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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

xandy.Class intervalFrequency0 − 10

5

10 − 20

x20 − 30

20

30 − 40

15

40 − 50

y50 − 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 distributionThe 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

marksand respective frequencies (16 letters)3)Graphical Representation (in rectangles) between class

intervalsand 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 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.

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

u_{i }?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 f

_{1}and f

_{2 }in the given dataSalary 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 workers12

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 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.

data

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.

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

p' if the mean of the given data is 15.45.pThe 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.

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 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

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 OGIVEand it'sCFinto 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. 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 ??

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 students30-40 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