# Following data gives Sales (in Lakh Rs.) and Advertisement expenditure (in Thousand Rs.) of 20 firms.

(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).

Construct a bivariate frequency distribution table for the above data by taking classes 115 – 125, 125 – 135, … etc. for sales and 60 – 62, 62 – 64, … etc. for advertisement expenditure.

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.

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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