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1.Find the missing frequency f, if the mode of the given data is 154.
Classes
120-130
130-140
140-150
150-160
160-170
170-180
Frequency
2
8
12
f
9
7
plzz give an answer fast....
IF THE N/2 IS EQUAL TO THE CUMULATIVE FREQUENCY THEN WE WOULD REGARD THAT CLASS AS THE MEDIAN CLASS OR NOT.
what is hit and trial method .
If the mean of the number 7,3,8,4,x,7,9,7 and 12 is 7, then the difference between the median and mode of the numbers 12,10,8,10,x,7,6,8 and 6 is:
(a)0. (b)1. (c)2. (d)3
If the median of the distribution is given below is 28.5, find the values of x and y.
Class interval
0 − 10
5
10 − 20
x
20 − 30
20
30 − 40
15
40 − 50
y
50 − 60
Total
60
The median from the table is
value 7, 8, 9, 10, 11, 12 ,13
frequency 2, 1, 4, 5, 6, 1, 3
if the median of the following frequency distribution is 24. find the missing frequency x :
Age Number of persons
0 - 10 5
10 - 20 25
20 - 30 x
30 - 40 18
40 - 50 7
find the missing frequencies if N=100 and median is 32.
marks obtained-0-10, 10-20, 20-30, 30-40, 40-50, 50-60, total
no. of students-10 ,? ,25 ,30 ,? ,10 ,100.
10. If median = 15 and mean = 16, find mode of the distribution
The median of the following distribution is 30. Find the missing frequencies f1 and f2:
Class frequency
0-10 10
10-20 10
20-30 f1
30-40 30
40-50 f2
50-60 10
Total 100
the mean of 10 numbers is 20.if 5 is subtracted from every number.what will be the new mean?
how do we find mean using step deviation method if the classes are unequal?
should we make the classes equal in such a case?
how to find the mean if the class intervals are unequal??
like fr e- 0-10,10-30,30-35....
how to calculate mode if two classes have same and highest frequency (bimodal) ?
The mean of 25 observations is 9 . if each observation is increased by 4 then the new mean is _______
the interest paid on each of the three different sums of money yeilding 1%, 2% and 4% simple interest per annum respectively, is the same. find the average yield percent on the total sum inverted.
how to find xi?
The following table shows the distribution of some families according to expenditure per week .the number of families for two class intervalsare missing. Median =Rs.25 , Mode = Rs.24. calculate the missing frequency and hence the find mean.
What is the formula to use median when more than ogive is given?
One word answer-VERY URGENT***
1) Difference between any 2 consecutive class marks (9 letters)
2)Graphical Representation (in rectangles) between class marks and respective frequencies (16 letters)
3)Graphical Representation (in rectangles) between class intervals and respective frequencies ( 9 letters)
4)Data in tabular form (7 letters)
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days
0 − 6
6 − 10
10 − 14
14 − 20
20 − 28
28 − 38
38 − 40
Number of students
11
10
4
3
1
3 Median = Mode + 2 Mean
please proof it.
In a continuous frequency distribution, the median of the data is 21.If each observation is increased by 5 find the new median
In the step deviation method,
What is the meaning of ui ?
The mean and median of same data are 24 and 26 respectively. The value of mode is ?
if the median salary of 100 employees of a factory is 24800, then find the missing frequencies f1
and f2 in the given data
Salary 10000-15000 15000-20000 20000-25000 25000-30000 30000-35000 35000-40000
frequen. 18 f1 f2 15 12 22
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
14
6
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
what is the formula for finding mean using assumed mean method?
Find the value of f1 from the following data if it's mode is 65:
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.
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)
Less than 38
0
Less than 40
Less than 42
Less than 44
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.
what is the difference between assumed mean method and step deviation method?
in which kind of sums do we use the 2 types of finding the mean of different data?
please explain as i am confused bout which method to use in which sum.
the median of the data is 525. find 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
frequeency 2 5 x 12 17 20 y 9 7 4
9. If the mean of the distribution is 57.6 and the sum of its observations is 50, find the missing
The mean of the following frequency dristibution is 50. Find the value of p
In a school 85 boys and 35 girls appeared in a public examination . The mean marks of the boys were found to be 40 % whereas the mean marks of the girls were 60% . Determine the average marks % of the school.
how can we find the median by only less than ogive curve?
The median of the following data is 32.5.find the value of x and y
class interval f
0-10 x
10-20 5
20-30 9
30-40 12
40-50 y
50-603
60-70 2
total 40
The following table shows the marks obtained by 100 students of class X in a school during a particular academic session. Find the mode of this distribution.
