Draw a histogram for the following distribution: Wt. in kg 40 - 44 45 - 49 50 - 54 55 - 59 60 - 64 65 - 69 Number of students 2 8 12 10 6 4 Hence, estimate the modal weight.

Wt in kg Number of students ( f ) 40 - 44 2 45 - 49 8 50 - 54 12 55 - 59 10 60 - 64 6 65 - 69 4 It is easy to see that the highest frequency is 12 corresponding to this class interval is 50 - 54 . The modal class is 50 - 54 Now constructing a histogram of given data: Since mode is dependent only on neighbourhood frequencies, drawing the rectangle corresponding to modal class and adjoining two classes. Joining the ends of opposite rectangles by straight lines - the point of intersection corresponds to mode. The Two lines meet at point A , which corresponds to value of 53 . Mode = 53 kg .

Find the mode of the following distribution by drawing a histogram Mid value 12 18 24 30 36 42 48 Frequency 20 12 8 24 16 8 12 Also state the modal class.

Mid value Frequency Class interval 12 20 9 - 15 18 12 15 - 21 24 8 21 - 27 30 24 27 - 33 36 16 33 - 39 42 8 39 - 45 48 12 45 - 51 It is easy to see that the highest frequency is 24 corresponding to this class interval is 27 - 33 . Hence, The modal class is 27 - 33 . We construct the relevant portion of histogram and estimate mode from it: Since mode is dependent only on neighbourhood frequencies, drawing the rectangle corresponding to modal class and adjoining two classes. Joining the ends of opposite rectangles by straight lines - the point of intersection corresponds to mode. The two lines meet at a point A , which corresponds to x - value of 31 . Hence, Mode = 31 .

Calculate the median and mode for the frequency distribution given below: Variable 10 - 13 13 - 16 16 - 19 19 - 22 22 - 25 25 - 28 28 - 31 31 - 34 34 - 37 37 - 40 Frequency 8 15 27 51 75 54 36 18 9 7

Variable Frequency f Cumulative Frequency 10 - 13 8 8 13 - 16 15 23 16 - 19 27 50 19 - 22 51 101 22 - 25 75 176 25 - 28 54 230 28 - 31 36 266 31 - 34 18 284 34 - 37 9 293 37 - 40 7 300 Here, N = Σ f = 300 So, N 2 = 300 2 = 150 We Notice That 150 th value lies in class 22 - 25 , Therefore, 22 - 25 is the median class. We know, Median = l + N 2 - C f × h l = the lower limit of median class = 22 , Frequency of the median class = f = 75 , Cumulative frequency of the class preceding the median class = C = 101 , Width of the median class = h = 3 , Sum of all frequencies = N , Median = 22 + 150 - 101 75 × 3 = 22 + 1 . 96 Median = 23 . 96 Here, the highest frequency is 75 , corresponding to this modal class is 22 - 25 . Lower limit of the modal class = l = 22 , Frequency of the modal class = f m = 75 , Frequency of the class preceding the modal class = f m - 1 = 51 , Frequency of the class succeeding the modal class = f m + 1 = 54 , h = 3 , Mode = l + f m - f m - 1 2 f m - f m - 1 - f m + 1 × h = 22 + 75 - 51 2 × 75 - 51 - 54 × 3 = 22 + 24 × 3 45 Mode = 23 . 6 Hence, median = 23 . 96 , Mode = 23 . 6 .

Estimate the median and mode of the following data : Wages ₹ Number of workers Less than 10 3 Less than 20 12 Less than 30 27 Less than 40 57 Less than 50 75 Less than 60 80

Wages Number of workers Cumulative frequency 0 - 10 3 3 10 - 20 9 12 20 - 30 15 27 30 - 40 30 57 40 - 50 18 75 50 - 60 5 80 Here, N = ∑ f = 80 So, N 2 = 80 2 = 40 We notice that 40 th value lies in class 30 - 40 . We know, Median = l + N 2 - C f × h Here, l = the lower limit of median class = 30 , Frequency of the median class = f = 30 , Cumulative frequency of the class preceding the median class = C = 27 , Width of the median class = h = 10 , Sum of all frequencies = N = 80 , Median = 30 + 40 - 27 30 × 10 = 30 + 4 . 33 Median = 34 . 33 Here, highest frequency is 30 , corresponding to this modal class is 30 - 40 . Lower limit of the modal class = l = 30 , Frequency of the modal class = f m = 30 , Frequency of the class preceding the modal class = f m - 1 = 15 , Frequency of the class succeeding the modal class = f m + 1 = 18 , h = 10 , Mode = l + f m - f m - 1 2 f m - f m - 1 - f m + 1 × h Mode = 30 + ( 30 - 15 ) × 10 ( 2 × 30 - 15 - 18 ) = 30 + 15 27 × 10 = 30 + 5 . 55 Mode = 35 . 55

E M B I B E

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>EXERCISE

M L Aggarwal Solutions for Chapter: Statistics, Exercise 4: EXERCISE

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Statistics, Exercise 4: EXERCISE

Attempt the practice questions on Chapter 1: Statistics, Exercise 4: EXERCISE with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 11 Volume 2 solutions are prepared by Experienced Embibe Experts.