In a continuous frequency distribution, the median of the data is 21.If each observation is increased by 5 find the new median
3 Median = Mode + 2 Mean
please proof it.
The mean and median of same data are 24 and 26 respectively. The value of mode is ?
Find the value of f1 from the following data if it's mode is 65:
Class frequency
0-20 6
20-40 8
40-60 f1
60-80 12
80-100 6
100-120 5
Where frequency 6,8, f1, and 12 are in ascending order.
how do we find mean using step deviation method if the classes are unequal?
should we make the classes equal in such a case?
The following distribution gives the daily income of 50 workers of a factory.
Daily income (in Rs)
100 − 120
120 − 140
140 − 160
160 − 180
180 − 200
Number of workers
12
14
8
6
10
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
the mean of first n odd natural numbers is n2/81,then n=
if the mean of the following frequency distribution is 91, find the missing frequency x and y :
classes frequency
0 - 30 12
30-60 21
60 - 90 x
90 -120 52
120 - 150 y
150 - 180 11
total 150
The mean of n observations is X . IF the first item is increased by 1, second by 2 and so on, then the new mean is:
1. X+n
2. X+n/2
3. X+(n+1)/2
4. X+(n-1)/2
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight (in kg)
Number of students
Less than 38
0
Less than 40
3
Less than 42
5
Less than 44
9
Less than 46
Less than 48
28
Less than 50
32
Less than 52
35
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
Find the sum of deviations of the variate values 3, 4, 6, 7, 8, 14 fromtheir mean.
In a continuous frequency distribution, the median of the data is 21.If each observation is increased by 5 find the new median
3 Median = Mode + 2 Mean
please proof it.
The mean and median of same data are 24 and 26 respectively. The value of mode is ?
marks number of students
0 and above 80
10 and above 77
20 and above 72
30 and above 65
40 and above 55
50 and above 43
60 and above 28
70 and above 16
80 and above 10
90 and above 8
100 and above 0
Find the value of f1 from the following data if it's mode is 65:
Class frequency
0-20 6
20-40 8
40-60 f1
60-80 12
80-100 6
100-120 5
Where frequency 6,8, f1, and 12 are in ascending order.
how do we find mean using step deviation method if the classes are unequal?
should we make the classes equal in such a case?
The following distribution gives the daily income of 50 workers of a factory.
Daily income (in Rs)
100 − 120
120 − 140
140 − 160
160 − 180
180 − 200
Number of workers
12
14
8
6
10
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
the mean of first n odd natural numbers is n2/81,then n=
if the mean of the following frequency distribution is 91, find the missing frequency x and y :
classes frequency
0 - 30 12
30-60 21
60 - 90 x
90 -120 52
120 - 150 y
150 - 180 11
total 150
The mean of n observations is X . IF the first item is increased by 1, second by 2 and so on, then the new mean is:
1. X+n
2. X+n/2
3. X+(n+1)/2
4. X+(n-1)/2
class frequency
40-50 5
50-60 x
60-70 15
70-80 2
80-90 7
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight (in kg)
Number of students
Less than 38
0
Less than 40
3
Less than 42
5
Less than 44
9
Less than 46
14
Less than 48
28
Less than 50
32
Less than 52
35
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
Class interval Frequency
0-10 10
10-20 20
20-30 x
30-40 40
40-50 y
50-60 25
60-70 15
Find the sum of deviations of the variate values 3, 4, 6, 7, 8, 14 from
their mean.