Marks No. of students
Less than 10 7
Less than 20 21
Less than 30 34
Less than 40 46
Less than 50 66
Less than 60 77
Less than 70 92
Less than 80 100
if mode=80 and mean=110,then the median=
Find the mean, mode and median for the following data.
The median of the following data is 20.75. Find the missing frequencies x and y, if the total frequency is 100.
Class interval:-0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Frequencies: -7 10 x 13 y 10 14 9
how to find the median class?
How to find missing frequency if the mode of the given data is given.
How to convert a LESS THAN OGIVE and it's CF into the Frequency distribution?
the value of the variable for ehich the frequency is maximum is known as ..????
Find the class marks of classes 10 - 25 and 35 - 55.
If the mean of the following distributions is 27, Find the value of p
If the less than ogive and more than ogive intersect each other at ( 20.5,15.5 ), then the median of the given data is :
A ) 36.0 B) 20.5 C ) 15.5 D ) 5.5
13. Find the mean, mode and median for the following data:
The mean of the following data is 53, find the missing frequencies.
age in years -- 0-20, 20-40 40-60 60-80, 80-100
number of people--- 15, f1, 21, f2, 17, 100
How is assumed mean calculated of even number of even number of class intervals ?
E.g.C.I. Fi
20-405140-606860-802480-10013100-12005120-14077140-16036160-18012
In this case what will be the assumed mean ? Please give reason also as to explain why u chose it ??
the class mark of classes in a distribution arer 6.10.14.18.22.26.30. find
a) class size
b) lower limit of second class
c) upper lmit of last class
d) third class
E.g: 9876543210, 01112345678
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Syllabus
1.Find the missing frequency f, if the mode of the given data is 154.
Classes
120-130
130-140
140-150
150-160
160-170
170-180
Frequency
2
8
12
f
9
7
plzz give an answer fast....
IF THE N/2 IS EQUAL TO THE CUMULATIVE FREQUENCY THEN WE WOULD REGARD THAT CLASS AS THE MEDIAN CLASS OR NOT.
what is hit and trial method .
If the mean of the number 7,3,8,4,x,7,9,7 and 12 is 7, then the difference between the median and mode of the numbers 12,10,8,10,x,7,6,8 and 6 is:
(a)0. (b)1. (c)2. (d)3
If the median of the distribution is given below is 28.5, find the values of x and y.
Class interval
Frequency
0 − 10
5
10 − 20
x
20 − 30
20
30 − 40
15
40 − 50
y
50 − 60
5
Total
60
The median from the table is
value 7, 8, 9, 10, 11, 12 ,13
frequency 2, 1, 4, 5, 6, 1, 3
if the median of the following frequency distribution is 24. find the missing frequency x :
Age Number of persons
0 - 10 5
10 - 20 25
20 - 30 x
30 - 40 18
40 - 50 7
find the missing frequencies if N=100 and median is 32.
marks obtained-0-10, 10-20, 20-30, 30-40, 40-50, 50-60, total
no. of students-10 ,? ,25 ,30 ,? ,10 ,100.
10. If median = 15 and mean = 16, find mode of the distribution
The median of the following distribution is 30. Find the missing frequencies f1 and f2:
Class frequency
0-10 10
10-20 10
20-30 f1
30-40 30
40-50 f2
50-60 10
Total 100
age (in years):
0-1 1-2 2-3 3-4 4-5 5-6 6-7
no. of pumps:
12 18 45 23 12 08 04
the mean of 10 numbers is 20.if 5 is subtracted from every number.what will be the new mean?
how do we find mean using step deviation method if the classes are unequal?
should we make the classes equal in such a case?
how to find the mean if the class intervals are unequal??
like fr e- 0-10,10-30,30-35....
how to calculate mode if two classes have same and highest frequency (bimodal) ?
The mean of 25 observations is 9 . if each observation is increased by 4 then the new mean is _______
the interest paid on each of the three different sums of money yeilding 1%, 2% and 4% simple interest per annum respectively, is the same. find the average yield percent on the total sum inverted.
how to find xi?
The following table shows the distribution of some families according to expenditure per week .the number of families for two class intervalsare missing. Median =Rs.25 , Mode = Rs.24. calculate the missing frequency and hence the find mean.
classes frequency
0-100 8
100- 200 12
200-300 x
300-400 20
400-500 14
500-600 7
What is the formula to use median when more than ogive is given?
One word answer-VERY URGENT***
1) Difference between any 2 consecutive class marks (9 letters)
2)Graphical Representation (in rectangles) between class marks and respective frequencies (16 letters)
3)Graphical Representation (in rectangles) between class intervals and respective frequencies ( 9 letters)
4)Data in tabular form (7 letters)
median.
