = 5 fod (5) = (2) (4) (7) 2 -4 -8 -3 -2 -3 -6 نيا 6 -1 7 49.5 - 54.5 54.5 - 59.5 59.5-64.5 64.5 - 69.5 69.5 - 74.5 74.5 - 79.5 79.5 - 84.5 84.5 - 89.5 89.5-94.5 --- 52 57 62 67 72=A 77 82 87 92 0 -6 0 11 2 3 6 12 19 30 40 63 77 1 11 10 23 20 2 3 69 56 14 4 = 77 Sf. d= 133 Στα Mean i = 4 + = 72 + 133 77 = 72 +1.7273 - 5 = 72 +8.6364 = 80.6364

mean = 80.6

To find Median Class = value of observation = value of 77th observation = value of 38th observation From the column of cumulative frequency of, we find that the 38th observation lies in the class 79.5 - 84.5. : The median class is 79.5 - 84.5. ..= Now, :: L = lower boundary point of median class 79.5 Total frequency = 77 :: cf = Cumulative frequency of the class preceding the median class = 30 : f= Frequency of the median class = 10 : c = class length of median class = 5 -of Median M=L+ = 79.5 = 38.5 - 30 .5 10 8.5 = 79.5 - 5 10 = 79.5-4.25 = 83.75

Median = 83.8

To find Mode Class Here, maximum frequency is 23. · The mode class is 84.5 - 89.5. : I = lower boundary point of mode class = 84.5 frequency of the mode class = 23 .fo = frequency of the preceding class = 10 - 13 = frequency of the succedding class = 14 :: c = class length of mode class = 5 Z= L - f1-fo 2. f - fo - fz = 84.5+ 23 - 10 2.23 - 10 - 14 .5 13 = 84.5+ .5 22 = 84.5 -2.9545 = 87.4545

Mode = 87.4