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Statistics

1.0Introduction

It is the branch of mathematics which deals with the collection, presentation, analysis and interpretation of numerical data.

In singular form, statistics is taken as a subject. And, in plural form, statistics means data. Class interval: Each group into which the raw data is condensed, is called a class-interval. Class limits: Each class is bounded by two figures, which are called class limits. The figure on the left side of a class is called its lower limit and that on its right is called its upper limit.

Exclusive form (or continuous interval form): A frequency distribution in which the upper limit of each class is excluded and lower limit is included, is called an exclusive form.

Example: Suppose the marks obtained by some students in an examination are given. We may consider the classes 0−10,10−20 etc. In class 0−10, we include 0 and exclude 10. In class 10−20, we include 10 and exclude 20.

Inclusive form (or discontinuous interval form): A frequency distribution in which each upper limit as well as lower limit is included, is called an inclusive form. Thus, we have classes of the form 0−10,11−20,21−30 etc.

In 0−10, both 0 and 10 are included.

2.0Important terms related to grouped data:

Class boundaries or true upper and true lower limits: (i) In the exclusive form, the upper and lower limits of a class are respectively known as the true upper limit and true lower limit. (ii) In the inclusive form, the number midway between the upper limit of a class and lower limit of the subsequent class gives the true upper limit of the class and the true lower limit of the subsequent class. Thus, in the above table of inclusive form, we have: true upper limit of class 1−10 is (210+11​)=10.5 and true lower limit of class 11−20 is 10.5 . Similarly, true upper limit of class 11−20 is (220+21​)=20.5, and true lower limit of class 21−30 is 20.5 .

Class size The difference between the true upper limit and the true lower limit of a class is called its class size.

Class mark of a class Class mark =(2 True upper limit + True lower limit ​)

  • The difference between any two successive class marks gives the class size.
  • Average is the statistic which describes the center of a set of data, a set of numbers which are measurements or counts.

The most commonly used averages are the mean (arithmetic average), mode (most frequent number), median (middle number when numbers are listed smallest to largest).

Numerical Ability 1 The class marks of a frequency distribution are 7,13,19,25,31,37,43. Find the class-size and all the class-intervals. Solution: Class size = Difference between two successive class-marks =(13−7)=6. Let the lower limit of the first-class interval be a. Then, its upper limit =(a+6). ∴2a+(a+6)​=7⇒2a=8⇒a=4 So, the first class-interval is 4-10. Let the lower limit of last class-interval be b. Then, its upper-class limit =(b+6). ∴2b+(b+6)​=43⇒2 b=80⇒ b=40. So, the last class-interval is 40-46. Hence, the required class-intervals are 4−10,10−16,16−22,22−28,28−34,34−40 and 40−46. An average tends to lie centrally with the values of the variable arranged in ascending order of magnitude. So, we call an average a measure of central tendency of the data. Three measures of central tendency are useful for analysing the data, namely (a) Mean (b) Median (c) Mode

  • We know that the mean of observations is the sum of the values of all the observations divided by the total number of observations i.e., if x1​,x2​,x3​,…...,xn​ are n observations, then mean xˉ=nx1​+x2​+x3​+….+xn​​ or xˉ=n∑i=1n​xi​​, where ∑i=1n​xi​ denotes the sum x1​+x2​+x3​+…..+xn​.

3.0Arithmetic mean

The arithmetic mean of grouped data may also be calculated by any one of the following methods:

  • Direct method
  • Assumed-mean method

Direct method

x1​,x2​,x3​,……..Xn​ are observations with respective frequencies f1​,f2​,f3​,……...fn​ then mean, (x) is defined by xˉ=f1​+f2​+f3​+….+fn​f1​x1​+f2​x2​+f3​x3​+….+fn​xn​​ or xˉ=∑i=1n​fi​∑i=1n​fi​xi​​, Where ∑i=1n​fi​=f1​+f2​+f3​+…..+fn​=N The following steps should be followed in finding the arithmetic mean of grouped data by direct method. Step-1: Find the class mark ( xi​ ) of each class using, xi​=2 Lowerlimit + Upperlimit ​ Step - 2: Calculate fiXi​ for each i Step - 3: Use the formula : mean xˉ=∑i=1n​fi​∑i=1n​fi​xi​​,

  • Sum of first ' n ' natural numbers is 2n(n+1)​

Numerical Ability 2 Find the mean of the following data:

