NCERT Class 10 Mathematics — Statistics, Exercise 13.2. All 6 questions solved.

This exercise is about the mode of grouped data. With grouped data we cannot see which single value occurs most often; we can only see which class does. That class is the modal class, and the mode is estimated inside it by

$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h$$

where $l$ is the lower limit of the modal class, $h$ the class size, $f_1$ the frequency of the modal class, and $f_0$ and $f_2$ the frequencies of the classes just before and just after it.

Questions 1, 3 and 4 also ask for the mean. That is done by the step-deviation method of Exercise 13.1, $\bar x = a + h\,\dfrac{\sum f_iu_i}{\sum f_i}$ with $u_i = \dfrac{x_i – a}{h}$.

Key insight. The formula pulls the mode towards whichever neighbouring class is bigger. Write the fraction as $\dfrac{f_1 – f_0}{(f_1 – f_0) + (f_1 – f_2)}$: if $f_0 > f_2$ it is less than $\tfrac12$ and the mode lies in the lower half of the modal class; if $f_2 > f_0$ it lies in the upper half. So before calculating anything you already know which half the answer must fall in — a free check that catches a swapped $f_0$ and $f_2$ every time. All six answers below pass it.

Question 1

The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution. Mode. The largest frequency is $23$, so the modal class is $35$–$45$. Its neighbours are $21$ (before) and $14$ (after); since $21 > 14$, the mode should lie in the lower half, below $40$.

$$l = 35,\quad h = 10,\quad f_1 = 23,\quad f_0 = 21,\quad f_2 = 14$$

$$\text{Mode} = 35 + \frac{23 – 21}{46 – 21 – 14} \times 10 = 35 + \frac{2}{11} \times 10 = 35 + 1.82 = 36.82 \approx 36.8$$

Mean. Class marks $10, 20, \ldots, 60$ are $10$ apart, so use step-deviation with $a = 30$ and $h = 10$.

$$\bar x = 30 + 10 \times \frac{43}{80} = 30 + 5.375 = 35.375$$

Interpretation. The mode says that the largest number of patients admitted were about $36.8$ years old; the mean says that the average age of a patient admitted was about $35.4$ years. The mean is a little lower because it is pulled down by the $17$ patients under $25$, while only $5$ are over $55$.

A note on the printed answer. NCERT prints the mean as $35.37$. The exact value is $\tfrac{283}{8} = 35.375$, which would normally round to $35.38$; the book has dropped the last digit rather than rounded it. The working is identical either way.

Question 2

The following data gives the information on the observed lifetimes (in hours) of $225$ electrical components:

Determine the modal lifetimes of the components.

Solution. The frequencies add to $225$, as stated. The largest is $61$, so the modal class is $60$–$80$, with neighbours $52$ and $38$. Since $52 > 38$, expect the mode below the class mark $70$.

$$l = 60,\quad h = 20,\quad f_1 = 61,\quad f_0 = 52,\quad f_2 = 38$$

$$\text{Mode} = 60 + \frac{61 – 52}{122 – 52 – 38} \times 20 = 60 + \frac{9}{32} \times 20 = 60 + 5.625 = 65.625$$

Question 3

The following data gives the distribution of total monthly household expenditure of $200$ families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution. Mode. The frequencies add to $200$. The largest is $40$, so the modal class is $1500$–$2000$, with neighbours $24$ and $33$. This time $f_2 > f_0$, so the mode lies in the upper half, above $1750$.

$$l = 1500,\quad h = 500,\quad f_1 = 40,\quad f_0 = 24,\quad f_2 = 33$$

$$\text{Mode} = 1500 + \frac{40 – 24}{80 – 24 – 33} \times 500 = 1500 + \frac{16}{23} \times 500 = 1500 + 347.83 = 1847.83$$

Mean. The class marks $1250, 1750, \ldots, 4750$ are large and $500$ apart, which is what step-deviation is for: $a = 2750$, $h = 500$.

