/ PESIT Bangalore South Campus
Hosur road, 1km before Electronic City, Bengaluru -100
Department of MBA
INTERNAL ASSESSMENT TEST – 1
Date: 07/11/2017 Max Marks: 40 Marks
Subject & Code:QUANTITATIVE METHODS (16MBA14) Section: Core
Name of faculty:Asst. Prof Roshny Unnikrishnan Time: 11:30– 1:00PM
Note: Answer all questions
1(a) / Find the Mode from the following dataWeight / 93-97 / 98-102 / 103-107 / 108-112 / 113-117 / 118-122 / 123-127 / 128-132
No:of students / 3 / 5 / 12 / 17 / 14 / 6 / 3 / 1
Answer:
Modal class = 108 – 112
Mode = 108 + (((17-12)/2(17)-12-14) * 5) = 110.125 / (2 marks)
(b) / Calculate First quartile (Q1) , Third Quartile (Q3), Quartile deviation and coefficient of Quartile deviation from the following data.
Wages / 32-34 / 35-37 / 38-40 / 41-43 / 44- 46
No: of wage earners / 14 / 62 / 99 / 18 / 7
Answer:
Wages / 32-34 / 35-37 / 38-40 / 41-43 / 44- 46
No: of wage earners / 14 / 62 / 99 / 18 / 7
CF / 14 / 76 / 175 / 193 / 200
Q1= size of N/4th item = 50
Q1 = 35+(50-14/62*2) = 36.16
Q3 = size of 3n/4th item = 150
Q3 = 38+(150-76/99*2) = 39.49
Q.D = 39.49-36.16/2 = 1.668
Co- efficient of Q.D = (Q3 –Q1)/ (Q3 +Q1) = 3.33/75.65 = 0.044 / (6 marks)
(c) / An Analysis of monthly wages of workers of two organizations C and D yielded the following results
Organization
C / D
No:of workers / 50 / 60
Average Monthly wages / Rs 60 / Rs 48
Variance / 100 / 144
Obtain the average monthly wage and standard deviation of all the workers in the two organizations taken together. Which organization is more equitable with respect to wages?
Answer:
σ12 =
= √159.7025 = 12.637
C.V of C = 10/60*100 = 16.67
C.V of D = 12/48 *100 = 25
C is more equitable / (8 marks)
2(a) / Define Median. What are the advantages and limitations of Median? / (2 marks)
(b) / Find Mode with the help of mean and median from the following frequency distribution
Marks / 40-45 / 45-50 / 50-55 / 55-60 / 60-65 / 65-70 / 70-75 / 75-80
No:of students / 8 / 20 / 50 / 60 / 65 / 21 / 12 / 6
Answer :
Marks / 40-45 / 45-50 / 50-55 / 55-60 / 60-65 / 65-70 / 70-75 / 75-80
No:of students / 8 / 20 / 50 / 60 / 65 / 21 / 12 / 6
CF / 8 / 28 / 78 / 138 / 203 / 224 / 236 / 242
A= 62.5
∑ fd = -208
Mean = 62.5 + (-208/242*5) = 58.21
Median = 55+(121-78/60*5) = 58.58
Mode = 60+((65-60/2(65)-60-21)*5) = 60.51 / (6 marks)
(c) / Frequency distribution of 100 families are given. Find the missing frequencies. Median = 50
X / 0-20 / 20-40 / 40-60 / 60-80 / 80-100
f / 14 / ? / 27 / ? / 15
Answer:
Let the missing frequencies be f1 and f2
X / 0-20 / 20-40 / 40-60 / 60-80 / 80-100
f / 14 / ? / 27 / ? / 15
CF / 14 / 14 +f1 / 41+ f1 / 41+ f1+ f2 / 56+ f1+ f2
Since median = 50 median class is 40-60
50 = 40+ ((50-14+ f1/27) * 20)
f1= 23
f2= 100 – (56+ f1) = 21 / (8 marks)
3 / Case Study
The scores of two batsman A and B in ten innings during a certain season are :
A / 32 / 28 / 47 / 63 / 71 / 39 / 10 / 60 / 96 / 14
B / 19 / 31 / 48 / 53 / 67 / 90 / 10 / 62 / 40 / 80
Find which of the two batsman is more consistent in performance?
Answer:
Batsman A
∑ X= 0
∑ X2 = 6500
Mean = 460/10=46
σ = = 25.495
C.V = 25.495/46*100 = 55.42
Batsman B is more consistent as C.V is lesser / (8 marks)
MBA ISEMESTER