Mathematics
Statement 1: Let m be the mid-value and x be the upper limit of a class in the continuous frequency distribution, then the lower limit of this class is 2m - x. Statement 2: For a given class: lower limit + upper limit/2 = mid-value of the class 1. Both the statements are true. 2. Both the statements are false. 3. Statement 1 is true, and statement 2 is false. 4. Statement 1 is false, and statement 2 is true.
Related Questions
In a frequency distribution, mid-values of the class is 10 and width of the class is 6, the lower limit of the class is :
6
7
8
9
Assertion (A): 30 children were asked about the number of hours they watched TV program everyday. The results are recorded as under.
Number of hours Frequency 0 - 5 8 5 - 10 16 10 - 15 4 15 - 20 2 Then the number of children who watched TV for 10 or more hours a day is 22.
Reason (R): The assertion is not correct as the required number is 4 + 2 = 6.
A is true, but R is false.
A is false, but R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.
Construct a frequency table from the following data :
Marks No. of students less than 10 6 less than 20 15 less than 30 30 less than 40 39 less than 50 53 less than 60 70