$$\bar x = 2750 + 500 \times \frac{-35}{200} = 2750 – 87.5 = 2662.5$$

Question 4

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution. Mode. The largest frequency is $10$, for $30$–$35$. Its neighbours are $9$ and $3$; since $9 > 3$ the mode is in the lower half, below $32.5$.

$$l = 30,\quad h = 5,\quad f_1 = 10,\quad f_0 = 9,\quad f_2 = 3$$

$$\text{Mode} = 30 + \frac{10 – 9}{20 – 9 – 3} \times 5 = 30 + \frac{1}{8} \times 5 = 30.625 \approx 30.6$$

Mean. Step-deviation with $a = 32.5$ and $h = 5$. The two empty classes must stay in the table: they keep the last class at $u = 4$, where it belongs.

$$\bar x = 32.5 + 5 \times \frac{-23}{35} = 32.5 – 3.29 = 29.21 \approx 29.2$$

Interpretation. The largest number of states and union territories have about $30.6$ students per teacher, while on average the ratio is about $29.2$.

Question 5

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Solution. The modal class is $4000$–$5000$, with frequency $18$. Its neighbours are $4$ and $9$; since $9 > 4$ the mode is in the upper half, above $4500$.

$$l = 4000,\quad h = 1000,\quad f_1 = 18,\quad f_0 = 4,\quad f_2 = 9$$

$$\text{Mode} = 4000 + \frac{18 – 4}{36 – 4 – 9} \times 1000 = 4000 + \frac{14}{23} \times 1000 = 4000 + 608.7 = 4608.7$$

Question 6

A student noted the number of cars passing through a spot on a road for $100$ periods each of $3$ minutes and summarised it in the table given below. Find the mode of the data:

Solution. The frequencies add to $100$, the number of periods. They rise and fall more than once — $60$–$70$ has $15$, more than either of its neighbours — but the modal class is the one with the largest frequency overall, $40$–$50$ with $20$. Its neighbours are $12$ and $11$, nearly equal, so the mode should be just below the class mark $45$.

$$l = 40,\quad h = 10,\quad f_1 = 20,\quad f_0 = 12,\quad f_2 = 11$$

$$\text{Mode} = 40 + \frac{20 – 12}{40 – 12 – 11} \times 10 = 40 + \frac{8}{17} \times 10 = 40 + 4.71 = 44.71 \approx 44.7$$

Common mistakes

• Swapping $f_0$ and $f_2$. In question 3 that turns $\tfrac{16}{23}$ into $\tfrac{7}{23}$ and moves the mode from $1847.83$ to $1652.17$ — into the wrong half of the modal class. The key-insight check above catches it before it is written down.
• Question 2, the wrong denominator. It is $2f_1 – f_0 – f_2$, not $f_1 – f_0 – f_2$. The second version gives $61 – 52 – 38 = -29$, a negative denominator that would put the mode below the modal class altogether.
• Question 4, dropping the empty classes from the mean table. Leaving out $40$–$45$ and $45$–$50$ because their frequency is $0$ makes $50$–$55$ look like $u = 2$ instead of $u = 4$, and the mean comes out too low.
• Question 5, using the upper limit as $l$. $l$ is the lower limit of the modal class, $4000$. Starting from $5000$ gives $5608.7$, outside the modal class — an impossible mode.
• Question 6, picking a local peak. $60$–$70$ stands out against its neighbours, but the modal class is the one with the greatest frequency in the whole table, $40$–$50$.
• Questions 1 and 4, mixing up the interpretations. The mode describes the most common value — where the largest group lies — and the mean describes the average. Saying “the average patient is $36.8$ years old” attaches the wrong number to the wrong idea.

Practise next

• Exercise 13.3 — the median of grouped data, which uses a cumulative frequency column in place of the modal class.
• Exercise 13.1 — worth revisiting for the choice between the direct, assumed mean and step-deviation methods used for the means on this page.