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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 .

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.

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 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.

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

1. Observe the information's in the following cumilative frequency table

no. of days absentless than 5

less than 10

less than 15

less than 20

less than 25

less than 30

less than 35

less than 40

less than 45

No. of students29224465582634644650653655Q.. Draw a 'less than' ogive and 'more than' ogive to represent the above data .

Q... Find median for the above data by interpreting the graph.

Q...why regularity is essential in students life (VBQ)

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

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....

In statistics Ex 14.1 Q8 of class 10..

if we solve the question by using step deviation, what value of h will

we take as in the book it is written that if the class sizes are

unequal, and xi are large numerically we can still apply step

deviation method taking h to be a suitable divisor of all the di "

But in Example 3 of statistics we have taken h to be 20 whereas it is

not a divisor to all the di's. It is not a divisor to -75...

Please explain this to me

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

frequency is 80.

how to find xi?

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

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?

In the step deviation method,

What is the meaning of

u_{i }?median.

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

frequency 10 8 12 24 6 25 15

Q. 27. School held sports day in which 150 students participated. Age of students are given in the following frequency distribution;

(in years)

For the above data, draw a more than type, ogive and from the curve, find median. Verify it by actual calculations.

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

(in years)

(A)

(B)

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

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

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.

3 Median = Mode + 2 Mean

please proof it.

CLASSFREQUENCY>20 50

>30 44

>40 28

>50 20

>60 15

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

How to make more than type and less than type frequency distribution?

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

3 12 156

4 20 ?

5 25 650

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.

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.

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

Q). Find the mean, median and mode of the following data:

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

…………………. Is called a positional average.

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

f_{0}= frequency of the class preceding the modal class andf_{2}= frequency of the class succeeding the modal class and how to use them to calculate mode when grouped data is given please tell its urgentclass frequency

40-50 5

50-60 x

60-70 15

70-80 2

80-90 7

If the mean of following distribution is 54 then find p.

Class Interval: 0-20 20-40 40-60 60-80 80-100

Frequeancy : 7 p 10 9 30

Pls provide the solution fast..

If the class marks of a continuous frequency distribution are 12,14,16,18,........., then find the class intervals corresponding to the class marks 16 and 22.

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

wrong I want to know the steps pls help me

C.I. F

45-55 7

55-65 12

65-75 17

75-85 30

85-95 32

95-105 6

105-115 10

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

their mean.

If mean is

166 9/26(mixed fraction) andsum of observations is 52.Findf1 and f2.C I FREQUENCY

140 - 150 5

150 -160 f1

160 -170 20

170-180 f2

180-190 6

190-200 2

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.

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

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

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

9. If the mean of the distribution is 57.6 and the sum of its observations is 50, find the missing

please tell how to find the modal class?

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

ppclass 0-20 20-40 40-60 60-80 80-100 100-120

frequency14 x 24 32

10 2

_{1},x_{2}.............x_{10}are replaced by x_{i}+5, i=1,2,3,4,5 and next five elements are replaced by x_{j}-5, j= 6,7,8,...........10 then the mean will change byhow can we find the median by only less than ogive curve?

the value of the variable for ehich the frequency is maximum is known as ..????

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

how to draw the more than and less than ogive for the same data ?

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

marks obtained by 70 students are given below

marks ---- 20 70 90 40 60 50 75

no. of ---- 8 12 5 12 6 18 9

students

FIND THE MEDIAN MARKS.

how to find the median class?

If the mean of x

_{1}and x_{2}is M_{1}and that of x_{1}, x_{2}, x_{3}, x_{4}is M_{2}, the mean of ax_{1}, ax_{1}.x_{3}/a, x_{4}/a isA) M

_{1}+ M_{2}/ 2 B) aM_{1}+ (M_{2}/a) / 2How 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.what is summision?

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

Class Frequency

5-10 5

10-15 6

15-20 15

20-25 10

25-30 5

30-35 4

If the mean of the following distributions is 27, Find the value of p

Class frequency

0-20. 6

20-40. 8

40-60. F

60-80. 12

80-100. 6

100-120. 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

Don't send link

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

Explain with an example.

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