What is a cumulative frequency table? A cumulative frequency is the running total of frequencies. It tells us "how many observations fall below (or above) a certain value," which is exactly the information the median and ogive formulas need. There are two types, and Class 10 problems use both.
Less-than cumulative frequency. Add frequencies from the lowest class upward and pair each running total with the upper class limit. The "less than 30" value answers "how many observations are below 30?" These totals always increase, and the final value equals the total frequency $N$.
More-than (greater-than) cumulative frequency. Add frequencies from the highest class downward and pair each running total with the lower class limit. The "more than 20" value answers "how many observations are at least 20?" These totals always decrease, starting at $N$ and ending at $0$.
Notice the symmetry: at any boundary the less-than cf and more-than cf add up to $N$. Here $13+22=35$ at the boundary 20, and $25+10=35$ at the boundary 30. That relationship, $(\text{less-than cf})+(\text{more-than cf})=N$, is a useful check.
Why this matters. Cumulative frequency is the bridge between a frequency table and three powerful tools:
- Finding the median class — the first class whose less-than cf is $\ge\dfrac{N}{2}$ — and then the median by $\text{Median}=l+\left(\dfrac{\frac{N}{2}-cf}{f}\right)h$.
- Drawing the ogive (cumulative frequency curve) to read the median graphically.
- Answering direct questions like "how many students scored less than 60?" or "how many earned at least ₹200?"
Recovering a frequency table from cumulative data. Examiners often give the cumulative frequencies and ask you to rebuild the simple frequencies. Just subtract each cf from the next: $f_i = (\text{cf}_i) - (\text{cf}_{i-1})$. For less-than cf $4,10,18,24$ the frequencies are $4,6,8,6$ and $N=24$. The first frequency equals the first cf.
Worked construction. Suppose marks are grouped as $0$–$10\ (f=4)$, $10$–$20\ (f=7)$, $20$–$30\ (f=10)$, $30$–$40\ (f=6)$, $40$–$50\ (f=3)$, with $N=30$. Reading downward, the less-than cf column is $4,11,21,27,30$ against upper limits $10,20,30,40,50$. Reading upward, the more-than cf column is $30,26,19,9,3$ against lower limits $0,10,20,30,40$. To find the median class you scan the less-than column for the first total that reaches $\dfrac{N}{2}=15$: that is $21$ (the class $20$–$30$). Every grouped-statistics question — median, ogive, "how many scored less than $x$" — starts from exactly this table, so building it accurately is the first habit to lock in.
Common mistakes to avoid. (1) Pairing a less-than cf with a lower limit (it must be the upper limit), and a more-than cf with an upper limit (it must be the lower limit). (2) Forgetting that the more-than cf for the lowest class equals $N$, not the first frequency. (3) Using $\dfrac{N+1}{2}$ instead of $\dfrac{N}{2}$ — grouped data uses $\dfrac{N}{2}$ because it is treated as continuous. (4) Stopping the less-than column before it reaches $N$, or the more-than column before it reaches $0$ — both are signs of an arithmetic slip.
Create a less-than cumulative frequency table for the data:
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|-------|------|-------|-------|-------|-------|
| Students | 4 | 7 | 10 | 6 | 3 |
Solution- Step 1: Pair each running total with the upper limit: $10,20,30,40,50$.
- Step 2: Less than 10: $4$.
- Step 3: Less than 20: $4+7=11$.
- Step 4: Less than 30: $11+10=21$.
- Step 5: Less than 40: $21+6=27$; less than 50: $27+3=30=N$.
Answer: $(10,4),(20,11),(30,21),(40,27),(50,30)$.
Using the table in Example 1, estimate how many students scored less than 35 marks.
Solution- Step 1: $35$ lies in the class $30$–$40$.
- Step 2: Students with marks below $30$ = $21$.
- Step 3: Assuming the $6$ students in $30$–$40$ are spread evenly, marks $30$–$35$ is half the class.
- Step 4: Half of $6$ is $3$.
- Step 5: Less than $35\approx21+3=24$.
Answer: About $24$ students scored less than $35$.
Create a more-than cumulative frequency table for the data in Example 1.
Solution- Step 1: Pair each running total with the lower limit: $0,10,20,30,40$.
- Step 2: More than 0: $30$ (all).
- Step 3: More than 10: $30-4=26$.
- Step 4: More than 20: $26-7=19$.
- Step 5: More than 30: $19-10=9$; more than 40: $9-6=3$.
Answer: $(0,30),(10,26),(20,19),(30,9),(40,3)$.
The less-than cumulative frequencies of a distribution are $4,10,18,24$ for classes $0$–$10,10$–$20,20$–$30,30$–$40$. Recover the frequency distribution and state $N$.
Solution- Step 1: First frequency $=$ first cf $=4$.
