Median of Grouped Data and the Empirical Relation, Class 10

The median class#

Work out $\displaystyle \frac{n}{2}$. The median class is the first class whose cumu­la­tive fre­quency is greater than or equal to $\displaystyle \frac{n}{2}$. The mid­dle obser­va­tion lies in this class.

The for­mula#

$$\displaystyle \text{Median} = l + \left( \dfrac{\frac{n}{2} - cf}{f} \right) \times h$$

where

• $l$ = lower limit of the median class,
• $n$ = total of all fre­quen­cies,
• $cf$ = cumu­la­tive fre­quency of the class before the median class,
• $f$ = fre­quency of the median class,
• $h$ = class size.

Why the for­mula works#

Up to the lower limit $l$ there are already $cf$ obser­va­tions. To reach the mid­dle, we need $\displaystyle \frac{n}{2} - cf$ more obser­va­tions, and these must come from the median class. The median class has $f$ obser­va­tions spread over a width $h$. Assum­ing they are spread evenly, each obser­va­tion takes up a width of $\displaystyle \frac{h}{f}$, so $\displaystyle \frac{n}{2} - cf$ obser­va­tions take up

$$\displaystyle \left(\dfrac{n}{2} - cf\right) \times \dfrac{h}{f}$$

Adding this to $l$ gives the for­mula. The "evenly spread" assump­tion is why the grouped median is an esti­mate.

Method in steps#

1. If the classes are inclu­sive (such as 11–15, 16–20), make them con­tin­u­ous: find half the gap between one upper limit and the next lower limit, sub­tract it from every lower limit and add it to every upper limit.
2. Make a cumu­la­tive fre­quency col­umn. Check that the last entry is $n$.
3. Work out $\displaystyle \frac{n}{2}$ and find the first class with $\displaystyle cf \ge \frac{n}{2}$. This is the median class.
4. Read $l$, $f$, $h$, and $cf$ of the class before it.
5. Sub­sti­tute and sim­plify. Check that the answer lies inside the median class.

Worked exam­ples#

Exam­ple 1: a stan­dard table#

Find the median of the marks of 50 stu­dents (our own data).

$n = 50$, so $\displaystyle \frac{n}{2} = 25$. The first $cf$ that reaches 25 is 38, so the median class is 40–50. Then $l = 40$, $cf = 24$, $f = 14$, $h = 10$.

$$\displaystyle \begin{aligned} \text{Median} &= 40 + \dfrac{25 - 24}{14} \times 10 \\ &= 40 + \dfrac{10}{14} \approx 40 + 0.71 = 40.71 \end{aligned}$$

So about half the stu­dents scored below 40.71 marks and half scored above.

Exam­ple 2: inclu­sive classes#

Find the median age of 50 peo­ple at a camp (our own data).

There is a gap between 15 and 16. Half the gap is $\displaystyle \frac{16 - 15}{2} = 0.5$, so sub­tract 0.5 from each lower limit and add 0.5 to each upper limit.

$\displaystyle \frac{n}{2} = 25$, so the median class is 20.5–25.5, with $l = 20.5$, $cf = 16$, $f = 15$, $h = 5$.

$$\displaystyle \text{Median} = 20.5 + \dfrac{25 - 16}{15} \times 5 = 20.5 + \dfrac{45}{15} = 20.5 + 3 = 23.5$$

The median age is 23.5 years. Using 21 instead of 20.5 as $l$ would have given 24, a wrong answer.

Exam­ple 3: two miss­ing fre­quen­cies#

The dis­tri­b­u­tion below has 60 obser­va­tions and its median is 25. Find $x$ and $y$.

The median 25 lies in 20–30, so that is the median class: $l = 20$, $f = 18$, $h = 10$, $\displaystyle \frac{n}{2} = 30$. The cumu­la­tive fre­quency before it is $cf = 4 + x$.

$$\displaystyle \begin{aligned} 20 + \dfrac{30 - (4 + x)}{18} \times 10 &= 25 \\ \dfrac{(26 - x) \times 10}{18} &= 5 \\ 26 - x &= 9 \\ x &= 17 \end{aligned}$$

From the total: $4 + 17 + 18 + y + 9 + 5 = 60$, so $53 + y = 60$ and $y = 7$.

Check: the cumu­la­tive fre­quen­cies are 4, 21, 39, 46, 55, 60. The first to reach 30 is 39, in 20–30, as assumed, and $\displaystyle 20 + \frac{30 - 21}{18} \times 10 = 20 + 5 = 25$. So $x = 17$, $y = 7$.

The empir­i­cal rela­tion­ship#

The rela­tion#

For many dis­tri­b­u­tions that are only mod­er­ately lop­sided, the three aver­ages are linked approx­i­mately by

$$3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$$

It is called empir­i­cal because it comes from obser­va­tion of real data, not from a proof. It lets you esti­mate one aver­age when the other two are known. In an exam, use it only when the ques­tion asks you to, or when only two aver­ages are given.

