CBSE Class 10 Mathematics Statistics Worksheet Set 10

Question. The mean of \( n \) terms is \( \bar{x} \). If first term is increased by 1, second term by 2 and so on, then new mean is
(a) \( \bar{x} + \frac{n+1}{2} \)
(b) \( \bar{x} + \frac{n-1}{2} \)
(c) \( \bar{x} + \frac{n}{2} \)
(d) \( \bar{x} + n \)
Answer: (a) \( \bar{x} + \frac{n+1}{2} \)

Question. The mean of a set of numbers is \( \bar{x} \). If each number is multiplied by \( k \), then the mean of the new set is
(a) \( \bar{x} + k \)
(b) \( \frac{\bar{x}}{k} \)
(c) \( \bar{x} - k \)
(d) \( k \cdot \bar{x} \)
Answer: (d) \( k \cdot \bar{x} \)

Question. Which of the following is most affected by extreme observations?
(a) mode
(b) median
(c) mean
(d) None of the options
Answer: (c) mean

Question. The mean and median of 100 items are 50 and 52 respectively. The value of the largest item is 100. It was later found that it is 110 not 100. The true mean and median are
(a) 50.10, 52
(b) 50, 52
(c) 50.20, 52
(d) None of the options
Answer: (a) 50.10, 52

Question. The class marks of a frequency distribution are 6, 10, 14, 18, 22, 26, 30. Find class size.
Answer: Class size \( = 10 - 6 = 4 \).

Question. Write the frequency distribution table for the following data:
Marks: Below 10, Below 20, Below 30, Below 40, Below 50, Below 60
No. of students: 0, 15, 20, 30, 35, 40
Answer:
Marks | No. of students
0 – 10 | 0
10 – 20 | 15
20 – 30 | 5
30 – 40 | 10
40 – 50 | 5
50 – 60 | 5

Question. For the following data, write the frequency distribution table:
Marks: Below 10, Below 20, Below 30, Below 40, Below 50, Below 60
No. of students: 0, 15, 20, 30, 35, 40
Answer:
The frequency distribution table is as follows:
Marks | No. of students
0 – 10 | 0
10 – 20 | 15
20 – 30 | 5
30 – 40 | 10
40 – 50 | 5
50 – 60 | 5

Question. Calculate mode of the following data:
Marks Obtained: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100
No. of students: 8, 10, 12, 6, 3
Answer:
Here, the maximum frequency is 12, which corresponds to the modal class 40 – 60.
\( l = 40, f_1 = 12, f_0 = 10, f_2 = 6, h = 20 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 40 + \left( \frac{12 - 10}{2(12) - 10 - 6} \right) \times 20 \)
\( = 40 + \left( \frac{2}{24 - 16} \right) \times 20 \)
\( = 40 + \frac{2}{8} \times 20 \)
\( = 40 + 5 = 45 \)

Question. The mean score of 25 observations is 80 and the mean score of another 55 observations is 65. Determine the mean score of the whole set of observations.
Answer:
Sum of first 25 observations \( = 25 \times 80 = 2000 \)
Sum of another 55 observations \( = 55 \times 65 = 3575 \)
Total sum of 80 observations \( = 2000 + 3575 = 5575 \)
Mean score of the whole set \( = \frac{5575}{80} = 69.6875 \)

Question. Arun scored 36 marks in English, 44 marks in Hindi, 75 marks in Mathematics and \( x \) marks in Science. If he has scored an average of 50 marks, find the value of \( x \).
Answer:
Average marks \( = \frac{36 + 44 + 75 + x}{4} = 50 \)
\( \implies 155 + x = 200 \)

\( \implies x = 200 - 155 = 45 \)
Therefore, marks in Science \( x = 45 \).

