Question 1

The class mark of the class interval $15$–$35$ is:

Accepted Answer

**Answer: (a)** $25$ Class mark $=\dfrac{	ext{lower limit}+	ext{upper limit}}{2}=\dfrac{15+35}{2}=25.$

Question 2

For a distribution, the median is $150$ and the mean is $148$. Using the empirical relationship, the mode is:

Accepted Answer

**Answer: (a)** $154$ $	ext{Mode}=3\,	ext{Median}-2\,	ext{Mean}=3(150)-2(148)=450-296=154.$

Question 3

Which measure of central tendency for grouped data can be obtained graphically as the $x$-coordinate of the point of intersection of the two ogives?

Accepted Answer

**Answer: (a)** Median The $x$-coordinate of the point where the 'less than' and 'more than' ogives intersect gives the **median** of the data.

Question 4

For the distribution below, the sum of the lower limits of the median class and the modal class is: | Class | 0–5 | 5–10 | 10–15 | 15–20 | 20–25 | |---|---|---|---|---|---| | Frequency | 10 | 15 | 12 | 20 | 9 |

Accepted Answer

**Answer: (a)** $25$ $N=66,\ \dfrac{N}{2}=33.$ Cumulative frequencies: $10,25,37,57,66.$ The c.f. first reaches $33$ in class $10$–$15$, so the **median class** is $10$–$15$ (lower limit $10$). The highest frequency $20$ is in $15$–$20$, so the **modal class** is $15$–$20$ (lower limit $15$). Sum $=10+15=25.$

Question 5

**Assertion (A):** The median of a grouped frequency distribution can be found graphically. **Reason (R):** The $x$-coordinate of the point of intersection of the 'less than' and 'more than' ogives gives the median.

Accepted Answer

**Answer: (a)** Both A and R are true and R is the correct explanation of A. The median can indeed be found graphically, precisely by the ogive-intersection method described in R, which is why R correctly explains A.

Question 6

The mean of $5$ observations is $18$. If one observation is removed, the mean of the remaining observations becomes $16$. Find the value of the removed observation.

Accepted Answer

Sum of $5$ observations $=5	imes 18=90.$ Sum of remaining $4$ observations $=4	imes 16=64.$ Removed observation $=90-64=26.$

Question 7

For the following distribution, write the median class and the modal class. | Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | |---|---|---|---|---|---| | Frequency | 5 | 8 | 15 | 7 | 5 |

Accepted Answer

$N=40,\ \dfrac{N}{2}=20.$ Cumulative frequencies: $5,13,28,35,40.$ The c.f. first reaches $20$ in class $20$ – $30$ , so the median class is $20$ – $30$ .

The highest frequency is $15$ (class $20$ – $30$ ), so the modal class is also $20$ – $30$ .

Question 8

Find the mean daily wage from the following data. | Daily wage (Rs) | 100–120 | 120–140 | 140–160 | 160–180 | 180–200 | |---|---|---|---|---|---| | No. of workers | 12 | 14 | 8 | 6 | 10 |

Accepted Answer

Class marks $x_i$: $110,130,150,170,190.$ $\sum f_ix_i=110(12)+130(14)+150(8)+170(6)+190(10)$ $=1320+1820+1200+1020+1900=7260.$ $N=\sum f_i=50.$ Mean $=\dfrac{7260}{50}=	ext{Rs }145.20.$

Question 9

Find the mode of the following distribution of the ages of patients admitted to a hospital. | Age (years) | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | |---|---|---|---|---|---| | No. of patients | 8 | 16 | 36 | 34 | 6 |

Accepted Answer

The highest frequency is $36$, so the modal class is $20$–$30$. Here $l=20,\ f_1=36,\ f_0=16,\ f_2=34,\ h=10.$ $	ext{Mode}=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}	imes h=20+\dfrac{36-16}{72-16-34}	imes 10=20+\dfrac{20}{22}	imes 10=20+9.09=29.09\ 	ext{years}.$

Question 10

The median of the following distribution is $28.5$ and the total frequency is $60$. Find the missing frequencies $x$ and $y$. | Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | |---|---|---|---|---|---|---| | Frequency | 5 | $x$ | 20 | 15 | $y$ | 5 |

Accepted Answer

Total: $5+x+20+15+y+5=60\Rightarrow x+y=15.$

$\dfrac{N}{2}=30$ ; since the median $28.5$ lies in $20$ – $30$ , that is the median class ( $l=20,\ f=20,\ h=10$ , $cf=5+x$ ).

