Class interval 1 ‐ 10 11 ‐ 20 21 ‐ 30 31 ‐ 40 41 ‐ 50 Frequency 4 7 12 5 2 What is total frequency and less than cumulative frequency? What do you infer?

Given: Class interval 1 ‐ 10 11 ‐ 20 21 ‐ 30 31 ‐ 40 41 ‐ 50 Frequency 4 7 12 5 2 The progressive total of frequencies from first class to the upper boundary of a particular class is called less than cumulative frequency. Class intervals Frequency Class boundaries Upper boundaries Less than cumulative frequency 1 ‐ 10 4 0 . 5 ‐ 10 . 5 10 . 5 4 11 ‐ 20 7 10 . 5 ‐ 20 . 5 20 . 5 7 + 4 = 11 21 ‐ 30 12 20 . 5 ‐ 30 . 5 30 . 5 11 + 12 = 23 21 ‐ 30 5 30 . 5 ‐ 40 . 5 40 . 5 23 + 5 = 28 31 ‐ 40 2 40 . 5 ‐ 50 . 5 50 . 5 28 + 2 = 30 50 Here, the total frequency is 50 and the less than cumulative frequency is also 50 . Therefore, we can say that the total and less than cumulative frequency is same.

Centuries scored and number of cricketers in the world are given below. Number of centuries 5 10 15 20 25 Number of cricketers 56 23 39 13 8 Find the mean, median and mode of the given data.

Number of centuries x Number of cricketers f x f Cumulative frequency c f 5 56 280 56 10 23 230 79 15 39 585 118 20 13 260 131 25 8 200 139 ∑ f = 139 ∑ x f = 1555 Mean = ∑ x f ∑ f = 1555 139 = 11 . 18 The mode is the value that appears most frequently in a data set. Therefore, mode = 5 Here, N = 139 , which is odd. Now, N + 1 2 observation = 139 + 1 2 th observation = 140 2 th observation = 70 th observation Now, we need to identify the median class, whose cumulative frequency just exceeds 70 th observation. Median class = 10 Therefore, median = 10 Hence, mean = 11 . 18 , mode = 5 and median = 10

The heights of boys and girls of IX class of a school are given below. Height cm 135 140 147 152 155 160 Boys 2 5 12 10 7 1 Girls 1 2 10 5 6 5 Compare the heights of the boys and girls

Height x 1 Boys f 1 Girls f 2 Cumulative frequency c f 1 Cumulative frequency c f 2 135 2 1 2 1 140 5 2 7 3 147 12 10 19 13 152 10 5 29 18 155 7 6 36 24 160 1 5 37 29 N 1 = 37 N 2 = 29 Here, N 1 = 37 → Odd Therefore, N 1 + 1 2 th observation = 37 + 1 2 th observation = 38 2 th observation = 19 th observation Now, we need to identify the median class, whose cumulative frequency just exceeds 19 th observation or equal to it. Median class = 147 ∴ Median = 147 Again, N 2 = 29 → Odd Therefore, N 2 + 1 2 th observation = 29 + 1 2 th observation = 30 2 th observation = 15 th observation Now, we need to identify the median class, whose cumulative frequencies just exceeds 15 th observation. Median class = 152 ∴ Median = 152 Hence, we can say from the following observations that the girls are taller than the boys.

Number of villages with respect to their population as per India census 2011 are given below. Population(in thousands) 12 5 30 20 15 8 Villages 20 15 32 35 36 7 Find the average population in each village.

Population x i Villages f i x i f i 12 20 240 5 15 75 30 32 960 20 35 700 15 36 540 8 7 56 ∑ f i = 145 ∑ f i x i = 2571 Mean = ∑ f i x i ∑ f i = 2571 145 = 17 . 73 Therefore, 17 . 73 is the average population in each village.

Number of families in a village in correspondence with the number of children are given below: Number of children 0 1 2 3 4 5 Number of families 11 25 32 10 5 1 Find the mean number of children per family.

Number of children x i Number of families f i x i f i 0 11 0 1 25 25 2 32 64 3 10 30 4 5 20 5 1 5 ∑ f i = 84 ∑ f i x i = 144 Mean = ∑ f i x i ∑ f i = 144 84 = 1 . 71 Therefore, the required mean is 1 . 71 .

Weights of parcels in a transport office are given below. Weight kg 50 65 75 90 110 120 Number of parcels 25 34 38 40 47 16 Find the mean weight of the parcels.

Weight x i Number of parcels f i x i f i 50 25 1250 65 34 2210 75 38 2850 90 40 3600 110 47 5170 120 16 1920 ∑ f i = 200 ∑ f i x i = 17000 Mean = ∑ f i x i ∑ f i = 17000 200 = 85 Therefore, the mean is 85 .

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 .

The median of the following data is 525 . Find the values of x and y , if total frequency is 100 . Class 0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 500 - 600 600 - 700 700 - 800 800 - 900 900 - 1000 Frequency 2 5 x 12 17 20 y 9 7 4

Cumulative frequency for the given data; Class interval Frequency f Cumulative frequency CF 0 - 100 2 2 100 - 200 5 7 200 - 300 x x + 7 300 - 400 12 x + 19 400 - 500 17 x + 36 500 - 600 20 x + 56 600 - 700 y x + y + 56 700 - 800 9 x + y + 65 800 - 900 7 x + y + 72 900 - 1000 4 x + y + 76 total Σ f = 76 + x + y = 100 Given that the median is 525 . So median class is 500 - 600 Now median = l + N 2 - C F f × h where l = 500 , h = 100 , N = 100 , C F = 36 + x , f = 20 so, 525 = 500 + 100 2 - ( 36 + x ) 20 × 100 ⇒ 525 - 500 = ( 50 - ( 36 + x ) ) × 5 ⇒ 25 = ( 14 - x ) × 5 ⇒ x = 9 Also, Σ f = 76 + x + y = 100 ⇒ 9 + y = 100 - 76 ⇒ y = 15 Hence, x = 9 and y = 15

If the median of the following frequency distribution is 32 . 5 . Find the values of f 1 and f 2 . Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency f 1 5 9 12 f 2 3 2 40

Find the Cumulative Frequency of the given data. class Frequency f Cumulative Frequency c f 0 - 10 f 1 f 1 10 - 20 5 f 1 + 5 20 - 30 9 f 1 + 14 30 - 40 12 f 1 + 26 40 - 50 f 2 f 1 + f 2 + 26 50 - 60 3 f 1 + f 2 + 29 60 - 70 2 f 1 + f 2 + 31 N = ∑ f = 40 Sum of Cumulative Frequency is f 1 + f 2 + 31 = 40 ⇒ f 1 + f 2 = 9 . . . . . . . ( 1 ) Now, The median 32 . 5 is lies between 30 - 40 . Hence, median class is 30 - 40 . ∴ l = 30 , N 2 = 40 2 = 20 , h = 10 , f = 12 , cumulative frequency of preceding class ( cf ) = 14 + f 1 . Now, median is given as, M = l + ( N 2 - cf ) f × h 32 . 5 = 30 + ( 20 - ( 14 + f 1 ) 12 × 10 2 . 5 = ( 60 - 10 f 1 ) 12 30 = 60 - 10 f 1 f 1 = 30 10 f 1 = 3 From equation ( 1 ) , we get f 2 = 9 - f 1 f 2 = 9 - 3 f 2 = 6 Hence, the value of f 1 is 3 and f 2 is 6 .