STATISTICS (LECTURE 7)

LESSON  STATISTICS

Dear Students,
Good Morning !!

Yesterday we covered the following learning outcomes:
1 . Recall ,measures  of central tendency.
2 .calculating Mean, Median and Mode.

Let us go through the guidelines for the  blog once again:
•              The text in Red, is to be written in your register
•              The text in blue is to be viewed by clicking on it
•              The text in green is to be practised for home work
General Instructions
•               take your SET-A Mathematics Register
.               Please write the learning outcomes as mentioned below:
                I will be able to  :
                   draw a less  than type  ogive and find its median
                draw a  more than type  ogive and find its median     

Cumulative

Cumulative means "how much so far".
Think of the word "accumulate" which means to gather together.
To have cumulative totals, just add up the values as you go.

Example: Jamie has earned this much in the last 6 months:

MonthEarned
March$120
April$50
May$110
June$100
July$50
August$20
To work out the cumulative totals, just add up as you go.
The first line is easy, the total earned so far is the same as Jamie earned that month:
MonthEarnedCumulative
March$120$120
But for April, the total earned so far is $120 + $50 = $170 :
MonthEarnedCumulative
March$120$120
April$50$170

And for May we continue to add up: $170 + $110 = $280
MonthEarnedCumulative
March$120$120
April$50$170
May$110$280

Do you see how we add the previous month's cumulative total to this month's earnings?
Here is the calculation for the rest:
  • June is $280 + $100 = $380
  • July is $380 + $50 = $430
  • August is $430 + $20 = $450
And this is the result
MonthEarnedCumulative
March$120$120
April$50$170
May$110$280
June$100$380
July$50$430
August$20$450
The last cumulative total should match the total of all earnings:
$450 is the last cumulative total ...
... it is also the total of all earnings:
$120+$50+$110+$100+$50+$20 = $450
So we got it right.

So that's how to do it, add up as you go down the list and you will have cumulative totals.

We could also call it a "Running Total"

so let us answer to the question given below









What is Cumulative Frequency ?


The frequency is the number of times an event occurs within a given scenario. Cumulative frequency is defined as the running total of frequencies. It is the sum of all the previous frequencies up to the current point. It is easily understandable through a Cumulative Frequency Table.
MarksFrequency
(No. of Students)
Cumulative Frequency
0 – 522
5 – 101012
10 – 15517
15 – 20522

Cumulative Frequency Curve


Cumulative Frequency Curve

Cumulative Frequency Curve
A curve that represents the cumulative frequency distribution of grouped data on a graph is called a Cumulative Frequency Curve or an Ogive. Representing cumulative frequency data on a graph is the most efficient way to understand the data and derive results.
There are two types of Cumulative Frequency Curves (or Ogives) :
  • More than type Cumulative Frequency Curve
  • Less than type Cumulative Frequency Curve

More Than Type Cumulative Frequency Curve

Here we use the lower limit of the classes to plot the curve.
How to plot a More than type Ogive:
  1. In the graph, put the lower limit on the x-axis
  2. Mark the cumulative frequency on the y-axis.
  3. Plot the points (x,y) using lower limits (x) and their corresponding Cumulative frequency (y)
  4. Join the points by a smooth freehand curve. It looks like an upside down S.
click the link for ogive  cumulative frequency curve or ogive

Less Than Type Cumulative Frequency Curve

Here we use the upper limit of the classes to plot the curve.
How to plot a Less than type Ogive:
  1. In the graph, put the upper limit on the x-axis
  2. Mark the cumulative frequency on the y-axis.
  3. Plot the points (x,y) using upper limits (x) and their corresponding Cumulative frequency (y)
  4. Join the points by a smooth freehand curve. It looks like an elongated S

CLICK THE LINK  TO FIND MEDIAN FROM AN OGIVE


Now quickly  answer the following questions :-
 Q1. Construction of a cumulative frequency table is useful determining the
        1.   mean
        2.   mode
        3.   median
        4.   all of the above

CLASS WORK 
Q3 .  Draw a less than type and more than type ogive for the given table and also find the median ( from the graph )  ALSO VERIFY  THE RESULT BY USING MEDIAN  FORMUL
ClassesFrequencies
0-1012
10-2016
20-3017
30-4013
40-5011
50-6019













EXERCISE 14.4 (HOME WORK)

1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Daily income in Rupees100-120120-140140-160160-180180-200
Number of workers12148610

SOLUTION 1

 Convert the given distribution table to a less than type cumulative frequency distribution, and we get

Daily incomeFrequencyCumulative Frequency
Less than 1201212
Less than 1401426
Less than 160834
Less than 180640
Less than 2001050

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

ncert solutions for class 10 maths chapter 14 fig 15

2. During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight in kgNumber of students
Less than 380
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235



Draw a less than type ogive for the given data.

Hence obtain the median weight from the graph and verify the result by using the formula.



Solution 2: 

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale.

Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.
ncert solutions for class 10 maths chapter 14 fig 16

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis.

The intersection point perpendicular to x-axis is the median of the given data.