$\text{Median}=l+\dfrac{\frac{N}{2}-cf}{f}\times h\Rightarrow 28.5=20+\dfrac{30-(5+x)}{20}\times 10.$

$8.5=\dfrac{25-x}{2}\Rightarrow 25-x=17\Rightarrow x=8.$ Then $y=15-8=7.$

$x=8,\ y=7.$

Question 11

For the following frequency distribution, find the mean and the median. Hence, using the empirical relationship, estimate the mode. | Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | |---|---|---|---|---|---| | Frequency | 5 | 8 | 15 | 16 | 6 |

Accepted Answer

**Mean:** class marks $5,15,25,35,45.$ $\sum f_ix_i=25+120+375+560+270=1350,\ N=50.$ Mean $=\dfrac{1350}{50}=27.$ **Median:** $\dfrac{N}{2}=25.$ Cumulative frequencies: $5,13,28,44,50.$ Median class $=20$–$30$ ($l=20,\ cf=13,\ f=15,\ h=10$). $	ext{Median}=20+\dfrac{25-13}{15}	imes 10=20+\dfrac{12}{15}	imes 10=20+8=28.$ **Mode (empirical):** $	ext{Mode}=3\,	ext{Median}-2\,	ext{Mean}=3(28)-2(27)=84-54=30.$

Question 12

The mean of the following distribution is $62.8$ and the sum of all frequencies is $50$. Find the missing frequencies $f_1$ and $f_2$. | Class | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 | 100–120 | |---|---|---|---|---|---|---| | Frequency | 5 | $f_1$ | 10 | $f_2$ | 7 | 8 |

Accepted Answer

Total: $5+f_1+10+f_2+7+8=50\Rightarrow f_1+f_2=20.$ …(1)

Class marks: $10,30,50,70,90,110.$

$\sum f_ix_i=5(10)+f_1(30)+10(50)+f_2(70)+7(90)+8(110)=2060+30f_1+70f_2.$

Mean $=\dfrac{\sum f_ix_i}{50}=62.8\Rightarrow 2060+30f_1+70f_2=3140\Rightarrow 3f_1+7f_2=108.$ …(2)

From (1), $f_1=20-f_2.$ Substituting: $3(20-f_2)+7f_2=108\Rightarrow 60+4f_2=108\Rightarrow f_2=12.$ Then $f_1=8.$

$f_1=8,\ f_2=12.$

Question 13

A company tested the life (in hours) of $200$ electric bulbs. The results are shown below.

Life (hours)	300–400	400–500	500–600	600–700	700–800
No. of bulbs	14	56	60	40	30

(i) Write the modal class.

(ii) Write the median class.

(iii) Find the median life of the bulbs.

Accepted Answer

**(i)** Highest frequency is $60$ (class $500$–$600$), so the **modal class is $500$–$600$**. **(ii)** $N=200,\ \dfrac{N}{2}=100.$ Cumulative frequencies: $14,70,130,170,200.$ The c.f. first reaches $100$ in $500$–$600$, so the **median class is $500$–$600$**. **(iii)** $l=500,\ cf=70,\ f=60,\ h=100.$ $	ext{Median}=500+\dfrac{100-70}{60}	imes 100=500+\dfrac{30}{60}	imes 100=500+50=550\ 	ext{hours}.$

Question 14

Are these Statistics important questions free?

Accepted Answer

Yes. All 13 CBSE Class 10 Maths important questions for Statistics are free, with full model answers and no login required.

Question 15

Do these Statistics questions follow the latest CBSE syllabus?

Accepted Answer

Yes — they are aligned to the NCERT 2026–27 syllabus for CBSE Class 10 Maths, so nothing here is outside the current course.

Question 16

How should I practise the Statistics important questions?

Accepted Answer

Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.

Question 17

What types of questions are covered for Statistics?

Accepted Answer

A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the CBSE paper is covered.

Question type	Count	Marks
MCQ	4	1
Assertion–Reason	1	1
Very Short	2	2
Short Answer	3	3
Long Answer	2	5
Case-based	1	4

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