# IGNOU Solved Question Papers: ECO - 07 (June' 2013)

BACHELOR'S DEGREE PROGRAMME
Term-End Examination. June, 2013
ELECTIVE COURSE: COMMERCE
ECO-7: ELEMENTS OF STATISTICS
Time: 2 hours Maximum Marks: 50
Weightage: 70%
Note: There (there are three sections and all are compulsory.
SECTION - A
1. Fill in the blanks with appropriate word (s) given in brackets: 5x1=5
(a) The sum of absolute deviations (deviations ignoring signs) measured from ____ is the least. (Mode/Median)
(b) The data collected from a report published in the Economic Times is an example of Secondary data. (primary/secondary)
(c) The goals scored by a football team is an example of Discrete variable. (discrete/ continuous)

(d) Squares are an example Two of dimensional diagram. (one/two)
(e) In a symmetrical distribution; the skewness is always equal to Zero. (zero/one)
2. State whether the statements given below are True or False. 5x1=5
(a) The study of only a part of population is called sample.  True
(h) The smallest and the largest possible measurements in each class is called the class limits.   True
(c) Negative values cannot be represented in a histogram. True
(d) The data should be arranged either in ascending or descending order before computing the median.  True
(e) The variance and co - efficient of variation mean the same thing.   False
SECTION - B
Attempt any two of the following:
3. (a) Draw a "less than" cumulative frequency distribution from the following data: 10+5
 Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of students 10 18 26 42 40 16 6
Solution:
Less than Cumulative Frequency Distribution
 Marks Frequency Marks less than Cumulative frequency 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 10 18 26 42 40 16 6 10 20 30 40 50 60 70 10 28 54 96 136 152 158

(b) Explain the process of constructing a histogram.
Ans: Histogram: A histogram in another kind of graph that uses bars in its display. This type of graph is used with quantitative data. Ranges of values, called classes, are listed at the bottom, and the classes with greater frequencies have taller bars. The technique of constructing histogram is now illustrated:
1. First for distribution having equal class-intervals, and
2. For distribution having unequal class-intervals.
1. First for distribution having equal class-intervals: When class-intervals are equal take frequency on the y-axis, the variable on the x-axis and construct adjacent rectangles. In such a case the heights of the rectangles will be proportional to the frequencies.
2. First for distribution having unequal class-intervals: When class intervals are unequal, the frequencies must be adjacent before constructing the histogram. For making the adjustment we take that class which has the lowest class-interval and adjust the frequencies of other classes in the following manner. If one class-interval is twice as wide as the one having lowest class-interval, we divide the height of its rectangle by two; if it is three times more, we divide the height of its rectangle by three etc. i.e., the heights will be proportional to the ratios of the frequencies to the widths of the classes.
4. (a) Find the value of Mode for the following frequency distribution: 10+5
 Central Sizes 1 2 3 4 5 6 7 8 9 10 Frequency 8 6 10 12 20 12 5 3 2 4
Solution:
Frequency Table
 X central size F F1 F2 F3 F4 F5 1 2 3 4 5 6 7 8 9 10 8 6 10 12 20 12 5 3 2 4 14 22 32 8 6 16 32 17 5 24 44 10 28 37 9 42 20
Here, F1 = Add two frequencies.
F2 = left 1st frequency and add two frequencies.
F4 = lest 1st frequency and add  3 frequencies.
F5 = lest 1st and 2nd frequency and add  3 frequencies.
Analysis of Frequency Table
 X F F1 F2 F3 F4 F5 Total 1 2 3 4 5 6 7 8 9 10 - - - - 1 - - - - - - - - - 1 1 - - - - - - - 1 1 - - - - - - - - 1 1 1 - - - - - - - - 1 1 1 - - - - - 1 1 1 - - - - - - - 1 3 6 3 1 - - -

Since, 5 has highest no of observations therefore required Mode is 5.

(b) Using the empirical relationship between mean, median and mode, estimate the median value of a distribution whose mode and mean are 32.1 and 35.4 respectively.
Solution: Given,
Mode = 32.1
Mean = 35.4
Median = ?
Now,
3 Median = 2 Mean + Mode
3 Median = 2 x 35.40 + 32.10
3 Median = 70.80 + 32.10
3 Median = 102.90
Median 5. (a) Calculate the standard deviation and co-efficient of variation for the following data:   10+5
 Class 0-4 4-8 8-12 12-16 F 4 8 2 1

Solution:  Calculation of Standard Deviation and variance
 Class f X d = ( x – A ) d1= d/4 d12 Fd1 Fd2 0 – 4 4 – 8 8 – 12 12 – 16 4 8 2 1 2 6 10 14 -8 -4 0 +14 -2 -1 0 +1 4 1 0 1 -8 -8 0 1 16 8 0 1 N = 15 -15 25

Assumed Mean = 10
Now, Mean   Co-efficient of variation (b) For a given data, Mode =47.07, Mean = 47.83 and standard deviation =14.8. Determine Karl Pearson's co-efficient of Skewness.

Ans: Coefficient of Skewness 6. (a) Distinguish between simple and complex statistical tables and give their examples. 10+5

(b) Write a note on Pie diagram.
Ans: This type of diagram is used to show the breakup of a total into component parts. A very common use of pie chart is to represent the division of a sum of money into its components. For example, the entire circle or pie may represent the budget of a family for a month and the sections may represent portions of the budget allotted to rent, food, clothing and so on. Similarly, through a pie diagram it can be shown as to how a rupee spent by a firm is distributed over various heads such as wages, raw materials, administrative expenses etc. The pie chart is so called because the entire graph looks like a pie and the components resemble slices cut from pie.
SECTION - C
7. Distinguish between any two of the following: 5+5
(a) Univariate and Bivariate Frequency Distribution
Ans: Univariate frequency distribution: The frequency distribution of a single variable is called a Univariate Distribution. For example height of students of a class. In such a data, we cannot compare a variable with other one. Bivariate frequency distribution A Bivariate Frequency Distribution is the frequency distribution of two variables. For example height and weight of students of a class. We can compare between two variables.
(b) Exclusive and Inclusive Class Intervals
Ans: Exclusive method: Under this method, the upper class limit is excluded but the lower class limit of a class is included in the interval. Thus an observation that is exactly equal to the upper class limit, according to the method, would not be included in that class but would be included in the next class. On the other hand, if it were equal to the lower class limit then it would be included in that class. For Example, In the class 800-900 the upper class limit 800 is included and the lower class limit 900 is excluded.
Inclusive method: The Inclusive Method does not exclude the upper class limit in a class interval. It includes the upper class in a class. Thus both class limits are parts of the class interval. For Example, In the class 800–899, both upper class limit and the lower class limit 800 and 899 are included.
(c) Less than and More than Ogives
Ans: Less than Method: In the less than method we start with upper limits of class and go on adding the frequencies. When these frequencies are plotted we get a rising curve.
More than Method: Here, we start with lower limit and go on subtracting the frequencies of each class. When these frequencies are plotted a decreasing curve will be obtained.
The basic difference between the less than and more than type is that in less than ogives, frequencies are added starting from the upper limit of the 1st class interval of the frequency distribution. On the other hand, in case of more than ogives, frequencies are added starting from the lower limit of 1st class interval of the frequency distribution. Accordingly, while in case of less than Ogive the cumulative total tends to increase, in case of more than Ogive, the cumulative total tends to decrease.
(d) Reasonable Accuracy and Spurious Accuracy