MEDIAN = 46.5





Now, to find the MEDIAN  by making a table.
Class intervalNumber of students(Frequency)Cumulative Frequency
Less than 3800
Less than 403-0=33
Less than 425-3=28
Less than 449-5=49
Less than 4614-9=514
Less than 4828-14=1428
Less than 5032-28=432
Less than 5235-22=335

Now the cumulative frequency just greater than Ncert Solutions Cbse Class 10 Mathematics Chapter - Statistics is 28 belonging to class interval 46 -  48

Median class = 46 - 48

Lower class limit (l) of median class = 46

Frequency (f) of median class = 14

Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Statistics

So median of this data is 46.5
Hence, value of median is verified.




3. The following tables gives production yield per hectare of wheat of 100 farms of a village.
Production Yield50-5555-6060-6565-7070-7575-80
Number of farms2812243816
Change the distribution to a more than type distribution and draw its ogive.

Solution:

 Converting the given distribution to a more than type distribution, we get
Production Yield (kg/ha)Number of farms
More than or equal to 50100
More than or equal to 55100-2 = 98
More than or equal to 6098-8= 90
More than or equal to 6590-12=78
More than or equal to 7078-24=54
More than or equal to 7554-38 =16

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.

ncert solutions for class 10 maths chapter 14 fig 18



AQAD

QUESTION
  The abscissa of the point of intersection of the ‘less than type’ and of the ‘more than type’ cumulative frequency curve of grouped data gives
      1.   mode
      2.   median

      3.   mean      4.   all of the above









    That's all for today  I know that this is quite a lot at once  but try to cover this in the double period. Tomorrow we will do more questions on this topic and it will become clearer.
please ask me if you have doubts in the topic.




!



   Good Morning and Thank you boys !  😊

Th

Comments

  1. Good morning ma'am
    Sargun Sethi X-B

    ReplyDelete
  2. Good morning ma'am
    Keshav gogia
    X-B

    ReplyDelete
  3. Good morning mam
    Shane Stephen
    10-B

    ReplyDelete
  4. Good morning ma’am
    Mathew Joe
    X-B

    ReplyDelete
  5. Good morning ma'am
    yashvardhan chaudhary
    XB

    ReplyDelete
  6. Good morning Ma'am
    Aaron De Menezes
    10-B

    ReplyDelete
  7. Good morning ma'am
    Sharad dubey
    10b

    ReplyDelete
  8. Is there a link for attendance in maths?

    ReplyDelete
  9. Good morning mam
    Singh Yashvardhan
    10B

    ReplyDelete
  10. Good morning ma'am
    Sanyam S Sahoo
    10B 🙏

    ReplyDelete
  11. GOOD MORNING MAM
    ANUGRAH SINGH
    AQAD QUESTION ANS- 2) MEDIAN

    ReplyDelete
  12. Utilise your double period and finish this topic, tomorrow's blog will give you more questions based on this chapter.
    Monday you will have a class test on Statistics...............
    today's maths attendance plz mark in the comments.
    thanks

    ReplyDelete
  13. Good Morning ma'am
    Sudhakar Simon Paul
    10-B

    ReplyDelete
  14. good morning ma'am
    krrish wadhawan 10-B

    ReplyDelete
  15. Good morning ma'am
    NAMAN MEHTA 10-B

    ReplyDelete
  16. Good Morning Ma'am
    Aniruddha Majumdar
    10 B

    ReplyDelete
  17. Good morning maam
    ANSH Aggarwal
    10-B

    ReplyDelete
  18. Good morning mam
    BOAZ LEPCHA
    10 B

    ReplyDelete
  19. Thank you ma'am 😊.Harsh Salodia.

    ReplyDelete
  20. Good morning maam
    Aniket Sharma
    10-B

    ReplyDelete
  21. thank you ma'am . -Karnaditya S. Solanki

    ReplyDelete
  22. Good morning mam.
    - Mohit Gogia 10-B

    ReplyDelete
  23. good morning maam
    hyam michael
    10 B

    ReplyDelete
  24. Maam I have a doubt regarding the previous blog in class work Q1:
    In the question it is shown that n=30, so n/2= 15
    therefore median class being 10-20(because 15 is b/w 10 & 20)
    but x is not involved anywhere in the equation
    How do we do this?
    1. maam in the blog there was an eg. showing that median class depends on whether the n/2 lies b/w its Lower limit and Upper limit.
    2. But in one video whose link you sent it showed that if the cf of the median class is bigger and closest to n/2 then we have to choose THAT median class
    maam which one is right? 1 or 2?

    ReplyDelete
    Replies
    1. please tell maam, i am very confused

      Delete
    2. This comment has been removed by the author.

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    3. This comment has been removed by the author.

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    4. joe , plz ignore the total frequency 30 , i have given you that question without total frequency ,then someone asked me without that he is not able to solve the qn.

      As per the formula 30+[ (21 +x)/2 - (5+x)]/ 6 x 10

      3 = 21 + x - 2 (5 + x)/2
      6= 21 + x - 10 -2x
      6= 11-x
      -5 =-x
      x= 5.

      So modal class is 30-40

      Delete
  25. good morning ma'am
    ibrahim farooqui
    10-B

    ReplyDelete

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