Find the mode of the given data: Class interval 0 - 20 20 - 40 40 - 60 60 - 80 Frequency 15 6 18 10

Class interval 0 - 20 20 - 40 40 - 60 60 - 80 Frequency 15 6 18 10 As the class 40 - 60 has the maximum frequency, so it is the modal class. x k = 40 , h = 20 , f k = 18 , f k - 1 = 6 and f k + 1 = 10 Now, Mode = x k + h f k - f k - 1 2 f k - f k - 1 - f k + 1 = 40 + 20 18 - 6 2 ( 18 ) - 6 - 10 = 40 + 20 × 12 20 = 40 + 12 = 52 .

In an interview with maximum marks 250 , the marks of top 200 students were distributed as below Marks 100 - 114 115 - 129 130 - 144 145 - 159 160 - 174 175 - 189 190 - 204 205 - 219 220 - 235 f 4 5 5 18 18 27 44 40 39 Calculate Median to the nearest integer

The distribution is not continuous, to make it continuous, we use adjustment factor = 115 - 114 2 = 0 . 5 The cumulative frequency distribution for given data is as under Marks 99 . 5 - 114 . 5 114 . 5 - 129 . 5 129 . 5 - 144 . 5 144 . 5 - 159 . 5 159 . 5 - 174 . 5 174 . 5 - 189 . 5 189 . 5 - 204 . 5 204 . 5 - 219 . 5 219 . 5 - 235 . 5 f 4 5 5 18 18 27 44 40 39 cumulative frequency 4 9 14 32 50 77 121 161 200 For Median = n 2 t h value i.e., = 200 2 t h value = 100 t h value. which lies in the class 189 . 5 - 204 . 5 l = 189 . 5 ; c = 77 ; f = 44 ; i = 15 ∴ M d = l + n 2 - C f × i = 189 . 5 + 100 - 77 44 × 15 = 189 . 5 + 23 × 15 44 = 197 . 34 Thus, M d = 197 (nearest to integer)

Find the median marks in the data. Marks 15 20 10 25 5 Number of students 10 8 6 4 1

First arrange the observations given in ascending. Marks Number of students Cumulative frequency 5 1 1 10 6 7 15 10 17 20 8 25 25 4 29 Total = 29 Here, N = 29 . 29 is odd. Median = N + 1 2 th observation = 29 + 1 2 th = 15 th observation From the table corresponding values of 15 th observations is the 15 . Hence, the median of the data is 15 .

Find the mean, median and mode of the following data: Classes 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 120 - 140 Frequency 6 8 10 12 6 5 3

Here, we have Class Frequency ( f i ) x i f i x i Cumulative frequency 0 - 20 6 10 60 6 20 - 40 8 30 240 14 40 - 60 10 50 500 24 60 - 80 12 70 840 36 80 - 100 6 90 540 42 100 - 120 5 110 550 47 120 - 140 3 130 390 50 N = ∑ f = 50 ∑ f i x i = 3120 Here, total frequency is N = 50 . So, median class is N 2 = 50 2 = 25 . The cumulative frequency just greater than 25 is 36 , then the median class is 60 - 80 , such that the lower limit ( l ) = 60 . Cumulative frequency of the class preceding median class is ( C ) = 24 , frequency of the median class = 12 , class size ( h ) = 20 . So, the median is l + N 2 - C f × h = 60 + 25 - 24 12 × 20 = 60 + 20 12 = 61 . 67 Thus, the median is 61 . 67 . Mean of the data = ∑ f i x i ∑ f Therefore, the mean = 3120 50 = 62 . 4 Now, mode of the = l + f k - f k - 1 2 f k - f k - 1 - f k + 1 × h l = lower limit of modal class h = width of the class interval f k = frequency of the modal class f k - 1 = frequency of the class preceding the modal class f k + 1 = frequency of the class succeeding the modal class Here, l = 60 , f k = 12 , f k - 1 = 10 , f k + 1 = 6 and h = 20 . So, the mode = 60 + 12 - 10 24 - 10 - 6 × 20 = 60 + 2 8 × 20 = 60 + 5 = 65 Hence, the required mode is 65 , median is 61 . 67 and mean is 62 . 4 .