NCERT Class 10 Mathematics — Statistics, Exercise 13.3. All 7 questions solved.

This exercise finds the median of grouped data. Everything runs through a cumulative frequency column. With $n = \sum f_i$ observations, the median class is the class whose cumulative frequency is greater than, and nearest to, $\tfrac n2$. Inside it,

$$\text{Median} = l + \left(\frac{\frac n2 – cf}{f}\right) \times h$$

where $l$ is the lower limit of the median class, $cf$ the cumulative frequency of the class before it, $f$ its own frequency and $h$ the class size.

Questions 1 and 6 also ask for the mean and the mode, done exactly as in Exercises 13.1 and 13.2.

Key insight. The formula is a proportion. Of the $f$ observations in the median class we need the $\left(\tfrac n2 – cf\right)$th, so we go that fraction of the way across the class. That picture explains the two rules students most often break: $cf$ counts only the observations already passed (the class before), and the classes must touch end to end, or “the way across” is measured on a ruler with gaps in it — which is why question 4 needs continuous classes. It also gives a check: the median must always lie inside the median class.

Question 1

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

Solution. One table carries all three calculations: the cumulative frequency for the median, and $u_i$ for the step-deviation mean with $a = 135$, $h = 20$.

Median. $\tfrac n2 = 34$. The cumulative frequency first passes $34$ at $42$, so the median class is $125$–$145$, with $l = 125$, $cf = 22$, $f = 20$, $h = 20$:

$$\text{Median} = 125 + \frac{34 – 22}{20} \times 20 = 125 + 12 = 137$$

Mean.

$$\bar x = 135 + 20 \times \frac{7}{68} = 135 + 2.0588\ldots = 137.0588\ldots$$

Mode. The modal class is also $125$–$145$ ($f_1 = 20$), with $f_0 = 13$ and $f_2 = 14$:

$$\text{Mode} = 125 + \frac{20 – 13}{40 – 13 – 14} \times 20 = 125 + \frac{7}{13} \times 20 = 125 + 10.769\ldots = 135.769\ldots$$

Comparison. The three measures are approximately the same, because the distribution is nearly symmetrical about the $125$–$145$ class.

A note on the printed answer. NCERT prints the mean as $137.05$ and the mode as $135.76$. The exact values are $\tfrac{2330}{17} = 137.0588\ldots$ and $\tfrac{1765}{13} = 135.769\ldots$, which round conventionally to $137.06$ and $135.77$; the book has cut off the digits rather than rounded. Nothing else differs.

Question 2

If the median of the distribution given below is $28.5$, find the values of $x$ and $y$.

Solution. Two unknowns need two equations. The total supplies one; the median supplies the other.

From the total, $45 + x + y = 60$, so

$$x + y = 15 \qquad (1)$$

The median $28.5$ lies in $20$–$30$, so that is the median class: $l = 20$, $h = 10$, $f = 20$, and the cumulative frequency before it is $cf = 5 + x$. With $\tfrac n2 = 30$,

$$28.5 = 20 + \frac{30 – (5 + x)}{20} \times 10 = 20 + \frac{25 – x}{2}$$

$$17 = 25 – x \;\Longrightarrow\; x = 8$$

and from $(1)$, $y = 7$. Check: the cumulative frequencies become $5, 13, 33, 48, 55, 60$; $30$ falls in $20$–$30$, and $20 + \tfrac{30 – 13}{20} \times 10 = 28.5$.

Question 3

A life insurance agent found the following data for distribution of ages of $100$ policy holders. Calculate the median age, if policies are given only to persons having age $18$ years onwards but less than $60$ year.

Solution. This is a less than table: the right-hand column is already cumulative. As in the book’s Example 7, each “below” figure is the upper limit of a class, and each class frequency is the difference of successive entries — for instance $6 – 2 = 4$ people are in $20$–$25$. The question’s condition that policies start at age $18$ tells us the first class is $18$–$20$.

The first class is narrower than the rest, but that does not matter: only the median class’s width enters the formula.

$\tfrac n2 = 50$. The cumulative frequency first passes $50$ at $78$, so the median class is $35$–$40$, with $l = 35$, $cf = 45$, $f = 33$, $h = 5$:

$$\text{Median} = 35 + \frac{50 – 45}{33} \times 5 = 35 + \frac{25}{33} = 35.76$$

Question 4

The lengths of $40$ leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to $117.5$–$126.5$, $126.5$–$135.5$, $\ldots$, $171.5$–$180.5$.)

