Answer
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Hint: Assume the Sales (in Lakh Rs.) as variable X and the Advertisement expenditure (in Thousand Rs.) as variable Y. Now, form the classes by checking the lowest and highest values of X and Y. In the horizontal column take the classes of values of X and in the vertical column take the classes of values of Y. Fill the boxes with the correct frequencies and take the sum f (X) = total frequency of X and f (Y) = total frequencies of Y to complete the table.
Complete step by step answer:
Here we have been provided with the data regarding the Sales (in Lakh Rs.) and Advertisement expenditure (in Thousand Rs.) of 20 firms and we are asked to form a bivariate frequency distribution table for this.
Now, in the bivariate frequency distribution table we consider two variables as per the given data. The date given is: -
(115, 61), (120, 60), (128, 61), (121, 63), (137, 62), (139, 62), (143, 63), (117, 65), (126, 64), (141, 65), (140, 65), (153, 64), (129, 67), (130, 66), (150, 67), (148, 66), (130, 69), (138, 68), (155, 69), (172, 68).
From the given data we assume that the Sales (in Lakh Rs.) is variable X and the Advertisement expenditure (in Thousand Rs.) is variable Y. So according to the question we have to consider the classes for X as 115 – 125, 125 – 135, … etc. and for Y as 60 – 62, 62 – 64, … etc. the highest value of X is 175 so we will go from X = 115 – 125 to X = 165 – 175. Similarly, we can see that the highest value of Y is 69 so we will go from Y = 60 – 62 to Y = 68 – 70.
Therefore, the table will be drawn as shown below.
Note: Note that in the univariate frequency distribution table we consider only one variable. Here the values f (X) = f (Y) = 20 is the proof that we haven’t missed any data and therefore the table is correct. You may interchange the columns in which we have considered the classes of X and Y as it will not have any effect on the frequencies.
Complete step by step answer:
Here we have been provided with the data regarding the Sales (in Lakh Rs.) and Advertisement expenditure (in Thousand Rs.) of 20 firms and we are asked to form a bivariate frequency distribution table for this.
Now, in the bivariate frequency distribution table we consider two variables as per the given data. The date given is: -
(115, 61), (120, 60), (128, 61), (121, 63), (137, 62), (139, 62), (143, 63), (117, 65), (126, 64), (141, 65), (140, 65), (153, 64), (129, 67), (130, 66), (150, 67), (148, 66), (130, 69), (138, 68), (155, 69), (172, 68).
From the given data we assume that the Sales (in Lakh Rs.) is variable X and the Advertisement expenditure (in Thousand Rs.) is variable Y. So according to the question we have to consider the classes for X as 115 – 125, 125 – 135, … etc. and for Y as 60 – 62, 62 – 64, … etc. the highest value of X is 175 so we will go from X = 115 – 125 to X = 165 – 175. Similarly, we can see that the highest value of Y is 69 so we will go from Y = 60 – 62 to Y = 68 – 70.
Therefore, the table will be drawn as shown below.
Y\X | 115 – 125 | 125 – 135 | 135 – 145 | 145 – 155 | 155 – 165 | 165 – 175 | Total f (Y) |
60 – 62 | || (2) | | (1) | - | - | - | - | 3 |
62 – 64 | | (1) | - | ||| (3) | - | - | - | 4 |
64 – 66 | | (1) | | (1) | || (2) | | (1) | - | - | 5 |
66 – 68 | - | || (2) | - | || (2) | - | - | 4 |
68 – 70 | - | | (1) | | (1) | - | | (1) | | (1) | 4 |
Total f (X) | 4 | 5 | 6 | 3 | 1 | 1 | 20 |
Note: Note that in the univariate frequency distribution table we consider only one variable. Here the values f (X) = f (Y) = 20 is the proof that we haven’t missed any data and therefore the table is correct. You may interchange the columns in which we have considered the classes of X and Y as it will not have any effect on the frequencies.
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