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CBSE Important Questions Class 10 Maths Chapter 13 – Statistics

Class 10 Maths Chapter 13 Statistics is part of the CBSE Class 10 Maths syllabus. It explains the three measures of central tendency: mean, median and mode. Students learn to find the mean of grouped data by the direct method, the assumed mean method and the step deviation method, and to find the median and mode of grouped data using their formulae.

These CBSE important questions for Class 10 Maths Chapter 13 cover finding the mean, median and mode of a raw data set, the range of a data set, the daily average sale, finding the upper and lower limits of a class from its mid value, and the median of an even number of observations. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. For the dataset 4, 8, 4, 7, 5, 4, 9, 14, 6, 6, 4, 13, 8, 11, 6, 6, 5, calculate the mean, median and mode. Which of these three has the greatest value?

Answer: Mean: Sum = 4+8+4+7+5+4+9+14+6+6+4+13+8+11+6+6+5 = 120
Number of observations = 17
Mean = 120/17 ≈ 7.06
Median: Arranging in order: 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 8, 8, 9, 11, 13, 14
Since n = 17 (odd), the median is the 9th value = 6
Mode: The value 4 occurs 4 times and 6 also occurs 4 times, so the data is bimodal with modes 4 and 6.
Comparing all three, the mean (7.06) has the greatest value.

Q2. The heights of 15 students are 165, 155, 168, 160, 163, 162, 165, 168, 156, 159, 160, 164, 163, 165, 160. Find (i) the range, (ii) the mode, (iii) the median.

Answer: Arranging in ascending order:
155, 156, 159, 160, 160, 160, 162, 163, 163, 164, 165, 165, 165, 168, 168
(i) Range = highest − lowest = 168 − 155 = 13
(ii) Mode: Both 160 and 165 occur 3 times, which is the highest frequency. So the data has two modes: 160 and 165.
(iii) Median: n = 15 (odd), so the median is the 8th value = 163

Q3. If the daily sale of kerosene (in litres) in a ration shop for six days is 75, 120, 12, 50, 70.5, 140.5, what is the daily average sale?
(a) 75 litres
(b) 76 litres
(c) 77 litres
(d) 78 litres

Answer: (d) 78 litres
Sum = 75 + 120 + 12 + 50 + 70.5 + 140.5 = 468
Number of days = 6
Average = 468/6 = 78 litres

Q4. The mid value of a class interval is 42. If the class size is 10, then the upper and lower limits of the class are
(a) 47.5 and 37.5
(b) 47 and 37
(c) 37 and 47
(d) 37.5 and 47.5

Answer: (b) 47 and 37
Half the class size = 10/2 = 5.
Lower limit = mid value − 5 = 42 − 5 = 37
Upper limit = mid value + 5 = 42 + 5 = 47
So the class is 37 to 47.

Q5. The mid value of a class interval is 48. If the class size is 10, then the upper and lower limits of the class are
(a) 53 and 43
(b) 43 and 53
(c) 43.5 and 53.5
(d) 53.5 and 43.5

Answer: (a) 53 and 43
Half the class size = 5.
Lower limit = 48 − 5 = 43
Upper limit = 48 + 5 = 53

Q6. The weights (in kg) of 11 students are 38, 45, 41, 32, 37, 35, 38, 40, 36, 46 and 39. What is the median?
(a) 38
(b) 39
(c) 37
(d) 40

Answer: (a) 38
Arranging in ascending order:
32, 35, 36, 37, 38, 38, 39, 40, 41, 45, 46
Here n = 11, which is odd, so the median is the 6th value.
Median = 38 kg

Q7. The mode of 1, 2, 3, 4, 0, 0, 2, 2, 3, 2 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer: (c) 2
The mode is the value that occurs most often.
Counting: 0 appears twice, 1 once, 2 appears four times, 3 twice, 4 once.
So the mode is 2.

Q8. The temperatures in degrees Celsius over a three-week period are 17, 20, 16, 24, 27, 16, 18, 17, 21, 24, 19, 21, 22, 19, 18, 17, 14, 12, 13, 19, 23. Compute the median.

Answer: Arranging in ascending order:
12, 13, 14, 16, 16, 17, 17, 17, 18, 18, 19, 19, 19, 20, 21, 21, 22, 23, 24, 24, 27
The total number of observations is 21, which is odd.
Median = the (21 + 1)/2 = 11th value
Counting to the 11th value, the median = 19°C

Q9. Find the median of 51, 26, 53, 18, 74, 45, 21, 52, 60 and 41.

Answer: Arranging in ascending order:
18, 21, 26, 41, 45, 51, 52, 53, 60, 74
Here n = 10, which is even. For an even number of observations the median is the average of the two middle values.
Median = (5th value + 6th value) / 2
= (45 + 51) / 2
= 96/2
= 48

Q10. The runs scored in a cricket match by 11 players are 17, 125, 54, 108, 92, 25, 35, 16, 18, 54, 51. Find the mean, mode and median. Are the three the same?

Answer: Arranging in ascending order:
16, 17, 18, 25, 35, 51, 54, 54, 92, 108, 125
Mean: Sum = 595, so mean = 595/11 = 54.09 (approximately)
Median: n = 11 (odd), so the median is the 6th value = 51
Mode: The value 54 appears twice, more than any other, so the mode = 54
No, the three are not the same. They are 54.09, 51 and 54 respectively. The three measures of central tendency usually differ unless the data is perfectly symmetrical.

Q11. Write the formulae for the mean, median and mode of grouped data.

