MODE OF A GROUPED DATA
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 =
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.
Ex.15 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
Lifetimes (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Sol. 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 =
=60 + � 20 = = 60 + 5.625 = 65.625
Hence, the modal lifetimes of the components is 65.625 hours.
Ex.16 Given below is the frequency distribution of the heights of players in a school.
Heights, (in cm) | 160-162 | 163-165 | 166-168 | 169-171 | 172-174 |
No. of students | 15 | 118 | 142 | 127 | 18 |
Find the average height of maximum number of students.
Sol. The given series is in inclusive form. We prepare the table in exculsive form, as given below :
Heights |in cm) | 159.5-162.5 | 162.5-165.5 | 165.5-168.5 | 168.5-171.5 | 171.5-174.5 |
No. of students | 15 | 118 | 142 | 127 | 18 |
We have to find the mode of the data.
Here, the class 165.5 -168.5 has maximum frequency, so it is the modal class.
Ex.17 The mode of the following series is 36. Find the missing frequency f in it.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 8 | 10 | f | 16 | 12 | 6 | 7 |
Sol. 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 =
⇒ 36 = 30 + x 10 ⇒ 6
⇒ 120 – 6f = 160 – 10f ⇒ 4f = 40 ⇒ f = 10
Hence, the value of the missing frequency f is 10. ˜
GRAPHICAL REPRESENTATION OF CUMULATIVE FREQUENCY DISTRIBUTION
= CUMULATIVE FREQUENCY POLYGON CURVE ( O GIVE )
Cumulative frequency is of two types and corresponding to these, the ogive is also of two types.
= LESS THAN SERIES = MORE THAN SERIES
= LESS THAN SERIES 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 SERIES 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 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.
Ex.18 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 |
No. of workers | 12 | 14 | 8 | 6 | 10 |
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Sol. From the given table, we prepare a less than type cumulative frequency distribution table, as given below:
Income less than (in Rs.) | 120 | 140 | 160 | 180 | 200 |
Cumulative frequency | 12 | 26 | 34 | 40 | 50 |
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.
Ex.19 The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield (m kg/ha) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |
No. offering | 2 | 3 | 12 | 24 | 33 | 16 |
Change the distribution to more than type distribution and draw its ogive.
Sol. From the given table, we may prepare more than type cumulative frequency distribution table, as given below:
Production more than (in kg/ ha) | 50 | 55 | 60 | 65 | 70 | 75 |
Cumulative frequency | 100 | 98 | 90 | 78 | 54 | 16 |
Now, plot the points (50,100), (55,98), (60,90), (65,78), (70,54) and (75,16).
Join these points by a freehand to get an ogive of 'more than' type.
Ex.20 The annual profits earned by 30 shops of a shopping complex in a locality gives rise to the following distribution:
Profit (in lakhs Rs.) | Ho. of shops (frequency) |
More than or equal to 5 | 30 |
More than or equal to 10 | 28 |
More than or equal to 15 | 16 |
More than or equal to 20 | 14 |
More than or equal to 25 | 10 |
More than or equal to 30 | 7 |
More than or equal to 35 | 3 |
Draw both ogives for the data above. Hence, obtain the median profit.
Sol. 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 | ’Less than’type | ||
Profit more than (Rs. in lakhs) | No ofshops | Profit less than (Rs. in lakhs) | No of shops |
5 | 30 | 10 | 2 |
10 | 28 | 15 | 14 |
15 | 16 | 20 | 11 |
20 | 14 | 25 | 20 |
25 | 10 | 30 | 23 |
30 | 7 | 35 | 27 |
35 | 3 | 40 | 30 |
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.
Ex.21 The following data gives the information on marks of 70 students in a periodical test:
Marks | Less than 10 | Less than 20 | Less than 30 | Less than 40 | Less than 50 |
No. of students | 3 | 11 | 28 | 48 | 70 |
Draw a cumulative frequency curve for the above data and find the median.
Sol. 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
∴ N/2 = 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, OM = 33.
Hence, the median marks is 33.
5 videos|292 docs|59 tests
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1. What is the mode of a grouped data in statistics? |
2. How can we find the mode of a grouped data? |
3. Can there be multiple modes in a grouped data? |
4. Why is it necessary to group data when finding the mode? |
5. How is the mode of a grouped data different from the mode of a raw data? |
5 videos|292 docs|59 tests
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