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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
Ans: Bar Diagram : It is very commonly used and is better for representation of qualitative data Bars are
simply vertical lines, where the lengths of the bars are proportional to there corresponding numerical
values. The width of the bar is un important but all bars should have the same width so as not to confuse
the reader of the diagram. Additionally the bar should be equally spaced.
With revenue represented by the vertical
axis and the year represented by the horizontal axis.
Eg.: The bar diagram can be constructed as follows :
120
100
Revenue
80
60
40
1980
1981
1982
Year
In statistics, a histogram is a graphical display of tabulated frequencies,
shown as bars. It shows what proportion of cases fall into each of several categories: it is
a form of data binning. The categories are usually specified as non-overlapping intervals
of some variable. The categories (bars) must be adjacent. The intervals are generally of
the same size.
Histogram :
Eg.:
In other words a histogram represents a frequency distribution by means of rectangles
whose widths represent class intervals and whose areas are proportional to the
corresponding frequencies. They only place the bars together to make it easier to compare
data.
Q-2) Calculate the Karl Pearson’s Coefficient of Skewness for the following
Class Interval
f
5-10
10-15
15-20
20-25
25-30
30-35
35-40
5
10
15
20
15
10
5
Ans: Calculation of coefficient of Skewness :
Values
5—10
10--15
15--20
20--25
25--30
30--35
35--40
Mid
Point
X
7.5
12.5
17.5
22.5
27.5
32.5
37.5
f
d
d/5
fd1
fd2
cf
5
10
15
20
15
10
5
N=80
-15
-10
-5
0
5
10
15
-3
-2
-1
0
1
2
3
-15
-20
-15
0
15
20
15
Efd1=0
45
40
15
0
15
40
45
200
5
15
30
50
65
75
80
15 marks
Karl Person’s Cofficient of Skewness = 3(Mean-Median)/Std. Deviation
Mean = A + E fd’/N x C
= 22.5 + 0/80 x 5
= 22.5
Median is the size of N/2 or 80/2 or 40th Item
Thus median lies in class 20-25
By interpolation :
Median = l1 + N/2-cf/f x c
= 20 + 40-30/20 x 5
= 22.5
Std. Deviation of z = 1.6 x 5 = 8
Cofficient of Skewness = 3(Mean- Median)/Std. Deviation
= 3(22.5-22.5)/8
=0
Coefficient of Skewness is 0, the distribuition 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”.
10 marks
Ans A) p= .4(since the drug is effective 40% of the time)
q = (l-p) = (l-.4) = .6
x= 20
n= 4
Then :
P(x) = (n/x) (p)2 (q)2
= (4/2) (.4)2 (.6)2
= 4!/2!2!(.4)2 (.6)2
= 6 X.16 X.36
= 0.3456.
The mean and the standard deviation for the distribution are given as follows :
μ = np
Sigma =( npq)2
B).Explain any with Example in brief
a) Addition Rule
OR
b) Multiplication Rule
OR
c) Conditional Probability
10 marks
Ans :B) Addition Rule : When two events are mutually exclusive, then the probability
that one or the other event will occur, is the sum of their separate probabilities
For example: if we roll a single die then probability that it will come up with face 5 or 6
where event A is face 5 and event B is face 6 is given by :
P[A or B] =P[A] + P[B]
P[5 or 6] = P[5] + P[6]
=1/6+1/6
=2/6
=1/3
P[A or B] is written as P[AUB] which is also known as Prob. [a union B]
b) Multiplication Rule : If event A and B are independent events then the probability
that they will both occur is the product of their seprate probabilities.
Events A and B are independent if and only if ,
P[AB] = P[A] x P[B]
Example : if we toss a coin twice the probability that the first results in head and
second toss results in a tail is given by :
P[H,T] = P[H] x P[T]
=½x½=¼
If events A and B are not independent meaning that the probability of occurrence
of one events is dependent or conditional upon the occurrence or non occurrence of the
other event then the probability that they will both occur is given by :
P [AB] = P[A]x P[B]
P[AB] = P[A] x P[B/A]
Where P[B/A] means the probability of event B on the condition of the out come of event
A.
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?
10 marks
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
Calculate Population Mean and Standard Deviation.
X4 = 8
X5 = 10
5 marks
OR
Explain
a) Null Hypothesis
OR
b) Alternative Hypothesis
5 marks
Ans : a)
Null Hypothesis : formally describes some aspect of the statistical behaviour of
a set of data; this description is treated as valid unless the actual behaviour of the data
contradicts this assumption. Thus, the null hypothesis is contrasted against another
hypothesis. Statistical hypothesis testing is used to make a decision about whether the
data contradicts the null hypothesis: this is called significance testing. A null hypothesis
is never proven by such methods, as the absence of evidence against the null hypothesis
does not establish it. In other words, one may either reject, or not reject the null
hypothesis; one cannot accept it. Failing to reject it gives no strong reason to change
decisions predicated on its truth, but it also allows for the possibility of obtaining further
data and then re-examining the same hypothesis.
Example: one may want to compare the test scores of two random samples of men and
women, and ask whether or not one group (population) has a mean score (which really is)
different from the other. A null hypothesis would be that the mean score of the male
population was the same as the mean score of the female population:
H0 : μ1 = μ2
where:
H0 = the null hypothesis
μ1 = the mean of population 1, and
μ2 = the mean of population 2.