Basic Tools For Economic Analysis Cont’d. Finding Median from A Grouped Frequency Distribution. Median for a Grouped frequency distribution is calculated using the formula below: Me=L1+(N/2-Fo)c Fe Where L1= the value of the lower class boundary of the median class; N=total number of items Fo= Cumulative frequency of the class just above the one containing the median Fe=Frequency of the median class C= Class size Eg. Use the table below to calculate the median mark.

Median from a grouped frequency distribution Class interval frequency Class boundary Cumulative frequency 1-10 6 0.5-10.5 11-20 8 10.5-20.5 14 21-30 20.5-30.5 20 31-40 4 30.5-40.5 24 41-50 40.5-50.5 28

From table above N=28, L1=10.5, C=10, Fe=8, Fo=6 Substituting we have Median= Me=L1+(N/2-Fo)c Fe Me=10.5+(28/2-6)10 8 =10.5+(14-6)10 =10.5+10 Median=20.5 The Median mark is 20.5 ans

Advantages of Median -The median could be determined by observation -It can be obtained graphically -It is easy to understand -It is easy to compute -It is not affected by extreme values Disadvantages of Median -It ignores extreme values -It is usually not representative of all values in distribution -It is not easy to arrange the data in order of magnitude when large population is involved

The Mode Mo=L1+ (F1-F0)c F1-F0+F1-F2 Where Mo= Mode The mode is the most frequent recurring number in a set of data. That is the number or value with the highest frequency 1)Eg. Find the mode of the following distribution: 3, 4, 5, 6, 9, 4, 6, 6, 7, 4, From the above numbers, the mode is 6. ie 6 has the highest appearance. 2) Finding mode using calculation method Mode can be calculated using the formula Mo=L1+ (F1-F0)c F1-F0+F1-F2 Where Mo= Mode L1=Lower class boundary of the modal class C= Class size F1= frequency of the modal group F0= frequency of the pre modal group F2= frequency of the post modal group

Mode continued Eg. Consider the distribution below Class Boundary Frequency 0.5-9.5 5 9.5-19.5 7 19.5-29.5 9---F0 29.5-39.5 12---F1 39.5-49.5 9---F2 49.5-59.5 6 59.5-69.5 69.5-79.5 3

Mode continued In calculating we use Mo=L1+(F1-F0/F1-F2+F1-F2)C L1=29.5, F1=12, F0=9, F2=9, C=10 So, Mo= 29.5+(12-9/12-9+12-9)10 = 29.5+(3/3+3)10 29.5+(3/6)10 =29.5+(30/6) =29.5+5 =34.5 ans

Assignment 1) Compare and contrast median and mode 2)Find the mode of the following a) 5, 7, 4, 5, 9, 4, 7, 8 b) 2, 4, 8, 5, 4, 3, 6, 7, 4, 2, 8, 4, 3, 5, 4, 2, 4.