Find the mean of the following frequency distribution by the assumed mean method: Number of apples 50 - 53 53 - 56 56 - 59 59 - 62 62 - 65 Number of boxes 20 150 115 95 20

Number of apples x i f i d i = x i - A f i d i 50 - 53 51 . 5 20 - 6 - 120 53 - 56 54 . 5 150 - 3 - 450 56 - 59 57 . 5 = A 115 0 0 59 - 62 60 . 5 95 3 285 62 - 65 63 . 5 20 6 120 ∑ f i = 400 ∑ f i d i = - 165 Mean x ¯ = A + ∑ f i d i ∑ f i = 57 . 5 - 165 400 = 57 . 5 - 0 . 4 = 57 . 1 Hence, the answer is 57 . 1 .

Find the median of the following data: Value of variable 12 13 14 15 16 17 18 19 20 Frequency 7 10 15 18 20 10 9 8 3

Value of variable Frequency c.f. 12 7 7 13 10 17 14 15 32 15 18 50 16 20 70 17 10 80 18 9 89 19 8 97 20 3 100 Here, N = 100 Median = 1 2 N 2 th term + N 2 + 1 th term = 1 2 100 2 th term + 100 2 + 1 th term = 1 2 50 th term + 51 th term = 1 2 × 15 + 16 = 15 . 5 Hence, the answer is 15 . 5 .

State the type for curve A and curve B

We know that, the frequencies of all preceding classes are added to the frequency of a class. This series is called the less than cumulative series. It is constructed by adding the first-class frequency to the second-class frequency and then to the third class frequency and so on. The downward cumulation results in the less than cumulative series. In the above graph, the rising curve represents the less than Ogive. The frequencies of the succeeding classes are added to the frequency of a class. This series is called the more than or greater than cumulative series. It is constructed by subtracting the first class, second class frequency from the total, third class frequency from that and so on. The upward cumulation result is greater than or more than the cumulative series. In the above graph, the falling curve represents the greater than Ogive. Hence, Curve A represents greater than Ogive and Curve B represents less than Ogive

Find the mode of the following data Class 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 Frequency 12 17 28 23 7 8 5

We have to find the mode of the following data Class 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 Frequency 12 17 28 23 7 8 5 Modal class is 50 - 60 as it has the highest frequency. l 1 = 50 f o = 17 f 1 = 28 f 2 = 23 c = 10 Mode = l 1 + f 1 − f o 2 f 1 − f o − f 2 × c = 50 + 11 56 - 17 - 23 × 10 = 50 + 11 16 × 10 = 50 + 6 . 875 = 56 . 875

Find the median of the following data: Class 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24 24 - 28 Frequency 9 16 12 7 15 1

We have to find the median of the following data: Class 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24 24 - 28 Frequency 9 16 12 7 15 1 Cumulative frequency(c.f.) 9 9 + 16 = 25 25 + 12 = 37 37 + 7 = 44 44 + 15 = 59 59 + 1 = 60 Here n = 60 ⇒ 2 n = 30 Cumulative frequency greater than 30 is 37 and the class corresponding to it is 12 − 16 &there4; Class 12 − 16 is the median class &there4; Lower limit ( l ) = 12 , 2 n = 30 , c . f . = 25 , f = 12 and h = 4 Substituting the values, we get M = l + ( 2 n − c . f . ) f × h M = 12 + 30 − 25 12 × 4 = 12 + 5 3 = 12 + 1 . 67 = 13 . 67 Hence, Median is 13 . 67 .

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. The median of the given frequency distribution is 137 units. Find the number of consumers whose monthly consumption is between 105 - 125 and also for 145 - 165 . Monthly consumption in units 65 - 85 85 - 105 105 - 125 125 - 145 145 - 165 165 - 185 185 - 205 Number of consumers 04 05 - 20 - 08 04

Let the missing frequencies be x and y Then, Class Intervals x i f i Cumulative frequency 65 - 85 75 4 4 85 - 105 95 5 9 105 - 125 115 x 9 + x 125 - 145 135 20 29 + x 145 - 165 155 y 29 + x + y 165 - 185 175 8 37 + x + y 185 - 205 195 4 41 + x + y N = 41 + x + y Given that, N = 68 and median = 137 Then, 68 = 41 + x + y x + y = 27 ....................(i) Now, Median = l + N 2 - c . f f × h , where l = Lower limit of median class N = Number of observations c . f = Cumulative frequency of class preceding the median class f = Frequency of the median class h = Class size Then, N 2 = 68 2 = 34 Now, 137 = 125 + 34 - 9 + x 20 × 20 12 = 34 - 9 - x 12 = 25 - x x = 13 Now putting this value of x in equation (i) we get, 13 + y = 27 y = 14 Therefore, the number of consumers whose monthly consumption is between 105 - 125 is 13 and for 145 - 165 is 14 .