Class Interval0−88−1616−2424−3232−40
Frequency671089

Solution: We may prepare the table as given below:

Class intervalFrequency (fi​)Class mark (xi​)fi​xi​
0−86424
8−1671284
16−241020200
24−32828224
32−40936324
Σfi​=40Σfi​xi​=856

∴ Mean, xˉ=∑i=1n​fi​∑i=1n​fi​xi​​=40856​=21.4

Numerical Ability 3 The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Daily pocket allowance (in Rs.)Number of children
11−137
13−156
15−179
17−1913
19−21f
21−235
23−254

Solution: We may prepare the table as given below:

Daily Pocket allowanceNumber of Children ( fi​ )Class mark ( xi​ )fi​xi​
11-1371284
13-1561484
15-17916144
17-191318234
19-21f2020 f
21-23522110
23-2542496
Σfi​=44+fΣfi​xi​=752+20f

∴ Mean xˉ=∑i=1n​fi​∑i=1n​fi​xi​​=44+f752+20f​ Given, mean =18∴18=44+f752+20f​ ⇒792+18f=752+20f ⇒2f=40 ⇒f=20

  • The sum of squares of ' n ' natural numbers is 6n(n+1)(2n+1)​

Numerical Ability 4 Find the missing frequencies f1​ and f2​ in the table given below, it is being given that the mean of the given frequency distribution is 50 .

Class0−2020−4040−6060−8080−100Total
Frequency17f1​32f2​19120

Solution: We may prepare the table as given below:

ClassFrequency ( fi​ )Class mark ( xi​ )fi​xi​
0-201710170
20-40f1​3030f1​
40-6032501600
60-80f2​7070f2​
80-10019901710
Σfi​=68+f1​+f2​Σfi​xi​=3480+30f1​+70f2​

∴ Mean, x=∑fi​∑fi​xi​​=68+f1​+f2​3480+30f1​+70f2​​ Given, mean =50 ∴50=68+f1​+f2​3480+30f1​+70f2​​ ⇒3400+50f1​+50f2​=3480+30f1​+70f2​ ⇒20f1​−20f2​=80 ⇒f1​−f2​=4 And Σfi​=68+f1​+f2​ ∴120=68+f1​+f2​[∵Σf1​=120] ⇒f1​+f2​=52 Adding (i) and (ii), we get 2f1​=56 ⇒f1​=28 ⇒f2​=24 Hence, the missing frequencies f1​ and f2​ are 28 and 24 respectively.

Numerical Ability 5 Find the mean of the following distribution.

Class50−7070−9090−110110−130130−150150−170
Frequency18121327822

Solution:

ClassFrequency ( fi​)Class mark (xi​)fi​xi​
50−7018601080
70−901280960
90−110131001300
110−130271203240
130−15081401120
150−170221603520
N=100Σfi​xi​=11220

∴ Mean,

x=∑fi​∑fii​​​=10011220​=112.2

Numerical Ability 6 Find the mean marks from the following data:

MarksNo. of Students
Below 105
Below 209
Below 3018
Below 4029
Below 5045
Below 6060
Below 7070
Below 8078
Below 9083
Below 10085

Solution: We may prepare the table as given below:

MarksNo. of StudentsClass Intervalfi​Class mark(xi​)fi​xi​
Below 1050−105525
Below 20910−2041560
Below 301820−30925225
Below 402930−401135385
Below 504540−501645720
Below 606050−601555825
Below 707060−701065650
Below 807870−80875600
Below 908380−90585425
Below 1008590−100295190
∑fi​=85∑fi​xi​=4105

∴ Mean,

x=∑fi​∑fi​xi​​=854105​=48.29

If less than type or more than type frequency distribution table is given then convert table is usual form to find the mean.

Numerical Ability 7 Find the mean marks of students from the adjoining frequency distribution table.