Question. Find the median wage of a worker engaged at a construction site whose data are given below:
Wages (in Rs.): 3500, 3800, 4100, 4500, 5500, 6500, 7000
Number of workers: 12, 13, 25, 17, 15, 12, 6
Answer:
Wages (in Rs.) | No. of workers (\( f \)) | Cumulative Frequency (\( cf \))
3500 | 12 | 12
3800 | 13 | 25
4100 | 25 | 50
4500 | 17 | 67
5500 | 15 | 82
6500 | 12 | 94
7000 | 6 | 100
Total \( n = 100 \). Since \( n \) is even, the median is the average of the \( \frac{n}{2} \)-th and \( (\frac{n}{2} + 1) \)-th observations.
50th observation \( = 4100 \)
51st observation \( = 4500 \)
Median wage \( = \frac{4100 + 4500}{2} = Rs. 4300 \)

Question. If the mean of the following data is 12, find \( p \).
\( x \): 4, 8, \( p \), 16, 20
\( y \): 5, 3, 12, 5, 4
Answer:
Mean \( \bar{x} = \frac{\sum x_i y_i}{\sum y_i} = 12 \)
\( \frac{(4 \times 5) + (8 \times 3) + (p \times 12) + (16 \times 5) + (20 \times 4)}{5 + 3 + 12 + 5 + 4} = 12 \)
\( \frac{20 + 24 + 12p + 80 + 80}{29} = 12 \)
\( 204 + 12p = 348 \)

\( \implies 12p = 144 \)

\( \implies p = 12 \)

Question. The mean of a set of 67 values is 35. If each of the 67 values is divided by 4, what will be the mean of new set of values?
Answer:
The mean of the new set will also be divided by 4.
New mean \( = \frac{35}{4} = 8.75 \)

Question. The average monthly salary of 15 workers in a factory is Rs. 285. If the salary of the manager is included, the average becomes Rs. 355. What is the manager’s salary?
Answer:
Total salary of 15 workers \( = 15 \times 285 = Rs. 4275 \)
Total salary of 16 people (including manager) \( = 16 \times 355 = Rs. 5680 \)
Manager's salary \( = 5680 - 4275 = Rs. 1405 \)

Question. 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:
Number of cars: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70, 70 – 80
Frequency: 7, 14, 13, 12, 20, 11, 15, 8
Answer:
Max frequency is 20, which belongs to modal class 40 – 50.
\( l = 40, f_1 = 20, f_0 = 12, f_2 = 11, h = 10 \)
Mode \( = 40 + \left( \frac{20 - 12}{2(20) - 12 - 11} \right) \times 10 \)
\( = 40 + \left( \frac{8}{40 - 23} \right) \times 10 \)
\( = 40 + \frac{80}{17} \approx 40 + 4.71 = 44.71 \)

Question. If the mean of the following distribution is 50, find the value of \( f_1 \):
Class: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100
Frequency: 17, 28, 32, \( f_1 \), 19
Answer:
Mid-values (\( x_i \)): 10, 30, 50, 70, 90
Mean \( = \frac{\sum f_i x_i}{\sum f_i} = 50 \)
\( \frac{(17 \times 10) + (28 \times 30) + (32 \times 50) + (f_1 \times 70) + (19 \times 90)}{17 + 28 + 32 + f_1 + 19} = 50 \)
\( \frac{170 + 840 + 1600 + 70f_1 + 1710}{96 + f_1} = 50 \)
\( 4320 + 70f_1 = 50(96 + f_1) \)
\( 4320 + 70f_1 = 4800 + 50f_1 \)

\( \implies 20f_1 = 480 \)

\( \implies f_1 = 24 \)

Question. Compute mean of the grouped data:
Income (in Rupees): 200 – 300, 300 – 400, 400 – 500, 500 – 600, 600 – 700, 700 – 800, 800 – 900
No. of workers: 5, 36, 24, 16, 9, 6, 4
Answer:
Income (\( x_i \)) | \( f_i \) | \( f_i x_i \)
250 | 5 | 1250
350 | 36 | 12600
450 | 24 | 10800
550 | 16 | 8800
650 | 9 | 5850
750 | 6 | 4500
850 | 4 | 3400
Total | 100 | 47200
Mean \( = \frac{47200}{100} = Rs. 472 \)