Classes 20-30 30-40 40-50 50-60 60-70 70-80 80-90
frequency 10 8 12 24 6 25 15
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days
0 − 6
6 − 10
10 − 14
14 − 20
20 − 28
28 − 38
38 − 40
Number of students
11
10
7
4
4
3
1
Marks Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80 Below 90 Below 100
No. of students 5 9 17 29 45 60 70 78 83 85
3 Median = Mode + 2 Mean
please proof it.
and y, if the total frequency is 50.
Marks No. of students
0-7 3
7-14 x
14-21 7
21-28 11
28-35 y
35-42 16
42-49 9
In a continuous frequency distribution, the median of the data is 21.If each observation is increased by 5 find the new median
In the step deviation method,
What is the meaning of ui ?
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
if the median salary of 100 employees of a factory is 24800, then find the missing frequencies f1
and f2 in the given data
Salary 10000-15000 15000-20000 20000-25000 25000-30000 30000-35000 35000-40000
frequen. 18 f1 f2 15 12 22
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.
what is the formula for finding mean using assumed mean method?
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.
the mean of first n odd natural numbers is n2/81,then n=
3 12 156
4 20 ?
5 25 650
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
frequency is 80.
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
The table given below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories of India find the mean percentage of female teacher by using deviation method.
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.
data
Class interval Frequency
10-20 12
20-30 30
30-40 f1
40-50 65
50-60 f2
60-70 25
70-80 18
no links plz.
what is the difference between assumed mean method and step deviation method?
in which kind of sums do we use the 2 types of finding the mean of different data?
please explain as i am confused bout which method to use in which sum.
the median of the data is 525. find 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
frequeency 2 5 x 12 17 20 y 9 7 4
9. If the mean of the distribution is 57.6 and the sum of its observations is 50, find the missing
The mean of the following frequency dristibution is 50. Find the value of p
In a school 85 boys and 35 girls appeared in a public examination . The mean marks of the boys were found to be 40 % whereas the mean marks of the girls were 60% . Determine the average marks % of the school.
Q). Find the mean, median and mode of the following data:
how can we find the median by only less than ogive curve?
The median of the following data is 32.5.find the value of x and y
class interval f
0-10 x
10-20 5
20-30 9
30-40 12
40-50 y
50-603
60-70 2
total 40
PLS SOON ANSWER PLSEASE
The following table shows the marks obtained by 100 students of class X in a school during a particular academic session. Find the mode of this distribution.
Marks No. of students
Less than 10 7
Less than 20 21
Less than 30 34
Less than 40 46
Less than 50 66
Less than 60 77
Less than 70 92
Less than 80 100
if mode=80 and mean=110,then the median=
Find the mean, mode and median for the following data.
The median of the following data is 20.75. Find the missing frequencies x and y, if the total frequency is 100.
Class interval:-0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Frequencies: -7 10 x 13 y 10 14 9
Life time in hrs No. of bulbs
400 - 499 24
500 - 599 47
600 - 699 39
700 - 799 42
800 - 899 34
900 - 999 14
marks <10 <20 <30 <40 <50 <60 <70 <80 <90 <100
no.of 5 9 17 29 45 60 70 78 83 85
students
how to find the median class?
How to find missing frequency if the mode of the given data is given.
How to convert a LESS THAN OGIVE and it's CF into the Frequency distribution?
the value of the variable for ehich the frequency is maximum is known as ..????
Find the class marks of classes 10 - 25 and 35 - 55.
If the mean of the following distributions is 27, Find the value of p
weight no of student
40-45 2
45-50 3
50-55 8
55-60 6
60-65 6
65-70 3
70-75 2
If the less than ogive and more than ogive intersect each other at ( 20.5,15.5 ), then the median of the given data is :
A ) 36.0 B) 20.5 C ) 15.5 D ) 5.5
13. Find the mean, mode and median for the following data:
The mean of the following data is 53, find the missing frequencies.
age in years -- 0-20, 20-40 40-60 60-80, 80-100
number of people--- 15, f1, 21, f2, 17, 100
How is assumed mean calculated of even number of even number of class intervals ?
E.g.C.I. Fi
20-4051
40-6068
60-8024
80-10013
100-12005
120-14077
140-16036
160-18012
In this case what will be the assumed mean ? Please give reason also as to explain why u chose it ??
Total = 50
height in cm frequency cumulative frequency
150-155 12 a
155-160 b 25
160-165 10 c
165-170 d 43
170-175 e 48
175-180 2 f
Marks No of students
30-40 8
40-50 7
50-60 10
60-70 15
70-80 5
80-90 8
90-100 7
By using the above information, calculate mean, median, and mode.
no links plz.
the class mark of classes in a distribution arer 6.10.14.18.22.26.30. find
a) class size
b) lower limit of second class
c) upper lmit of last class
d) third class