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Mathematics

Assertion (A): A survey was conducted by a group of student as a part of their environment awareness program in which they collected the following data regarding the number of plants in 10 houses in a locality.

Number of plantsNumber of houses
2 - 42
4 - 63
6 - 81
8 - 103
10 - 121

The mean of the data is 6.9.

Reason (R): If the observation x1, x2, x3,………, xk has frequencies f1, f2, f3, …….., fk, then mean = f1x1+f2x2+f3x3+........+fkxkf1+f2+f3+........+fk\dfrac{f1x1 + f2x2 + f3x3 + …….. + fkxk}{f1 + f2 + f3 + …….. + fk}

  1. Assertion (A) is true, but Reason (R) is false.

  2. Assertion (A) is false, but Reason (R) is true.

  3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

  4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Measures of Central Tendency

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Answer

Number of plantsClass mark(xi)Number of houses(fi)fi.xi
2 - 4326
4 - 65315
6 - 8717
8 - 109327
10 - 1211111
TotalΣfi = 10Σfixi = 66

Mean = fixifi=f1x1+f2x2+f3x3+........+fkxkf1+f2+f3+........+fk\dfrac{∑fixi}{∑fi} = \dfrac{f1x1 + f2x2 + f3x3 + …….. + fkxk}{f1 + f2 + f3 + …….. + f_k}

∴ Reason (R) is true.

Mean = 6610\dfrac{66}{10}​ = 6.6

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

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