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MID-TERM QUESTION PAPER
QUANTITATIVE TECHNIQUES
Maximum Marks: 50
Attempt all questions :
(It is an open book test. Please show complete working notes)
Q.1) Write the difference between Bar diagram and Histogram
ANS:
BAR DIAGRAM: The bar diagram is very commonly used and is better for the representation of qualitative
data and bars are vertical lines, the length of the bars are propositional to the bars their corresponding
numerical values. The width of the bar is unimportant but all bars should have the same width so as not to
confuse the readers of the diagram and these bars should be equally spaced and it is one dimensional For
example
100%
80%
North
60%
W est
40%
East
20%
0%
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
In the above bar chart there is X axis represent the different region like East, Weast and South and rain
fall in the four subsequent quarters and y axis represent the intensity of rain fall (cm) .
The bars can be sub divided into component depending up on the type of information to be shown in the
diagram.
HISTOGRAM: In the histogram the give information are plotted in the from of a series of rectangles. Class
intervals are marked along the X axis and the frequencies along the Y axis according to a suitable scale
and it is a two dimensional in which the length and width are both important. It was constructed from a
frequency distribution of grouped data and the height of the rectangle is proportional to the respective
frequency and the width represents the class intervals. Each rectangle is joined with the other and the
blank space between the rectangle would mean that the category is empty and there are no values in that
class interval.
100
80
60
East
40
West
20
North
0
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
Q-2) Calculate the Karl Pearson’s Coefficient of Skewness for the following
15 marks
ANS: Calculation of coefficient of Skewness.
Values
Mid Point
X
f
7.5
5
12.5
10
17.5
15
22.5
20
27.5
15
32.5
10
37.5
5
N= 80
5 – 10
10- 15
15- 20
20- 25
25- 30
30- 35
35- 40
X- 22.5
d
- 15
- 10
-5
0
5
10
15
Karl Pearson’s coefficient of skewness =
Mean
=
d/5
fd
fd2
cf
-3
-15
-2
-20
-1
-15
0
0
1
15
2
20
3
15
SEMESON fd=0
45
40
15
0
15
40
45
200
5
15
30
50
65
75
80
3(mean –Median)
--------------------Std. deviation
22.5
N
80
Median = ------ = ---- = 40th item
2
2
Std . Deviation = 8
So
coefficient of skewness =
3(mean –Median)
--------------------Std. deviation
3(22.5 – 22.5)
= -----------------8
= 0
Since the coefficient of skewness is 0, the distribution is symentrical.
Q-3) Do Any Two A OR B OR C
20 marks
b) Multiplication Rule
ANS: If the events A and B are the independent events than the probability that will both
occur is the product of their separate probabilities. Events A and B are independent ,if
and only if
P[AB]=P[A] X P[B]
For example if we toss a coin twice, the probability that the first toss results in a head and
the second toss results in a tail is
P[H,T] = P[H] X P[T]
=½X½
= ¼.
If events A and B are not a independent, meaning that the probability of occurrence of
one event is dependent or conditional upon the occurrence or non occurrence of other
event,than the probability that they will both occur is
P[AB]= P[A] X P[B]
It will be
P[AB] = P[A] X P[B/A].Where P[B/A] mean the probability of event B on the condition
of the outcome of event A.
Example : A basket has 6 black balls and 4 white balls, a ball is drawn at random from
the basket. Than the second ball is drawn without replacement of the first ball back in the
basket.the probability of the second ball being black or white would dependent on the
results of the first draw,where the first ball was black or white. The probability that both
balls will be black is
P[two black balls]
= P[black on the first draw] x P[black on the second draw/black on the first draw]
= 6/10 x 5/9
= 30/90
= 1/3
Q-4)
a) Null Hypothesis
ANS: A Null hypothesis is an assertion about the value of population parameter that is
being tested by the sample results is called Null hypothesis and it is written as a Ho.For
example, imagine flipping a coin three times for three heads and then forming the opinion
that we have used a two-headed trick coin. Clearly this opinion is based on the premise
that such a sequence is unlikely to have arisen using a normal coin. In fact, such
sequences (three consecutive heads or three consecutive tails) occur a quarter of the time
on average when using normal unbiased coins. Therefore, the opinion that this coin is
two-headed has little support. Formally, the hypothesis to be tested in this example is
"this is a two-headed coin." One tests it by assessing whether the data contradict the null
hypothesis that "this is a normal, unbiased coin." Since the observed data arise reasonably
often by chance under the null hypothesis, we cannot reject the null hypothesis as an
explanation for the data, and we conclude that we cannot assert our hypothesis on the
basis of the observed sequence.