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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....

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

what is hit and trial method .

_{1 }, f_{2}and f_{3}in the following frequency distribution , when it is given taht f_{2 }: f_{3}= 4 : 3 , and mean = 5020-40

40-60

60-80

80-100

17

f

_{1}f

_{2}f

_{3 19}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

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.

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

monthly consumption of electricity of some consumers is given below as a distribution. Find the missing frequency(x), if mode is given to be 200 units.

Monthly cosumption(in units ) 90-120 120-150 150-180 180-210 210-240

Number of consumers 20 15 X 75 30

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.

13. Find the mean, mode and median for the following data:

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

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

the mean of 10 numbers is 20.if 5 is subtracted from every number.what will be the new mean?

In the step deviation method,

What is the meaning of

u_{i }?how to find the mean if the class intervals are unequal??

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

IF THE N/2 IS EQUAL TO THE CUMULATIVE FREQUENCY THEN WE WOULD REGARD THAT CLASS AS THE MEDIAN CLASS OR NOT.

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

Find the median of the following data.

Wages ( in rupees ) :More than 150 , More than 140 , More than 130 , More than 120 , More than 110 , More than 100 , More than 90 . More than 80.No.of workers : NIL ,12,27,60,105,124,141,150.how to find xi?

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

Cost of living index /number of weeks

1400-1500—8

1550-1700—15

1700-1850—21

1850-2000—8

median.

Classes 20-30 30-40 40-50 50-60 60-70 70-80 80-90

frequency 10 8 12 24 6 25 15

d

_{i}is the deviation of x_{i }from assumed mean a.If mean = x + ∑f

_{i}d_{i}/∑f_{i}, then x isA 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

class frequency

40-50 5

50-60 x

60-70 15

70-80 2

80-90 7

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 years)

(A)

(B)

Find the modal age of employees in factory A and B.

The mean and median of same data are 24 and 26 respectively. The value of mode is ?

53 – 55

56 – 58

59 – 61

62 – 64

120

105

125

30

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.

Height No. of students.

125-130 2

130-135 4

135-140 x

140-145 y

145-150 8

150-155 9

155-160 5

Mode and mean of a data are 12k and 15k. Median of the data is

the mean of first n odd natural numbers is n2/81,then n=

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.

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?

CLASSFREQUENCY>20 50

>30 44

>40 28

>50 20

>60 15

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 frequency

25-30 x

30-35 22

35-40 y

40-45 8

45-50 7

50-55 3

55-60 2

Class interval Frequency0-10 10

10-20 20

20-30 x

30-40 40

40-50 y

50-60 25

60-70 15

Class 0 – 20 20 – 40 40 – 60 60 – 80 80 –100 100 –120

Frequency 6 8 f 12 6 5

8. The mean of the following distribution is 53. Calculate missing frequencies

Find the sum of deviations of the variate values 3, 4, 6, 7, 8, 14 from

their mean.

calculate the median from the following data

class interval frequency

130-1405

140-150 9

150-160 17

160-170 28

170-180 24

180-190 10

190-200 7

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 central value of the set of observations is called -------

The mean of the following frequency dristibution is 50. Find the value of

pp^{3}and 1/y^{3 }is :In a given data if the MODAL CLASS is at the bottom(last) and has nothing Preceeding it then how will we find the MODE?

how can we find the median by only less than ogive curve?

Marks No.of students

Below 10 0

Below 20 12

Below 30 20

Below 40 28

Below 50 33

Below 60 40

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

How is assumed mean calculated of even number of even number of class intervals ?

E.g.C.I. F

_{i}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 ??

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

a) ogive b) histogram c) frequency polygon d) frequency curve

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

Compute the median from the following:Mid Value: 115 125 135 145 155 165 175 185 195

Frequency: 6 25. 48. 72. 116 60 38. 22. 3

classes frequency

0-100 8

100- 200 12

200-300 x

300-400 20

400-500 14

500-600 7

how to find the median class?

How to find missing frequency if the mode of the given data is given.

(in percent)

is the ans median class 60-70 and modal class 50-60??

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

in a health check up , the nuber of heart beats of 40 women were recorded in the following table:

-no.of heart beats per minute | 65-69 | 70-74 | 75-79 | 79-84 |

- no of women | 2 | 18 | 16 | 4 |

find the mean of the data.

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

height 150-156 156-162 162-168 168-174 174-180

no.students 4 7 15 8 6

find mean height.

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

Q1) among the following which is not the measure of central tendency?

A) Mean B) Median C) Mode D) range

Q2) among the following which is the best measure of central tendency?

A) Mean B) Median C) Mode

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

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

The mode of the following data is 36. ffind the missing frequency .

class interval frequency0-10 810-20 1020-30 x30-40 1640-50 1250-60 660-70 7