EXERCISE 14.3

Median of Grouped Data:

To find the median of a grouped data, we need to find the cumulative frequency and n/2
Then we have to find the median class, which is the class of the cumulative frequency near or greater than the value of n/2.
Cumulative Frequency is calculated by adding the frequencies of all the classes preceding the given class.
Then substitute the values in the formula
Median = 1 + (n/2− cf)/f ×h
where l = lower limit of median class
n = no: of observations
cf = cumulative frequency of the class preceding to the median class
f = frequency of the median class
h = size of class
Example : Find the median of the given table

Solution: Let’s find the n/2.
n = 20, ∴ n/2 = (20 )/2 = 10
The median class is 11 − 15 as its cumulative frequency is 13 which is greater than 10.
Median = 1 + (n/2− cf)/f ×h
Median = 11 + (10− 7)/6 ×5
= 13.5
Remark: The empirical relation between the three measures of central tendency is
3 Median = Mode + 2 Mean

EXERCISE 14.3
The following frequency distribution gives the monthly consumption of electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

Class Interval Frequency Cumulative frequency65 − 854485 − 10559105 − 1251322125 − 1452042145 − 1651456165 − 185864185 − 205468
N=68

From the table, it is observed that, n = 68 and hence n/2 =34
Hence, the median class is 125−145 with cumulative frequency = 42
Where, l = 125,   n = 68,  Cf = 22,   f = 20,  h = 20
Median is calculated as follows:
Median = l + [(n )/2−  cf]/f ×h
= 125 + (34 − 22)/20 × 20
=125 + 12 = 137
Therefore, median = 137
To calculate the mode:
Modal class = 125−145,
h =20, f1 = 20 ,f0 = 13, f2 = 14
Mode = l + (f1 − f0)/(2f1 − f0 −  f2 ) ×h
= 125 + [(208 − 13)/(40 − 13  − 14) ]×20
= 125 + 140/13
=125+10.77
=135.77
Therefore, mode = 135.77
Calculate the Mean:

x = a + h (∑fi ui )/(∑fi )
= 135 + 20 ×( 7/68 )
Mean = 137.05
In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5 then, find the value of x & y.

Solution:: Given data, n = 60
Median of the given data = 28.5
Where, n/2 = 30
Median class is 20 – 30 with a cumulative frequency = 25 + x
Lower limit of median class, l = 20,
Cf = 5 + x,
f = 20 & h = 10
Median = l + ((n )/2 − cf)/f ×h
Substitute the values
28.5 = 20 + (30 − 5 − x)/20 × 10
8.5 = (25 − x)/2
17 = 25 − x
Therefore, x = 8
Now, from cumulative frequency, we can identify the value of x + y as follows:
Since,  60 = 5 + 20 + 15 + 5 + x + y
Now, substitute the value of x, to find y
60 = 5 + 20 + 15 + 5 + 8 + y
y = 60 − 53
y = 7
Therefore, the value of x = 8 and y = 7.

3. The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.

Solution:

Given data: n = 100 and n/2 = 50
Median class = 35−45
Then, l = 35, Cf = 45, f = 33 and  h = 5
Median = l + ((n )/2  − cf)/f ×h
= [35 + (50 − 45)/33] × 5
35 + 5/33 × 5
= 35.75
Therefore, the median age = 35.75 years.

4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:

Find the median length of leaves.             

Solution: Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

 

So, the data obtained are:
n = 40 and n/2 = 20
Median class = 144.5  − 153.5
then, l = 144.5,
Cf = 17, f = 12 and  h = 9
Median = l + [(n )/2 − cf ]/f  × h
= 144.5 + (20 − 17)/12 × 9
144.5 + 9/4
= 146.75 mm
Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a life time of 400 neon lamps.

 

Find the median lifetime of a lamp.

Solution:

Data:

n = 400 & n/2= 200
Median class = 3000 – 3500
Therefore, l = 3000, Cf = 130,
f = 86 & h = 500
Median = l + ((n )/2 − cf)/f × h
= 3000 + [(200 − 130)/12 ] × 500
3000 + 35000/86
= 3000 + 406.97
= 3406.97
Therefore, the median life time of the lamps = 3406.97 hours

6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

 

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution:  To calculate median:

Given: n = 100 & n/2 = 50
Median class = 7 − 10
Therefore, l = 7, Cf = 36, f = 40 & h = 3
Median = 7 + [(50 − 36)/40] × 3
= 7 + 42/40
Median=8.05
Calculate the Mode:
Modal class = 7-10,
Where, l = 7, h =3, f1 = 40 ,f0 = 30, f2 = 16
Mode formula:
Mode = l + (f1 −  f0)/(2f1 −  f0 − f2 ) × h
= 7 + [(40  − 30)/(2 × 40 − 30 −   16) ] × 3
= 7 + 30/34
= 7.88
Therefore mode = 7.88
Calculate the Mean:

Mean = x̄ = (∑fi xi )/(∑fi )
= ( 825/100 ) = 8.25
Therefore, mean = 8.25

7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

Solution:

Given: n = 30 and n/2 = 15
Median class = 55 − 60
l = 55, C_f = 13, f = 6 & h = 5
Median = 55 +[ (15  − 13)/6] × 5
= 55 + 10/6
= 55 + 1.666
Median = 56.67
Therefore, the median weight of the students = 56.67