Solution. A length recorded as $126$ mm could really be anything from $125.5$ to $126.5$, so the printed classes leave gaps — nothing is shown between $126$ and $127$. Following the hint, subtract $0.5$ from every lower limit and add $0.5$ to every upper limit. The classes now touch, and each is $9$ wide.

$\tfrac n2 = 20$, which falls in $144.5$–$153.5$: $l = 144.5$, $cf = 17$, $f = 12$, $h = 9$.

$$\text{Median} = 144.5 + \frac{20 – 17}{12} \times 9 = 144.5 + 2.25 = 146.75$$

Question 5

The following table gives the distribution of the life time of $400$ neon lamps:

Find the median life time of a lamp.

Solution. The frequencies add to $400$. Build the cumulative column:

$\tfrac n2 = 200$, which falls in $3000$–$3500$: $l = 3000$, $cf = 130$, $f = 86$, $h = 500$.

$$\text{Median} = 3000 + \frac{200 – 130}{86} \times 500 = 3000 + \frac{35000}{86} = 3000 + 406.98 = 3406.98$$

Question 6

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

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Solution. The frequencies add to $100$. As in question 1, one table serves all three measures, with $a = 8.5$ and $h = 3$ for the mean.

Median. $\tfrac n2 = 50$, in $7$–$10$: $l = 7$, $cf = 36$, $f = 40$, $h = 3$.

$$\text{Median} = 7 + \frac{50 – 36}{40} \times 3 = 7 + 1.05 = 8.05$$

Mean.

$$\bar x = 8.5 + 3 \times \frac{-6}{100} = 8.5 – 0.18 = 8.32$$

Mode. The modal class is again $7$–$10$, with $f_1 = 40$, $f_0 = 30$, $f_2 = 16$:

$$\text{Mode} = 7 + \frac{40 – 30}{80 – 30 – 16} \times 3 = 7 + \frac{10}{34} \times 3 = 7 + 0.88 = 7.88$$

Question 7

The distribution below gives the weights of $30$ students of a class. Find the median weight of the students.

Solution. The frequencies add to $30$. Cumulative frequencies:

$$2,\ 5,\ 13,\ 19,\ 25,\ 28,\ 30$$

$\tfrac n2 = 15$. The cumulative frequency first passes $15$ at $19$, so the median class is $55$–$60$ — not $50$–$55$, which has the largest frequency but is the modal class. Here $l = 55$, $cf = 13$, $f = 6$, $h = 5$:

$$\text{Median} = 55 + \frac{15 – 13}{6} \times 5 = 55 + \frac{5}{3} = 56.67$$

Common mistakes

• Taking $cf$ from the median class itself. In question 1 the cumulative frequency before the median class is $22$. Using $42$ — the median class’s own running total — gives $125 + \tfrac{34 – 42}{20} \times 20 = 117$, which lies outside the median class and so cannot be the median.
• Question 7, confusing the median class with the modal class. The median class is found from the cumulative frequency and $\tfrac n2$, not from the largest frequency. Here the two are different classes, $55$–$60$ and $50$–$55$.
• Question 4, skipping the continuity correction. With the printed classes a student takes $l = 145$ and, reading $145$–$153$ as $8$ wide, $h = 8$ — giving $147$ instead of $146.75$. The true class runs from $144.5$ to $153.5$ and is $9$ wide.
• Question 3, treating the “below” column as frequencies. Those figures are already cumulative. Adding them up again gives a total far above $100$, which is the giveaway.
• Question 2, writing $cf = 5$. The cumulative frequency before $20$–$30$ includes the unknown class $10$–$20$, so it is $5 + x$. Leaving out $x$ makes the median equation independent of $x$ and the question unsolvable.
• Questions 1 and 6, rounding before the end. Using $\tfrac{7}{13} \approx 0.54$ in question 1’s mode gives $135.8$; carry the fraction through and round once.

Practise next

• Exercise 13.2 — the mode formula used in questions 1 and 6 here, practised on its own.
• Class 11 Statistics, Exercise 13.1 — mean deviation about the median, whose grouped-data questions begin with exactly the median calculation on this page.