Assertion (A): If the class marks of two overlapping intervals of equal size in a distribution are 94 and 104 then the corresponding intervals are 89-99, 99-109. Reason (R): The class mark of a class interval = actual lower limit + actual upper limit/2 1. A is true, R is false. 2. A is false, R is true. 3. Both A and R are true. 4. Both A and R are false.

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KnowledgeBoat · Accepted Answer

Both A and R are true. Explanation Class mark = Class mark for the first interval: Let the lower limit be L1 and the upper limit be U1. Using the formula, 94 = ⇒ L1 + U1 = 94 x 2 ⇒ L1 + U1 = 188 .................(1) Class mark for the second interval: Let the lower limit of the second interval be L2 and the upper limit be U2. Using the formula, 104 = ⇒ L2 + U2 = 104 x 2 ⇒ L2 + U2 = 208 .................(2) Since the intervals are of equal size and overlap, we can assume that the difference between the upper limit of the first interval and the lower limit of the second interval is the size of the interval. U2 - L2 = U1 - L1 = Interval Size From eq (1) and (2), the difference between the two class marks is: 104 - 94 = 10 Thus, the interval size is 10. For the first interval: L1 + U1 = 188, U1 - L1 = 10 Adding these two equations: (L1 + U1) + (U1 - L1) = 188 + 10 ⇒ L1 + U1 + U1 - L1 = 198 ⇒ 2U1 = 198 ⇒ U1 = ⇒ U1 = 99 Thus, L1 = 188 - 99 = 89 For the second interval: L2 + U2 = 208 U2 - L2 = 10 Adding these two equations: (L2 + U2) + (U2 - L2) = 208 + 10 ⇒ L2 + U2 + U2 - L2 = 218 ⇒ 2U2 = 218 ⇒ U2 = ⇒ U2 = 109 Thus, L2 = 208 - 109 = 99 The corresponding intervals are 89−99 and 99−109. ∴ Assertion (A) is true. The class mark of a class interval is the midpoint of the interval, given by: Class Mark = ∴ Reason (R) is true. Hence, both Assertion (A) and Reason (R) are true.

Class interval	Frequency	Cumulative Frequency
0 - 5	5	5
5 - 10	9	14
10 - 15	a	22
15 - 20	6	28
20 - 25	10	b

Mathematics

Statistics

Related Questions

Assertion (A):
Class interval Frequency Cumulative Frequency
0 - 5 5 5
5 - 10 9 14
10 - 15 a 22
15 - 20 6 28
20 - 25 10 b
⇒ a = 22 and b = 10
Reason (R):
14 + a = 22 ⇒ a = 8
28 + 10 = b ⇒ b = 38
A is true, R is false.
A is false, R is true.
Both A and R are true.
Both A and R are false.

Mathematics

Statistics

Related Questions

Assertion (A):Class intervalFrequencyCumulative Frequency0 - 5555 - 1091410 - 15a2215 - 2062820 - 2510b⇒ a = 22 and b = 10Reason (R):14 + a = 22 ⇒ a = 828 + 10 = b ⇒ b = 38A is true, R is false.A is false, R is true.Both A and R are true.Both A and R are false.

Assertion (A):
Class interval Frequency Cumulative Frequency
0 - 5 5 5
5 - 10 9 14
10 - 15 a 22
15 - 20 6 28
20 - 25 10 b
⇒ a = 22 and b = 10
Reason (R):
14 + a = 22 ⇒ a = 8
28 + 10 = b ⇒ b = 38
A is true, R is false.
A is false, R is true.
Both A and R are true.
Both A and R are false.