MarksNo. of Students
Above 080
Above 1077
Above 2072
Above 3065
Above 4055
Above 5043
Above 6023
Above 7016
Above 8010
Above 908
Above 1000

Solution: We may prepare the table as given below:

MarksNo. of StudentsClass Intervalfi​Class mark(xi​)fi​xi​
Above 0800−103515
Above107710−2051575
Above207220−30725175
Above306530−401035350
Above405540−501245540
Above504350−6020551100
Above602360−70765455
Above 701670−80675450
Above 801080−90285170
Above 90890−100895760
Above 1000100−11001050
∑fi​=80∑fi​xi​=4090

∴ Mean, x=∑fi​∑fi​xi​​=804090​=51.125=51.1 (approx)

Assumed mean method

In this case, to calculate the mean, we follow the following steps: Step - 1: Find the class mark xi​ of each class using xi​=2 Lower limit + Upper limit ​ Step - 2 : Choose a suitable value of xi​ in the middle as the assumed mean and denote it by 'a'. Step-3: Find di​=xii​−a for each i Step-4: Find fi​×di​ for each i Step-5: Find N=∑fi​ Step - 6 : Calculate the mean, (xˉ) by using the formula xˉ=a+N∑fi​di​​

Numerical Ability 8 The following table gives the marks scored by 100 students in a class test:

Mark0−1010−2020−3030−4040−5050−60
No. of Students121827201760

Find the mean marks scored by a student in class test.

Solution: We may prepare the table with assumed mean, a=35 as given below:

MarksNo. of StudentsClass mark ( xi​)di​=xi​−a=xi​−35fi​di​
0−10125-30-360
10−201815-20-360
20−302725-10-270
30−402035=a00
40−50174510170
50−6065520120
N=100∑fi​di​=−700

∴ Mean, x=a+N∑fi​di​​ =35+100(−700)​=35−7=28

Numerical Ability 9 Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, by using assumed-mean method.

No. of heart beats per minute65−6868−7171−7474−7777−8080−8383−86
Frequency2438742

Solution: We may prepare the table with assumed mean, a=75.5 as given below:

No. of heart beats per minuteNo. of women (fi​)Class mark (xi​)di​=xi​−a=xi​−75.5fi​di​
65−68266.5-9-18
68−71469.5-6-24
71−74372.5-3-9
74−77875.5=a00
77−80778.5321
80−83481.5624
83−86284.5918
N=30∑fi​di​=12

∴ Mean, x=a+N∑fi​di​​=75.5+3012​=75.5+52​=75.9

Numerical Ability 10 Find the arithmetic mean of the following frequency distribution.

Class25−2930−3435−3940−4445−4950−5455−59
Frequency1422166534

Solution: The given series is in inclusive form. We will prepare the table in exclusive form with assumed mean a =42 as given below:

ClassFrequency (fi​)Class mark (xi​)di​=xi​−a=xi​−42fi​di​
24.5−29.51427-15-210
29.5−34.52232-10-220
34.5−39.51637-5-80
39.5−44.5642=a00
44.5−49.5547525
49.5−54.53521030
54.5−59.54571560
N=70∑fi​di​=−395

∴ Mean, x=a+ N∑fi​ di​​=42+70(−395)​=702940−395​=702545​=36.36 (approx.) 

  • It is not necessary to convert the data in continuous interval form for finding mean.

Step-deviation method or short-cut method

Sometimes, the values of x and f are so large that the calculation of mean by assumed mean method becomes quite inconvenient. In this case, we follow the following steps: Step-1 : Find the class mark xi​ of each class by using xi​=2 lower limit + Upper limit ​ Step-2 : Choose a suitable values of xi​ in the middle as the assumed mean and denote it by 'a'. Step-3 : Find h= (upper limit -lower limit) for each class. Step-4 : Find ui​=hxi​−a​ for each class. Step-5 : Find fi​ui​ for each i. Step-6 : Calculate, the mean by using the formula =a+{NΣfi​×ui​​}×h, where N=∑fi​.

Numerical Ability 11 Find the mean of the following distribution by step-deviation method:

Class50−7070−9090−110110−130130−150150−170
Frequency18121327822

Solution: We may prepare the table with assumed mean a=120 and h=20 as given below:

ClassFrequency (fi​)Class mark (xi​)ui​=hxi​−a​=20xi​−120​fi​ui​
50−701860-3-54
70−901280-2-24
90−11013100-1-13
110−13027120=a00
130−150814018
150−17022160244
N=100∑fi​ui​=−39

∴ Mean, x=a+ N∑fi​ui​​ h=120+100(−39)×20​=120−539​=5561​=112.2

  • If class size is different then h is taken as HCF of the class marks.

Numerical Ability 12 Calculate the mean for the following frequency distribution (By step deviation method).

