Consider the following frequency distribution: array|c|c|c|c|c|c|c|c| Value & 4 & 5 & 8 & 9 & 6 & 12 & 11 \ Frequency & 5 & f1 & f2 & 2 & 1 & 1 & 3 \ array Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6. For the given frequency distribution, let denote the mean deviation about the mean, denote the mean deviation about the median, and 2 denote the variance. Match each entry in List-I to the correct entry in List-II and choose the correct option.
- (P) → (5), (Q) → (3), (R) → (2), (S) → (4)
- (P) → (5), (Q) → (2), (R) → (3), (S) → (1)
- (P) → (5), (Q) → (3), (R) → (2), (S) → (1)
- (P) → (3), (Q) → (2), (R) → (5), (S) → (4)
Answer
(3) (P) → (5), (Q) → (3), (R) → (2), (S) → (1)
Solution
12 + f1 + f2 = 19 f1 + f2 = 7.
Ascending order: 4(5), 5(f1), 6(1), 8(f2), ; median = 10th value = 6 needs 5 + f1 < 10 6 + f1 f1 = 4, f2 = 3. (P) 7f1 + 9f2 = 28 + 27 = 55.
Mean = 20 + 20 + 6 + 24 + 18 + 33 + 1219 = 13319 = 7.
(Q) 19 = f|x - 7| = 15 + 8 + 1 + 3 + 4 + 12 + 5 = 48.
(R) 19 = f|x - 6| = 10 + 4 + 0 + 6 + 6 + 15 + 6 = 47.
(S) 19 2 = f(x - 7)2 = 45 + 16 + 1 + 3 + 8 + 48 + 25 = 146.
P → 5, Q → 3, R → 2, S → 1.
Answer: (C)