Elementary Statistics and Computer Application

Ungrouped or Raw Data

Computation of the mode

• For ungrouped data or a series of individual observations, mode is often found by mere inspection.

Example 8
Find the mode for the following seed weight 2 , 7, 10, 15, 10, 17, 8, 10, 2 gms
.^..Mode = 10
In some cases the mode may be absent while in some cases there may be more than one mode.

Example 9
(1) 12, 10, 15, 24, 30 (no mode)
(2) 7, 10, 15, 12, 7, 14, 24, 10, 7, 20, 10
the modal values are 7 and 10 as both occur 3 times each.

Grouped Data

• For Discrete distribution, see the highest frequency and corresponding value of x is mode.

Example

Weight of sorghum in gms (x)	No. of ear head (f)
50	4
65	6
75	16
80	8
95	7
100	4

Solution

• The maximum frequency is 16. The corresponding x value is 75.
• .^.. mode = 75 gms.

Continuous distribution

• Locate the highest frequency the class corresponding to that frequency is called the modal class. Then apply the formula.

Where = lower limit of the model class

= the frequency of the class preceding the model class
= the frequency of the class succeeding the model class

and c = class interval
Example 10
For the frequency distribution of weights of sorghum ear-heads given in table below. Calculate the mode

Weights of ear heads (g)	No of ear heads (f)
60-80	22
80-100	38
100-120	45	f
120-140	35
140-160	20
Total	160


Solution

Here 100, f = 45 , c = 20, m =60, =38, =35


= 109.589
Geometric mean

• The geometric mean of a series containing n observations is the nth root of the product of the values.
• If x1,x2…, xn are observations then







For grouped data

GM is used in studies like bacterial growth, cell division, etc.

Example 11
If the weights of sorghum ear heads are 45,60,48,100 , 65 gms . Find the Geometric mean for the following data

Weight of ear head x (g)	Log x
45	1.653
60	1.778
48	1.681
100	2.000
65	1.813
Total	8.925




= 60.95
Grouped Data

Example 12

Find the Geometric mean for the following

Weight of sorghum (x)	No. of ear head(f)
50	4
65	6
75	16
80	8
95	7
100	4
Here n= 50





= Antilog 1.9842 = 96.43