Answer: Mean (direct method): x̄ = Σfᵢxᵢ / Σfᵢ
Mean (step deviation method): x̄ = a + h × (Σfᵢuᵢ / Σfᵢ), where a is the assumed mean and h the class size
Median = l + [(n/2 − cf) / f] × h
where l = lower limit of the median class, n = total frequency, cf = cumulative frequency of the class before the median class, f = frequency of the median class, h = class size
Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
where l = lower limit of the modal class, f₁ = frequency of the modal class, f₀ = frequency of the class before it, f₂ = frequency of the class after it

Q12. Write the empirical relationship between the mean, median and mode. If the mean and median of a data set are 27 and 25 respectively, find the mode.

Answer: The empirical relationship between the three measures of central tendency is:
3 Median = Mode + 2 Mean
This can also be written as Mode = 3 Median − 2 Mean.
Calculation: Given Mean = 27 and Median = 25.
Mode = 3(25) − 2(27)
= 75 − 54
= 21

Related Study Material on Extramarks

Q.1 Water in a canal, 30 dm. wide and 12 dm. deep is flowing with a velocity of 10 km/hr. How much area will it irrigate in 30 minutes, if 8 cm of standing water is required for irrigation?

Marks:3
Ans

The width of the canal = 30 dm
= 3 m [10 dm = 1m]
Depth of the canal = 12 dm
= 1.2 m
Length of flowing water 1/2 hour = 5 km
= 5000 m
Volume of the water flowing in 1/2 hour = 5000 — 3 — 1.2
Let x m² area is irrigated in 1/2 hr. Then,
x — (8/100) = 5000 — 3 — 1.2
x = 1800000/8
= 225000 m²
Hence, the canal irrigates 225000 m² area in 1/2 hour.

Q.2 A toy is in the form of a cone mounted on a hemisphere of diameter 7 cm. The total height of the toy is 14.5 cm. Find the volume and total surface area of the toy.

Marks:4
Ans

$\begin{array}{l} Total height of the toy = 14 .5 cm \\ Diameter of the hemisphere = 7 cm \\ Radius of the hemisphere = \frac{7}{2} = 3 .5 cm \\ Height of the conical portion = 14 .5 âˆ’ 3 .5 = 11 cm \\ Volume of solid = \frac{2}{3} r^{3} + \frac{1}{3} r^{2} h \\ = \frac{1}{3} r^{2} (2 r + h) \\ = \frac{1}{3} — \frac{22}{7} — {(\frac{7}{2})}^{2} (2 — \frac{7}{2} + 11) \\ = 231 {cm}^{3} \\ Total surface area of the toy \\ = (\begin{array}{l} Surface area of a conical part + Surface \\ area of a hemispherical part \end{array}) \\ = (rl + 2 r^{2}) \\ = r (l + 2 r) \\ = \frac{22}{7} — \frac{7}{2} — (\sqrt{h^{2} + r^{2}} + 2 r) {cm}^{2} \\ = \frac{22}{7} — \frac{7}{2} — (\sqrt{{(11)}^{2} + {(\frac{7}{2})}^{2}} + 2 (\frac{7}{2})) {cm}^{2} \\ = 203 .94 {cm}^{2} \end{array}$

Q.3 The slant height of the frustum of a cone is 4 cm and the circumferences of its circular ends are 18 cm and 6cm.Find the curved surface area of the frustum.

Marks:4
Ans

$\begin{array}{l} Given, \\ Slant height of the frustum (l) = 4 cm \\ Let radii of the top and bottom of the frustum \\ of the cone be r_{1} and r_{2} . \\ By the question circumference of the top (2 r_{1}) \\ and bottom (2 r_{2}) of the frustum and 6 cm and \\ 18 cm respectively . \\ â‡’ r_{1} = \frac{6}{2} cm and r_{2} = \frac{18}{2} cm \\ Since, \\ Curved surface area of the frustum \\ = (r_{1} + r_{2}) — l \\ = (\frac{6}{2} + \frac{18}{2}) — 4 {cm}^{2} \\ = (3 + 9) — 4 {cm}^{2} \\ = 48 {cm}^{2} \end{array}$

Q.4 A circus tent is cylindrical upto a height of 3 m and conical above it. If the diameter of the base is 105 m and the vertical height of the conical part is 7.26 m, find the total canvas used in making the tent.

Marks:5
Ans

$\begin{array}{l} Total canvas used = Total curved surface area of tent \\ Total curved surface area of the tent \\ = Curved surface area of the cylindrical portion + Curved surface area of the conical portion \\ Radius of the cylindrical portion = \frac{D}{2} = \frac{105}{2} = 52 .5 m \\ Height of the cylindrical portion = 3 m \\ Curved surface â€‹â€‹ area of the cylindrical portion = 2 rH = 2 — \frac{22}{7} — 52 .5 — 3 m^{2} \\ For the conical part : \\ l = slant height = \sqrt{h^{2} + r^{2}} = \sqrt{{(7 .26)}^{2} + {(52 .5)}^{2}} = 52 .99 53 m \\ Curved surface area of the conical part = rl = \frac{22}{7} — 52 .5 — 53 \\ Total curved surface area = 2 — 22 — 7 .5 — 3 + 22 — 7 .5 — 53 \\ = 22 — 7 .5 — (6 + 53) \\ = 22 — 7 .5 — 59 = 9735 m^{2} \end{array}$

Q.5 Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.

Marks:2
Ans

$\begin{array}{l} Let the edge of the cube be x cm \\ Radius of the Sphere inside the cube = \frac{x}{2} \\ Volume of the cube = x^{3} \\ Volume of the sphere = \frac{4}{3} r^{3} = \frac{4}{3} — \frac{22}{7} — \frac{x^{3}}{8} \\ Ratio of the volume = \frac{x^{3}}{\frac{11}{21} x^{3}} = \frac{21}{11} = 21 : 11 \end{array}$

CBSE Important Questions Class 10 Maths Chapter 13 – Statistics

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