Question. Given below is the expenditure (in rupees) on water consumption of 70 houses in a locality. Find the average expenditure (in rupees) per house.
Expenditure on water (in rupees): 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40, 40 – 45, 45 – 50, 50 – 55, 55 – 60, 60 – 65, 65 – 70
Number of houses: 7, 5, 7, 8, 9, 11, 7, 5, 4, 4, 3
Answer:
Mid-values (\( x_i \)): 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5, 57.5, 62.5, 67.5
\( \sum f_i x_i = (7 \times 17.5) + (5 \times 22.5) + (7 \times 27.5) + (8 \times 32.5) + (9 \times 37.5) + (11 \times 42.5) + (7 \times 47.5) + (5 \times 52.5) + (4 \times 57.5) + (4 \times 62.5) + (3 \times 67.5) \)
\( = 122.5 + 112.5 + 192.5 + 260 + 337.5 + 467.5 + 332.5 + 262.5 + 230 + 250 + 202.5 = 2770 \)
Total houses \( \sum f_i = 70 \)
Average expenditure \( = \frac{2770}{70} \approx Rs. 39.57 \)

Question. Find the average marks of students from the following table:
Marks: Above 0, Above 10, Above 20, Above 30, Above 40, Above 50, Above 60, Above 70, Above 80, Above 90, Above 100
No. of students: 80, 77, 72, 65, 55, 43, 23, 16, 10, 8, 0

Answer: We first convert the given "above" distribution into class intervals:
Class | Frequency (\( f \)) | Class Mark (\( x \)) | \( f \times x \)
0 – 10 | 3 | 5 | 15
10 – 20 | 5 | 15 | 75
20 – 30 | 7 | 25 | 175
30 – 40 | 10 | 35 | 350
40 – 50 | 12 | 45 | 540
50 – 60 | 20 | 55 | 1100
60 – 70 | 7 | 65 | 455
70 – 80 | 6 | 75 | 450
80 – 90 | 2 | 85 | 170
90 – 100 | 8 | 95 | 760
Total | 80 | | 4090
Average marks \( = \frac{\sum fx}{\sum f} = \frac{4090}{80} = 51.125 \)

Question. The mean monthly wages of a group of 15 workers in a factory was Rs. 3351. 4 workers whose mean monthly wage was Rs. 3310 left the factory and a new worker at a monthly wage of Rs. 3025 was appointed. Find the mean wage of the remaining group.

Answer: Total monthly wages of 15 workers \( = 15 \times 3351 = \text{Rs. } 50265 \)
Total wages of 4 workers who left \( = 4 \times 3310 = \text{Rs. } 13240 \)
Total wages of remaining 11 workers \( = 50265 - 13240 = \text{Rs. } 37025 \)
Wages of the new group of 12 workers \( = 37025 + 3025 = \text{Rs. } 40050 \)
Mean wage of the group of 12 workers \( = \frac{40050}{12} = \text{Rs. } 3337.5 \)

Question. The following table shows marks secured by 140 students in an examination:
Marks | No. of students
0 — 10 | 20
10 — 20 | 24
20 — 30 | 40
30 — 40 | 36
40 — 50 | 20
Calculate mean marks by using all the three methods, i.e. direct method, assumed mean method and step deviation method.

