Answer:
a. Mean = 35.2 ≈ 35
b. Median = 35.6 ≈ 36
c. Mode = 36.6 ≈ 37
Explanation:
==>Given:
Class of ages in yrs
No. of cases of each class = f
Midpoint of each class = x
Product of midpoint and no. of cases of each class = fx
==>Required:
a. Mean
b. Median
c. Mode
==>SOLUTION:
a. Mean = (Σfx)/Σf
Σf = sum of no. of cases = 5+10+20+22+13+5 = 75
Σfx = 47.5+195+590+869+643.5+297.5 = 2,642.5
Mean = 2,642.5/75 = 35.2 ≈ 35
b. Median = Lm + [((Σf/2) - Cfb)/fm]Cw
Our median is between the 37th and the 38th term, which can be found in the class interval 35-44. This is our median class.
Lm = Lower class boundary of the median class = lower limit of the Medina class + upper limit of the class before the median class ÷ 2 = (35+34)/2 = 34.5
Σf/2 = 75/2 = 37.5
Cfb = Cumulative frequency of class before the median class = 5+10+20 = 35
fm = frequency of the Medina class = 22
Cw = Class width = upper class boundary - lower class boundary = 44.5-34.5 = 10
Median = 34.5 + [(37.5-35)/22] × 10
= 34.5 + [2.5/22] × 10
= 34.5 + [25/22]
= 34.5 + 1.1
= 35.6 ≈ 36
c. Mode = Lm + [∆¹/(∆¹+∆²)]Cw
Modal class = (35-44) [i.e. the class with the highest frequency, which is where our mode falls in]
Lm = lower class boundary of the modal class = lower limit of the modal class + upper limit of the class before the modal class ÷ 2 = (35+34)/2 = 34.5
∆¹ = difference between the frequency of the modal class & the frequency of the class before the modal class = 22 - 20 = 2
∆² = difference between the frequency of the modal class & the frequency of the class after the modal class = 22 - 13 = 9
Cw == Upper class boundary - Lower class boundary = 44.5 - 34.5 = 10
Mode = 34.5 + [2/(2+9)] × 10
= 34.5 + [2/11] × 10
= 34.5 + [20/11]
= 34.5 + 1.8
Mode = 36.6 ≈ 37