Class interval0−8080−160160−240240−320320−400
Frequency2235442524

Solution:

Class IntervalMid - value (xi​)fi​ui​=(xi​−a)/hfi​ui​
0−804022-2-44
80−16012035-1-35
160−240200(a)4400
240−32028025125
320−40036024248
Total∑fi​=150∑fi​ui​=−6

Mean (x)=a+Σfi​Σfi​ui​​×h=200+(150−6​)×80 =200−(52×8​)=200−516​=51000−16​=5984​=196.8

4.0Median

Median: It is a measure of central tendency which gives the value of the middle most observation in the data. In a grouped data, it is not possible to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves.

Median Class: The class whose cumulative frequency is greater than 2N​ is called the median class.

To calculate the median of a grouped data, we follow the following steps.

Step - 1: Prepare the cumulative frequency table corresponding to the given frequency distribution and obtain N=Σfi​

Step-2: Find 2N​

Step - 3: Look at the cumulative frequency just greater than 2N​ and find the corresponding class (Median class).

Step - 4: Use the formula Median =ℓ+{f2N​−cf​}×h Where, ℓ= Lower limit of median class. f= Frequency of the median class. cf= Cumulative frequency of the class preceding the median class. h= Size of the median class. N=∑fi​.

  • Data must be in continuous interval form to find median and mode of grouped data.

Numerical Ability 13 Find the median of the following frequency distribution:

Mark0−1010−2020−3030−4040−50Total
No. of Students820362412100

Solution: At first, we prepare a cumulative frequency distribution table as given below:

MarksNo. of students (fi​)Cumulative frequency
0−1088
10−202028
20−303664
30−402488
40−5012100
N=100

Here, N = 100 ∴2N​=50 The cumulative frequency just greater than 50 is 64 and the corresponding class is 20−30. So, the median class is 20−30. ∴ℓ=20, N=100,cf=28,f=36 and h=10 Therefore, median =ℓ+{f2N​−cf​}×h=20+(3650−28​)×10=20+3622×10​=20+955​=9180+55​=9235​=26.1

  • The median is the middle of a distribution: half the scores are above the median and half are below the median.

Numerical Ability 14 A health insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age(in years)No. of policy holders
Below 204
Below 258
Below 3024
Below 3545
Below 4078
Below 4589
Below 5090
Below 5596
Below 60100

Solution: From the given table we can find the frequency and cumulative frequencies as given below :

Age in(years)No. of holders ( fi​ )Cumulative frequency
15−2044
20−2548
25−301624
30−352145
35−403378
40−451189
45−50190
50−55696
55−604100
N=100

Here, N=100 ∴2N​=50 The cumulative frequency just greater than 50 is 78 and the corresponding class is 35−40. So, the median class is 35−40. ∴ℓ=35 N=100,cf=45,f=33 and h=5 Therefore, median =ℓ+{f2N​−cf​}×h =35+(3350−45​)×5 =35+335×5​ =331155+25​=331180​=35.76 Hence, the median age is 35.76 years.

Numerical Ability 15 From the following frequency distribution, calculate the median.

MarksNo. of students
5−107
10−1515
15−2024
20−2531
25−3042
30−3530
35−4026
40−1515
45−5010

Solution: Here 2N​=2200​=100 The median lies in the class 25−30. ℓ=25, c.f. =77,f=42 and h=5

Calculation of Median

Marksfc.f.
5−1077
10−151522
15−202446
20−253177
25−3042119
30−3530149
35−4026175
40−4515190
45−5010200

Applying the formula, Median =ℓ+f2N​− c.f. ​×h We get, Median =25+42100−77​×5 =25+42115​=25+2.74=27.74 So, above half the student have scored marks less than 27.74 and the other half scored marks more than 27.74.

Numerical Ability 16 Calculate the missing frequency 'a' from the following distribution, it is being given that the median of the distribution is 24 .

Age (in years)0−1010−2020−3030−4040−50
No. of persons525a187

Solution: At first we prepare a cumulative frequency distribution table as given below :

Age (in years)0−1010−2020−3030−4040−50Total
No. of persons(fi)525a18755+a
Cumulative frequency53030+a48+a55+a

Since the median is 24, therefore, the median class will be 20−30. Hence, ℓ=20, N=55+a,cf=30,f=a and h=10 Therefore, median =ℓ+{f2N​−cf​}×h ⇒24=20+(a255+a​−30​)×10 ⇒24=20+2a(a−5)​×10 ⇒4=a(a−5)​×5 ⇒4a=5a−25 ⇒a=25 Hence, the value of missing frequency a is 25 .