- Step 2: Second frequency $=10-4=6$.
- Step 3: Third frequency $=18-10=8$.
- Step 4: Fourth frequency $=24-18=6$.
- Step 5: $N=$ last cf $=24$.
Answer: Frequencies $4,6,8,6$ and $N=24$.
From the data in Example 1, find the median class and then the median.
Solution- Step 1: $N=30$, $\dfrac{N}{2}=15$.
- Step 2: Less-than cf: $4,11,21,27,30$.
- Step 3: First $cf\ge15$ is $21$, so median class is $20$–$30$.
- Step 4: $l=20$, $cf=11$, $f=10$, $h=10$.
- Step 5: $\text{Median}=20+\left(\dfrac{15-11}{10}\right)10=20+4=24$.
Answer: Median class $20$–$30$; median $=24$.
A more-than cumulative frequency table reads: more than 0 → 50, more than 10 → 46, more than 20 → 34, more than 30 → 20, more than 40 → 8. Find the frequency of each class.
Solution- Step 1: More-than cf decreases; subtract consecutive values.
- Step 2: Class $0$–$10$: $50-46=4$.
- Step 3: Class $10$–$20$: $46-34=12$.
- Step 4: Class $20$–$30$: $34-20=14$; class $30$–$40$: $20-8=12$.
- Step 5: Class $40$–$50$: $8-0=8$ (the last cf drops to $0$). Total $4+12+14+12+8=50=N$.
Answer: Frequencies $4,12,14,12,8$, with $N=50$.
The daily incomes (in ₹) of 100 workers are tabulated. Find how many earn at least ₹250.
| Income (₹) | 150-200 | 200-250 | 250-300 | 300-350 | 350-400 |
|-------|------|------|------|------|------|
| Workers | 18 | 25 | 30 | 17 | 10 |
Solution- Step 1: "At least ₹250" means income in $250$–$300$, $300$–$350$ or $350$–$400$.
- Step 2: Add those frequencies: $30+17+10$.
- Step 3: $=57$.
- Step 4: As a check, more-than cf for $250$ also equals $N-(18+25)=100-43=57$.
Answer: $57$ workers earn at least ₹250.
For the income data of Example 7, build the less-than cumulative frequency table.
Solution- Step 1: Upper limits: $200,250,300,350,400$.
- Step 2: Less than 200: $18$.
- Step 3: Less than 250: $18+25=43$.
- Step 4: Less than 300: $43+30=73$.
- Step 5: Less than 350: $73+17=90$; less than 400: $90+10=100=N$.
Answer: $(200,18),(250,43),(300,73),(350,90),(400,100)$.
Find the median income for the data in Example 7.
Solution- Step 1: $N=100$, $\dfrac{N}{2}=50$.
- Step 2: Less-than cf: $18,43,73,90,100$.
- Step 3: First $cf\ge50$ is $73$, so median class is $250$–$300$.
- Step 4: $l=250$, $cf=43$, $f=30$, $h=50$.
- Step 5: $\text{Median}=250+\left(\dfrac{50-43}{30}\right)50=250+\dfrac{7}{30}\times50=250+11.67=261.67$.
Answer: Median income $\approx$ ₹$261.67$.
A class of 40 students has the less-than cf table: less than 10 → 3, less than 20 → 12, less than 30 → 27, less than 40 → 35, less than 50 → 40. Find the frequency of the class $20$–$30$.
Solution- Step 1: Frequency of a class $=$ its less-than cf minus the previous one.
- Step 2: cf at less than 30 $=27$; cf at less than 20 $=12$.
- Step 3: Frequency of $20$–$30$ $=27-12$.
- Step 4: $=15$.
Answer: The frequency of class $20$–$30$ is $15$.
Using the table in Example 10, find the median.
Solution- Step 1: $N=40$, $\dfrac{N}{2}=20$.
- Step 2: Less-than cf: $3,12,27,35,40$.
- Step 3: First $cf\ge20$ is $27$, so median class is $20$–$30$.
- Step 4: $l=20$, $cf=12$, $f=27-12=15$, $h=10$.
- Step 5: $\text{Median}=20+\left(\dfrac{20-12}{15}\right)10=20+\dfrac{80}{15}=20+5.33=25.33$.
Answer: Median $\approx 25.33$.
Verify, for the data of Example 1, that the less-than cf and more-than cf at the boundary 30 add up to $N$.
Solution- Step 1: Less-than cf at 30 (from Example 1) $=21$.
- Step 2: More-than cf at 30 (from Example 3) $=9$.
- Step 3: Sum $=21+9=30$.
- Step 4: This equals $N=30$, confirming $(\text{less-than cf})+(\text{more-than cf})=N$ at any boundary.
Answer: $21+9=30=N$ — the relationship holds.