Answer: By Direct Method:
Class marks (\( x_i \)): 5, 15, 25, 35, 45
\( \sum f_i x_i = (20 \times 5) + (24 \times 15) + (40 \times 25) + (36 \times 35) + (20 \times 45) \)
\( = 100 + 360 + 1000 + 1260 + 900 = 3620 \)
Mean \( = \frac{3620}{140} = 25.86 \text{ marks} \)

Question. In the following frequency distribution, the frequency of the class-interval (40—50) is missing. It is known that the mean of the distribution is 52. Find the missing frequency.
Wages (in Rs.) | No. of Workers
10 — 20 | 5
20 — 30 | 3
30 — 40 | 4
40 — 50 | \( f \)
50 — 60 | 2
60 — 70 | 6
70 — 80 | 13

Answer: Total sum of \( f_i x_i = (5 \times 15) + (3 \times 25) + (4 \times 35) + (f \times 45) + (2 \times 55) + (6 \times 65) + (13 \times 75) \)
\( = 75 + 75 + 140 + 45f + 110 + 390 + 975 = 1765 + 45f \)
Total frequency \( \sum f_i = 33 + f \)
Mean \( = \frac{1765 + 45f}{33 + f} = 52 \)

\( \implies 1765 + 45f = 1716 + 52f \)

\( \implies 7f = 49 \)

\( \implies f = 7 \)

Question. Calculate the median for the following data:
Class interval | Frequency
130 — 140 | 5
140 — 150 | 9
150 — 160 | 17
160 — 170 | 28
170 — 180 | 24
180 — 190 | 10
190 — 200 | 7

Answer: Cumulative Frequencies are 5, 14, 31, 59, 83, 93, 100.
\( n = 100, \frac{n}{2} = 50 \)
Median class = 160 – 170
\( l = 160, cf = 31, f = 28, h = 10 \)
Median \( = 160 + \frac{50 - 31}{28} \times 10 = 160 + 6.785 = 166.785 \)

Question. Calculate the median for the following data:
Marks obtained | Number of students
Less than 20 | 0
Less than 30 | 4
Less than 40 | 16
Less than 50 | 30
Less than 60 | 46
Less than 70 | 66
Less than 80 | 82
Less than 90 | 92
Less than 100 | 100

Answer: \( n = 100, \frac{n}{2} = 50 \).
The 50th observation lies in the "Less than 70" group (class 60 – 70).
\( l = 60, cf = 46, f = 20, h = 10 \)
Median \( = 60 + \frac{50 - 46}{20} \times 10 = 60 + 2 = 62 \)

Question. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heart beats per minute | Number of women
65 — 68 | 2
68 — 71 | 4
71 — 74 | 3
74 — 77 | 8
77 — 80 | 7
80 — 83 | 4
83 — 86 | 2

Answer: Class Marks \( x_i \): 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5.
\( \sum f_i x_i = (2 \times 66.5) + (4 \times 69.5) + (3 \times 72.5) + (8 \times 75.5) + (7 \times 78.5) + (4 \times 81.5) + (2 \times 84.5) \)
\( = 133 + 278 + 217.5 + 604 + 549.5 + 326 + 169 = 2277 \)
Mean \( = \frac{2277}{30} = 75.9 \text{ beats per minute} \)

Question. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes | Number of boxes
50 — 52 | 15
53 — 55 | 110
56 — 58 | 135
59 — 61 | 115
62 — 64 | 25
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer: Since class intervals are not continuous, we first make them continuous: 49.5 – 52.5, 52.5 – 55.5, etc.
Class Marks \( x_i \): 51, 54, 57, 60, 63
Total boxes \( \sum f_i = 400 \)
\( \sum f_i x_i = (15 \times 51) + (110 \times 54) + (135 \times 57) + (115 \times 60) + (25 \times 63) \)
\( = 765 + 5940 + 7695 + 6900 + 1575 = 22875 \)
Mean \( = \frac{22875}{400} = 57.1875 \). Assumed mean method is suitable here.

Question. To find out the concentration of \( SO_2 \) in the air (in parts per million, i.e. ppm), the data was collected for 30 localities in a certain city and is presented below:
Concentration of \( SO_2 \) (in ppm) | Frequency
0.00 — 0.04 | 4
0.04 — 0.08 | 9
0.08 — 0.12 | 9
0.12 — 0.16 | 2
0.16 — 0.20 | 4
0.20 — 0.24 | 2
Find the mean concentration of \( SO_2 \) in the air.