Numerical Ability 17 The median of the following data is 525 . Find the values of x and y, if the total frequency is 100.

Class IntervalFrequency( fi​ )
0-1002
100-2005
200-300x
300-40012
400-50017
500-60020
600-700y
700-8009
800-9007
900-10004
N=100

Solution: At first we prepare a cumulative frequency distribution table as given below :

Class intervalFrequency (fi)Cumulative frequency
0−10022
100−20057
200−300x7+x
300−4001219+x
400−5001736+x
500−6002056+x
600−700y56+x+y
700−800965+x+y
800−900772+x+y
900−1000476+x+y
N=100

We have N=100 ∴76+x+y=100 ⇒x+y=24 Since the median is 525 , so, the median class is 500-600 ∴ℓ=500, N=100,cf=36+x,f=20 and h=100 Therefore, median =ℓ+{f2N​−cf​}×h ⇒525=500+(2050−36−x​)×100 ⇒25=(14−x)×5 ⇒5=14−x ⇒x=9 Also, putting x=9 in (i), we get 9+y=24 ⇒y=15 Hence, the values of x and y are 9 and 15 respectively.

5.0Mode

Mode : Mode is that value among the observations which occurs most often i.e. the value of the observation having the maximum frequency.

In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies.

Modal Class : The class of a frequency distribution having maximum frequency is called modal class of a frequency distribution.

The mode is a value inside the modal class and is calculated by using the formula. Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h Where ℓ= Lower limit of the modal class. h = Size of class interval f1​= Frequency of modal class f0​= Frequency of the class preceding the modal class f2​= Frequency of the class succeeding the modal class.

  • A disadvantage of the mode is that many distributions have more than one mode. These distribution are called "multi modal" and is therefore not recommended to be used as the only measure of central tendency.

Numerical Ability 18 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Lifetimes (in hours)0−2020−4040−6060−8080−100100−120
Frequency103552613829

Determine the modal lifetime of the components.

Solution: Here the class 60-80 has maximum frequency, so it is the modal class. ∴ℓ=60, h=20,f1​=61,f0​=52 and f2​=38 Therefore, mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h =60+(2×61−52−3861−52​)×20 =60+329​×20 =60+5.625 =65.625 Hence, the modal lifetime of the components is 65.625 hours.

Numerical Ability 19 Given below is the frequency distribution of the heights of players in a school.

Heights (in cm)160−162163−165166−168169−171172−174
No. of students1511814212718

Find the modal class.

Solution: The given series is in inclusive form. We prepare the table in exclusive form, as given below :

Heights (in cm)159.5−162.5162.5−165.5165.5−168.5168.5−171.5171.5−174.5
No. of students1511814212718

Here, the class 165.5-168.5 has maximum frequency, so it is the modal class.

Numerical Ability 20 The mode of the following series is 36 . Find the missing frequency f in i.

Class0−1010−2020−3030−4040−5050−6060−70
Frequency810f161267

Solution: Since the mode is 36 , so the modal class will be 30−40 ∴ℓ=30, h=10,f1​=16,f0​=f and f2​=12 Therefore, mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h ⇒36=30+(2×16−f−1216−f​)×10 ⇒6=(20−f)(16−f)​×10 ⇒120−6f=160−10f ⇒4f=40 ⇒f=10 Hence, the value of the missing frequency f is 10 .

  •  Mode = 3 Median - 2 Mean 

Numerical Ability 21 The mean and mode of a frequency distribution are 28 and 19 respectively. Then find the median. Solution: Mean =28 and Mode =19 Median = ? Mode =3 Median -2 Mean 19=3 Median - 2(28) 3 Median = 19+56 Median =375​ Median =25

6.0Graphical representation of cumulative frequency distribution

Cumulative frequency polygon curve (Ogive) Cumulative frequency is of two types and corresponding to these, the ogive is also of two types. • Less than ogive • More than ogive

Less than ogive: To construct a cumulative frequency polygon and an ogive, we follow these steps : Step-1: Mark the upper class limit along x -axis and the corresponding cumulative frequencies along y-axis. Step-2: Plot these points successively by line segments. We get a polygon, called cumulative frequency polygon. Step-3: Plot these points successively by smooth curves, we get a curve called cumulative frequency curve or an ogive.