Exam­ple 4: using the rela­tion#

For a dis­tri­b­u­tion, the median is 26 and the mean is 27.5. Esti­mate the mode.

$$\begin{aligned} \text{Mode} &= 3\,\text{Median} - 2\,\text{Mean} \\ &= 3(26) - 2(27.5) = 78 - 55 = 23 \end{aligned}$$

The mode is about 23.

Exam­ple 5: how close is it?#

For the marks data of Exam­ple 1, the mean is 40.4 and the mode (by the mode for­mula) is about 42.22. We found the median to be about 40.71. Check the rela­tion.

$$\begin{aligned} 3\,\text{Median} &\approx 3 \times 40.71 = 122.14 \\ \text{Mode} + 2\,\text{Mean} &\approx 42.22 + 80.8 = 123.02 \end{aligned}$$

The two sides dif­fer by less than 1, so the rela­tion holds approx­i­mately. If we had used it to esti­mate the mode, we would have got $122.14 - 80.8 = 41.34$, against the for­mula value of 42.22. Good enough as an esti­mate, but not a replace­ment for the mode for­mula.

Choos­ing the right aver­age#

• Mean: when every value mat­ters and there are no extreme val­ues, for exam­ple the aver­age marks of a class or aver­age rain­fall.
• Median: when a few very large or very small val­ues would pull the mean away from the typ­i­cal value, for exam­ple incomes, or when the last class is open-ended.
• Mode: when you want the most com­mon or most pop­u­lar value, for exam­ple the most sold shoe size.

Com­mon mis­takes#

• Using the cumu­la­tive fre­quency of the median class for $cf$. It must be the cumu­la­tive fre­quency of the class before it.
• Choos­ing the class with the high­est fre­quency as the median class. That is the modal class; the median class comes from $\displaystyle \frac{n}{2}$.
• For­get­ting to con­vert inclu­sive classes, so $l$ is wrong by 0.5.
• Using $\displaystyle \frac{n+1}{2}$. For grouped data, the for­mula uses $\displaystyle \frac{n}{2}$.
• Writ­ing the empir­i­cal rela­tion as $3\,\text{Mean} = \text{Mode} + 2\,\text{Median}$. The median car­ries the 3.
• Not check­ing that the last cumu­la­tive fre­quency equals $n$.

Try these#

1. Find the median: classes 0 to 10, 10–20, 20–30, 30–40, 40–50 with fre­quen­cies 5, 8, 12, 10, 5. Answer: about 25.83
2. Find the median daily wage: ₹400–450, 450–500, 500–550, 550–600, 600–650 with 5, 12, 18, 10, 5 work­ers. Answer: about ₹522.22
3. Find the median: classes 1 to 10, 11–20, 21–30, 31–40 with fre­quen­cies 4, 9, 5, 2. Answer: about 17.17
4. The mean of a dis­tri­b­u­tion is 34 and its median is 32. Esti­mate the mode. Answer: 28

Key terms#

Median

The mid­dle value when data is arranged in order; half the obser­va­tions lie on each side of it.

Cumu­la­tive fre­quency

The run­ning total of fre­quen­cies up to and includ­ing a class.

Median class

The first class whose cumu­la­tive fre­quency is at least $\displaystyle \frac{n}{2}$.

Inclu­sive classes

Classes such as 11–15, 16–20 where both lim­its are included and there are gaps between classes.

Con­tin­u­ous classes

Classes such as 10.5–15.5, 15.5–20.5 with no gaps, where the upper limit of one is the lower limit of the next.

Class bound­aries

The lim­its of the con­tin­u­ous classes obtained from inclu­sive classes.

Empir­i­cal rela­tion­ship

The approx­i­mate rule $3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$, based on obser­va­tion.

Com­mon ques­tions#

Why do we use n/2 and not (n + 1)/2 for grouped data?#

For grouped data we treat the obser­va­tions as spread con­tin­u­ously through each class, so the point that splits the total in half is at $\displaystyle \frac{n}{2}$. The $\displaystyle \frac{n+1}{2}$ rule is for list­ing indi­vid­ual ungrouped val­ues.

What if n/2 equals a cumu­la­tive fre­quency exactly?#

Then that class is the median class, and the for­mula gives its upper limit as the median. For exam­ple, if $cf = 25$ for 30–40 and $\displaystyle \frac{n}{2} = 25$, the median is 40.

Do I have to con­vert inclu­sive classes for the median?#

Yes. The lower limit $l$ must be a true bound­ary, so con­vert before read­ing $l$. The class width $h$ is unchanged.

Is the empir­i­cal rela­tion exact?#

No. It is an approx­i­ma­tion that works well for mod­er­ately lop­sided data, as Exam­ple 5 shows. Use it only when asked or when two aver­ages are given.

Which aver­age is best when the last class is "60 and above"?#

The median, because it does not need the class mark of the open class, whereas the mean does.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.