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TRIMESTER – 5
MID-TERM QUESTION PAPER
QUANTATIVE TECHNIQUES
TIME ALLOWED
: 2 hours
Maximum Marks: 50
Attempt all questions :
(It is an open book test. Please show complete working notes)
Q-1) Construct a sub-divided bar chart for the 3 types of expenditures in dollars of a
10 marks
family of four for the years 2002, 2003, 2004 and 2005 as given below:
Year
2002
2003
2004
2005
Expenditure
Food
3000
3500
4000
5000
Education
2000
3000
3500
5000
Other
3000
4000
5000
6000
Total
8000
10500
12500
16000
OR
Construct “Less than ogive” and “More than ogive”
10 marks
Class Interval
Frequency
Cumulative Frequency
Less than
Cumulative Frequency
More than
15 Upton 25
25 “ 35
35 “ 45
45 “ 55
55 “ 65
65 “ 75
5
3
7
5
3
7
5
8
15
20
23
30
30
25
22
15
10
7
OR
Write the difference between Bar diagram and Histogram
BAR DIAGRAM:
- It is a diagrammatic representation of data.
- It is one of the ways of representing qualitative data.
- The bar diagram is unidimentional, that is, the length of the bar
defines the value of the data it represents.
- The lengths of the bars are proportional to their corresponding
numerical value.
- While constructing the bar diagram, care should be taken to keep the
width of the bars uniform to avoid ambiguity.
- Also while constructing the bar diagram, they should be equally
placed and the bars have space in between them.
HISTOGRAM:
-
It is a graphic representation of data.
-
Here the data are plotted in the form of rectangles in a graph.
-
The histogram is two dimensional, because the length and width
of the rectangle signifies the values it represents.
-
The height of the rectangle gives the frequency and the width
gives the class interval of the frequency distribution of the
grouped data.
-
The rectangles are joined together with the other and empty
space between the rectangles signifies that the category is
empty and there are no values in that class interval.
Q-2) Calculate the Karl Pearson’s Coefficient of Skewness for the following
15 marks
Class Interval
f
5-10
10-15
15-20
20-25
25-30
30-35
35-40
5
10
15
20
15
10
5
Values
X
f
5-10
10-15
15-20
20-25
25-30
30-35
35-40
7.5
12.5
17.5
22.5
27.5
32.5
37.5
5
10
15
20
15
10
5
d
d’
-15
-10
-5
0
5
10
15
-3
-2
-1
0
1
2
3
N =80
Where d= X – 22.5,
fd’
fd
cf
-15
-20
-15
0
15
20
15
45
40
15
0
15
40
45
5
15
30
50
65
75
80
Sigma fd’=0
200
d’ = d/5
Karl Pearson’s Coefficient of Skewness = 3 (Mean- Median)/ Standard deviation
Mean = A + Sigma fd’/N multiplied by C
= 22.5 + 0/80 X 5 = 22.5
Median = Size of N/2 = 80/2 = 40th item
Thus median lies in the class interval 20-25
Then,
Median = l1 + N/2 – cf / f X C
= 20 + 10/20 X 5 = 22.5
Standard deviation sigma = sq. rt. Of Sigma fd’2/N – ( sigma fd’/N ) 2 X C
= sq.rt. of 200/80 – (0/80)2 X 5
= sq.rt. of 2.5 X 5 = 1.6 X 5 = 8
Coefficient of Skewness = 3 (Mean – Median) / Standard deviation
= 3 ( 22.5 – 22.5)/8
=0
As the co-efficient of skewness is 0, the distribution is symmetrical.
OR
Calculate the i) Quartile Deviation iii) Coefficient of Skewness
15 marks
Class Interval
f
cf
5-10
10-15
15-20
20-25
25-30
30-35
35-40
5
10
15
20
15
10
5
5
15
30
50
65
75
80
Q-3) Do Any Two A OR B OR C
20 marks
A). If a new drug has been found to be effective 40% of the time, then what is the
probability that in a random sample of 4 patients, it will be effective on 2 of them?
Find “Effective” as success and “Non-effective”.
Probability of success p = 40/100 = 0.4
Probability of failure q = 1-p = 1- 0.4 = 0.6
x=2
n=4
10 marks
P (x) = (n/x) (p)2 (q)2
= 4 !/ 2! 2! (0.4) 2 (0.6) 2
= 4X3X2X1 / 2X1 2X1 X 0.16 X 0.36
= 6 X0.16 X 0.36
= 0.3456
Hence, the probability for the new drug to be effective on 2 patients is 34.56 %
B).Explain any with Example in brief
a) Addition Rule
OR
b) Multiplication Rule
OR
c) Conditional Probability
10 marks
C.) The probability that a student passes a test in Statistics is 2/3, and the probability that
he passes both a test in statistics and a test in mathematics is 14/45. The probability that
he passes the test in at least one test is 4/5. What is the probability that he passes the test
in Mathematics?
- Let the event that the student passes in statistics be A
- Let the event that the student passes in Mathematics be B
Given that,
P(A) = 2/3
P( A and B) = 14/45
P(A or B) =
4/5,
10 marks
Then,
P(B) = is derived from the formula P(A or B) = P(A) + P(B) – P(A and B)
Therefore,
4/5 = 2/3 + P(B) – 14/15
P(B) = 4/5 + 14/45 – 2/3
36 +14 – 30
= __________
45
= 20/45 = 4/9
Therefore, the probability that the student passes in Mathematics is 4/9.
Q-4)
A babysitter has 5 children under her supervision with average age of 6 years. But
individually, the age of each child be as follows:
X1 = 2
X2 = 4
X3 = 6
X4 = 8
Calculate Population Mean and Standard Deviation.
Population size N = 5
Population mean mu = Sigma X/N
= 2 +4+6+8+10 / 5 = 30/5
=6
Standard Deviation, Sigma = Sq.rt. of Sigma (X- mu)2/N
Let the corresponding values be tabulated as follows:
X5 = 10
5 marks
X
mu
(X-mu)2
2
6
16
4
6
4
6
6
0
8
6
4
10
6
16
Total = Sigma (X-mu)2 =
40
Therefore, Standard deviation = Sq. rt. 40/5
= Sq. rt. Of 8
= 2.83.
OR
Explain
a) Null Hypothesis
OR
b) Alternative Hypothesis
5 marks