More than ogive: To construct a cumulative frequency polygon and an ogive, we follow these steps: Step-1: Mark the lower class limits along x-axis and the corresponding cumulative frequencies along y-axis. Step-2: Plot these points successively by line segments, we get a polygon, called cumulative frequency polygon. Step-3: Plot these points successively by smooth curves, we get a curve, called cumulative frequency curve or an ogive.

Application of an ogive

Ogive can be used to find the median of a frequency distribution. To find the median, we follow these steps.

Method-I

Step-1: Draw anyone of the two types of frequency curves on the graph paper. Step-2: Compute 2N​(N=Σfi​) and mark the corresponding points on the y-axis. Step-3: Draw a line parallel to x-axis from the point marked in step 2, cutting the cumulative frequency curve at a point P. Step-4: Draw perpendicular PM from P on the x-axis. The x -coordinate of point M gives the median.

Method-II

Step-1: Draw less than type and more than type cumulative frequency curves on the graph paper. Step-2: Mark the point of intersecting (P) of the two curves drawn in step 1. Step-3: Draw perpendicular PM from P on the x-axis. The x - coordinate of point M gives the median.

Numerical Ability 22 The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs.)100−120120−140140−160160−180180−200
No. of workers12148610

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Solution: From the given table, we prepare a less than type cumulative frequency distribution table, as given below:

Income less than (in Rs.)120140160180200
Cumulative frequency1226344050

Now, plot the points (120,12),(140,26),(160,34),(180,40) and (200,50). Join these points by a freehand curve to get an ogive of 'less than' type.

Numerical Ability 23 The following table gives the weight of 120 articles :

Weight (in kg)0−1010−2020−3030−4040−5050−60
Number of Students141722262318

Change the distribution to a 'more than type' distribution and draw its ogive.

Solution:

Weight (in Kg)Cumulative Frequency
More than or equal to 0120
More than or equal to 10106
More than or equal to 2089
More than or equal to 3067
More than or equal to 4041
More than or equal to 5018

Plotting the points :

Numerical Ability 24 The annual profits earned by 30 shops of a shopping complex in a locality gives rise to the following distribution:

Profit (in lakhs Rs.)No. of shops (frequency)
More than or equal to 530
More than or equal to 1028
More than or equal to 1516
More than or equal to 2014
More than or equal to 2510
More than or equal to 307
More than or equal to 353

Draw both ogives for the data above. Hence, obtain the median profit.

Solution: We have a more than type cumulative frequency distribution table. We may also prepare a less than type cumulative frequency distribution table from the given data, as given below:

More than type
Profit more than (Rs. in lakhs)No. of shops
530
1028
1516
2014
2510
307
353
Less than type
Profit less than (Rs. in lakhs)No. of shops
102
1514
2016
2520
3023
3527
4030

Now, plot the points A (5,30), B (10,28), C (15,16),D(20,14),E(25,10),F(30,7) and G (35,3) for the more than type cumulative frequency and the points P(10,2),Q(15,14),R (20,16),S(25,20),T(30,23),U(35,27) and V(40,30) for the less than type cumulative frequency distribution table. Join these points by a freehand to get ogives for 'more than' type and 'less than' type.

The two ogives intersect each other at point (17.5, 15). Hence, the median profit is Rs. 17.5 lakhs.

Numerical Ability 25 The following data gives the information on marks of 70 students in a periodical test: Draw a cumulative frequency curve for the given data and find the median.

MarksLess than 10Less than 20Less than 30Less than 40Less than 50
No. of students311284870

Solution: We have a less than cumulative frequency table. We mark the upper class limits along the x -axis and the corresponding cumulative frequencies (no. of students) along the y -axis. Now, plot the points (10,3),(20,11),(30,28),(40,48) and (50,70). Join these points by a freehand curve to get an ogive of 'less than' type.

Here, N=70 ∴2N​=35 Take a point A(0,35) on the y-axis and draw AP∥x-axis, meeting the curve at P. Draw PM⊥x-axis, intersecting the x -axis, at M . Then, 0M=33. Hence, the median marks is 33 .

7.0Memory map

On this page


  • 1.0Introduction
  • 2.0Important terms related to grouped data:
  • 3.0Arithmetic mean
  • 3.1Direct method
  • 3.2Assumed mean method
  • 3.3Step-deviation method or short-cut method
  • 4.0Median
  • 5.0Mode
  • 6.0Graphical representation of cumulative frequency distribution
  • 6.1Application of an ogive
  • 6.1.1Method-I
  • 6.1.2Method-II
  • 7.0Memory map

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