Answer: Class Marks \( x_i \): 0.02, 0.06, 0.10, 0.14, 0.18, 0.22.
\( \sum f_i x_i = (4 \times 0.02) + (9 \times 0.06) + (9 \times 0.10) + (2 \times 0.14) + (4 \times 0.18) + (2 \times 0.22) \)
\( = 0.08 + 0.54 + 0.90 + 0.28 + 0.72 + 0.44 = 2.96 \)
Mean \( = \frac{2.96}{30} \approx 0.099 \text{ ppm} \)

Question. Find the mean, mode and median for the following data:
Marks | Frequency
0 — 10 | 8
10 — 20 | 16
20 — 30 | 36
30 — 40 | 34
40 — 50 | 6
Total | 100

Answer:
Mean \( = \frac{(8 \times 5) + (16 \times 15) + (36 \times 25) + (34 \times 35) + (6 \times 45)}{100} = \frac{40 + 240 + 900 + 1190 + 270}{100} = 26.4 \)
Median: \( n = 100, \frac{n}{2} = 50 \). Median class = 20 – 30. \( l = 20, cf = 24, f = 36, h = 10 \)
Median \( = 20 + \frac{50 - 24}{36} \times 10 = 20 + 7.22 = 27.22 \)
Mode: Modal class = 20 – 30. \( l = 20, f_1 = 36, f_0 = 16, f_2 = 34, h = 10 \)
Mode \( = 20 + \frac{36 - 16}{2(36) - 16 - 34} \times 10 = 20 + \frac{20}{22} \times 10 = 29.09 \)

Question. If the mean of the following frequency distribution is 65.6, find the missing frequencies (\( f_1, f_2 \)):
Class | Frequency
10 — 30 | 5
30 — 50 | 8
50 — 70 | \( f_1 \)
70 — 90 | 20
90 — 110 | \( f_2 \)
110 — 130 | 2
Total | 50

Answer: \( \sum f = 5 + 8 + f_1 + 20 + f_2 + 2 = 50 \implies f_1 + f_2 = 15 \)
Mean \( = \frac{(5 \times 20) + (8 \times 40) + (f_1 \times 60) + (20 \times 80) + (f_2 \times 100) + (2 \times 120)}{50} = 65.6 \)

\( \implies 100 + 320 + 60f_1 + 1600 + 100f_2 + 240 = 3280 \)

\( \implies 60f_1 + 100f_2 = 1020 \)
Solving the two equations: \( f_1 = 12, f_2 = 3 \)

Question. If the mean of the following frequency distribution is 91, find the missing frequencies (\( f_1, f_2 \)):
Class | Frequency
0 — 30 | 12
30 — 60 | 21
60 — 90 | \( f_1 \)
90 — 120 | 52
120 — 150 | \( f_2 \)
150 — 180 | 11
Total | 150

Answer: \( f_1 + f_2 = 150 - (12 + 21 + 52 + 11) = 54 \)
Mean \( = \frac{(12 \times 15) + (21 \times 45) + (f_1 \times 75) + (52 \times 105) + (f_2 \times 135) + (11 \times 165)}{150} = 91 \)

\( \implies 180 + 945 + 75f_1 + 5460 + 135f_2 + 1815 = 13650 \)

\( \implies 75f_1 + 135f_2 = 5250 \)
Solving the two equations: \( f_1 = 34, f_2 = 20 \)

Question. The median of the following data is 32.5.
Class interval | Frequency
0 — 10 | \( x \)
10 — 20 | 5
20 — 30 | 9
30 — 40 | 12
40 — 50 | \( y \)
50 — 60 | 3
60 — 70 | 2
Total | 40
Find the values of \( x \) and \( y \).

