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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....

what is hit and trial method .

If the median of the distribution is given below is 28.5, find the values of

xandy.Class intervalFrequency0 − 10

5

10 − 20

x20 − 30

20

30 − 40

15

40 − 50

y50 − 60

5

Total

60

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.

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 to find the mean if the class intervals are unequal??

like fr e- 0-10,10-30,30-35....

The mean of 25 observations is 9 . if each observation is increased by 4 then the new mean is _______

how to find xi?

how to calculate mode if two classes have same and highest frequency (bimodal) ?

What is the formula to use median when more than ogive is given?

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 days0 − 6

6 − 10

10 − 14

14 − 20

20 − 28

28 − 38

38 − 40

Number of students11

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.

In a continuous frequency distribution, the median of the data is 21.If each observation is increased by 5 find the new median

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

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 workers12

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=

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 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

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

how do we find mean using step deviation method if the classes are unequal?

should we make the classes equal in such a case?

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 studentsLess 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 Frequency0-10 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.

Class interval Frequency10-20 12

20-30 30

30-40 f1

40-50 65

50-60 f2

60-70 25

70-80 18

no links plz.

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

ppIn 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 studentsLess 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

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 OGIVEand it'sCFinto the Frequency distribution?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

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

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

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