(a) Both Assertion (A) and reason (R) are true and Reason (R) is a correct explanation of Assertion (A). (b) Both Assertion (A) and reason (R) are true and Reason (R) is not a correct explanation of...

(a) Both Assertion (A) and reason (R) are true and Reason (R) is a correct explanation of Assertion (A). (b) Both Assertion (A) and reason (R) are true and Reason (R) is not a correct explanation of...

(a) 13 (b) 14 (c) 15 (d) 16 Answer: (c) 15 Sol: Median of 6 numbers is the average of 3rd and 4th term. ∴ 13 = (????−1)+(????−3) 2 ⇒ 26 = 2x – 4 ⇒ 2x = 30 ⇒ x = 15 Thus, x is equal to...

(a) 0 (b) 1 (c) 10 (d) 19 Answer: (d) 19 Sol: It is given that mean of 20 numbers is zero. i.e., average of 20 numbers is zero. i.e., sum of 20 numbers is zero. Thus, at most, there can be 19...

(a) 7 (b) 9 (c) 11 (d) 13 Answer: (b) 9 Sol: First 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19. Median of 8 numbers is average of 4th and 5th terms.

Answer: (c) 13 Sol: Converting the given data into a frequency table, we get: Hence, the number of families having an income range of Rs. 20,000 – Rs. 25,000 is 13. The correct option is...

(a) 27.5 (b) 24.5 (c) 28.4 (d) 25.8 Answer: (b) 24.5 Sol: Mode = (3 × median) – (2 × mean) ⇒ (2 × mean) = (3 × median) – mode ⇒ (2 × mean) = 3 × 26 – 29 ⇒ (2 × mean) = 49 ⇒ Mean = 49 2 ∴ Mean = 24.5...

(a) 22     (b) 23.5      (c) 24    (d) 24.5 Answer: (c) 24 Sol: Mode = (3 × median) – (2 × mean) ⇒ (3 × median) = (mode + 2 mean) ⇒ (3 × median) = 16 + 56 ⇒ (3 × median) = 72 ⇒ Median = 72 3 ∴...

(a) 7.2 (b) 8.2 (c) 9.2 (d) 10.2 Answer: (c) 9.2 Sol: It is given that the mean and median are 8.9 and 9, respectively, ∴ Mode = (3 × Median) – (2 × Mean) ⇒ Mode = (3 × 9) – (2 × 8.9) = 27 – 17.8 =...

(a) 5.5            (b) 15.5 (c) 20.5          (d) 36.0 Answer: (c) 20.5 Sol: The x- coordinate represents the median of the given data. Thus, median of the given data is 20.5.

(a) mode=(3 * mean) – (2 * median) (b) mode=(3 *median) – (2 *mean) (c) median=(3 * mean) – (2 * mode) (d) mean=(3 * median) – (2 *mode) Answer: (b) mode=(3 * median) – (2 *mean) Sol: mode=(3 *...

(a) evenly distributed over the classes (b) centred at the class marks of the classes (c) centred at the lower limits of the classes (d) centred at the upper limits of the classes Answer: (b)...

(a) lower limits of the classes (b) upper limits of the classes (c) midpoints of the classes (d) none of these Answer: (c) midpoints of the classes Sol: The ???????? ′???? are the deviations from A...

(a) Mean          (b) Median (c) Mode          (d)None of these Answer: (b) Median Sol: The abscissa of the point of intersection of the ‘less than type’ and that of the ‘more than type’ cumulative...

(a) Mean (b) Median (c) Mode (d) all of these Answer: (b) Median Sol: The cumulative frequency table is useful in determining the median.

(a) a histogram                        (b) a frequency curve (c) a frequency polygon          (d) ogives Answer: (d) ogives Sol: This because median of a frequency distribution is found graphically...

(a) a frequency curve (b) a frequency polygon (c) a histogram (d) an ogive Answer: (c) a histogram Sol: The mode of a frequency distribution can be obtained graphically from a histogram.

(a) Mean         (b) Median (c) Mode         (d) None of these Answer: (a) Mean Sol: Mean is influenced by extreme values.

(a) Mean             (b) Median (c) Mode             (d) None of these Answer: (a) Mean Sol: The mean cannot be determined graphically because the values cannot be summed.

(a) Mean          (b) Mode (c) Median       (d) Standard Deviation Answer: (d) Standard Deviation Sol: The standard deviation is a measure of dispersion. It is the action or process of distributing...

Sol: Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged Thus, the median of 21 observations taken together is 30.

Sol: Upper class boundary = Lowest class boundary + width × number of classes = 8.1 + 2.5×12 = 8.1 + 30 = 38.1 Thus, upper class boundary of the highest class is 38.1.