Answer: \( \sum f = x + y + 31 = 40 \implies x + y = 9 \)
Median class = 30 – 40. \( l = 30, cf = x + 14, f = 12, h = 10, \frac{n}{2} = 20 \)
Median \( = 30 + \frac{20 - (x + 14)}{12} \times 10 = 32.5 \)

\( \implies \frac{6 - x}{12} \times 10 = 2.5 \)

\( \implies 6 - x = 3 \implies x = 3 \)
Then, \( y = 9 - 3 = 6 \)

Question. An incomplete distribution is given below:
Variable | Frequency
10 — 20 | 12
20 — 30 | 30
30 — 40 | \( x \)
40 — 50 | 65
50 — 60 | \( y \)
60 — 70 | 25
70 — 80 | 18
You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula and fill up missing frequencies.
(ii) Calculate the A.M. of the completed distribution.

Answer: (i) \( x + y = 230 - 150 = 80 \).
Median class = 40 – 50. \( l = 40, cf = x + 42, f = 65, h = 10, \frac{n}{2} = 115 \)
\( 46 = 40 + \frac{115 - (x + 42)}{65} \times 10 \implies 6 = \frac{73 - x}{6.5} \implies x = 34 \)
Then, \( y = 80 - 34 = 46 \)
(ii) Using the calculated frequencies, A.M. \( = \frac{\sum f_i x_i}{N} = \frac{(12 \times 15) + (30 \times 25) + (34 \times 35) + (65 \times 45) + (46 \times 55) + (25 \times 65) + (18 \times 75)}{230} = \frac{10550}{230} = 45.87 \)

Question. The median of the following data is 525. Find the values of \( x \) and \( y \), if the total frequency is 100.
Class interval | Frequency
0 — 100 | 2
100 — 200 | 5
200 — 300 | \( x \)
300 — 400 | 12
400 — 500 | 17
500 — 600 | 20
600 — 700 | \( y \)
700 — 800 | 9
800 — 900 | 7
900 — 1000 | 4

Answer: \( \sum f = x + y + 76 = 100 \implies x + y = 24 \)
Median class = 500 – 600. \( l = 500, cf = x + 36, f = 20, h = 100, \frac{n}{2} = 50 \)
\( 525 = 500 + \frac{50 - (x + 36)}{20} \times 100 \implies 25 = 5(14 - x) \implies x = 9 \)
Then, \( y = 24 - 9 = 15 \)

ASSERTION AND REASON QUESTIONS

Question. Assertion (A): If the median and mode of a frequency distribution are 8.9 and 9.2 respectively, then its mean is 9.
Reason (R): Mean, median and mode of a frequency distribution are related as: mode = 3 median – 2 mean.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (d) A is false but R is true.

Question. Assertion (A): If for a certain frequency distribution, \( l = 24.5, h = 4, f_0 = 14, f_1 = 14, f_2 = 15 \), then the value of mode is 25.
Reason (R): Mode of a frequency distribution is given by: mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: (d) A is false but R is true.

CASE-BASED QUESTIONS

Electricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption).
A survey is conducted for 56 families of a Colony A. The following tables gives the weekly consumption of electricity of these families.
Weekly consumption (in units) | No. of families
0-10 | 16
10-20 | 12
20-30 | 18
30-40 | 6
40-50 | 4
50-60 | 0
The similar survey is conducted for 80 families of Colony B and the data is recorded as below:
Weekly consumption (in units) | No. of families
0-10 | 0
10-20 | 5
20-30 | 10
30-40 | 20
40-50 | 40
50-60 | 5

Question. Refer to data received from Colony B: (i) Find the modal weekly consumption.

Answer: For Colony B, the maximum frequency is 40, which corresponds to the class interval 40 – 50.
Modal class = 40 – 50
\( l = 40, f_1 = 40, f_0 = 20, f_2 = 5, h = 10 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)

\( \implies \text{Mode} = 40 + \left( \frac{40 - 20}{2(40) - 20 - 5} \right) \times 10 \)

\( \implies \text{Mode} = 40 + \frac{20}{55} \times 10 \)

\( \implies \text{Mode} = 40 + 3.64 = 43.64 \text{ units} \)