CBSE Class 10 Mathematics Statistics Worksheet Set 09

SECTION A

Question. Choose and write the correct option in the following questions.
(i) A data set is shown.
Class Interval: 4–6, 6–8, 8–10, 10–12, 12–14, 14–16, 16–18
Frequency: 2, 6, 20, 28, 12, 10, 2
What is the mean and mode of the data shown?
(a) Mean: 11; Mode: 10.33
(b) Mean: 11; Mode: 10.67
(c) Mean: 10; Mode: 10.33
(d) Mean: 10; Mode: 10.67
Answer: (b) Mean: 11; Mode: 10.67

Question. A grouped data is shown below:
Class Interval: 0–15, 15–30, 30–45, 45–60, 60–75, 75–90
Frequency: 2, 26, 32, 42, 28, 30
Which of the following is the most effective measure of central tendency?
(a) Mean because the data has extreme data points.
(b) Mean because the data has no extreme data points.
(c) Median because the data has extreme data points.
(d) Median because the data has no extreme data points.
Answer: (c) Median because the data has extreme data points.

Question. In the formula \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} \) for finding the mean of a grouped data, \( d_i \)'s are deviations from \( a \) of
(a) lower limits of the classes
(b) upper limits of the classes
(c) mid-points of the classes
(d) frequencies of the class marks
Answer: (c) mid-points of the classes

Question. Solve the following questions. (2 x 1 = 2)
(i) The mean of 11 observations is 50. If the mean of first 6 observations is 49 and that of the last six observations is 52, what is the value of 6th observation?
(ii) Find the median of first 10 prime numbers.
Answer: (i) Sum of 11 observations \( = 11 \times 50 = 550 \)
Sum of first 6 observations \( = 6 \times 49 = 294 \)
Sum of last 6 observations \( = 6 \times 52 = 312 \)
Value of 6th observation \( = (294 + 312) - 550 \)
\( \implies \) \( 606 - 550 = 56 \)
(ii) First 10 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Median \( = \frac{11 + 13}{2} = 12 \)

SECTION B

Question. Find the mean of the following distribution:
\( x \): 4, 6, 9, 10, 15
\( f \): 5, 10, 10, 7, 8
Answer: \( \sum f_i = 40 \)
\( \sum f_i x_i = (4 \times 5) + (6 \times 10) + (9 \times 10) + (10 \times 7) + (15 \times 8) \)
\( = 20 + 60 + 90 + 70 + 120 = 360 \)
Mean \( = \frac{360}{40} = 9 \)

Question. If \( x_i \)'s are the mid-points of the class intervals of a grouped data. \( f_i \)'s are the corresponding frequencies and \( \bar{x} \) is the mean, then find \( \sum f_i(x_i - \bar{x}) \).
Answer: \( \sum f_i(x_i - \bar{x}) = \sum f_i x_i - \bar{x} \sum f_i \)
\( = \sum f_i x_i - \left( \frac{\sum f_i x_i}{\sum f_i} \right) \sum f_i = \sum f_i x_i - \sum f_i x_i = 0 \)

Question. If the mean of the following distribution is 27, find the value of \( p \).
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 8, \( p \), 12, 13, 10
Answer: \( p = 7 \)

Question. Following table shows the weight of 12 students:
Weight (in kgs): 67, 70, 72, 73, 75
Number of students: 4, 3, 2, 2, 1
Find the mean weight of the students.
Answer: Total weight \( = (67 \times 4) + (70 \times 3) + (72 \times 2) + (73 \times 2) + (75 \times 1) \)
\( = 268 + 210 + 144 + 146 + 75 = 843 \)
Mean weight \( = \frac{843}{12} = 70.25 \text{ kg} \)

Question. Calculate the median from the following data:
Rent (in Rs.): 1500–2500, 2500–3500, 3500–4500, 4500–5500, 5500–6500, 6500–7500, 7500–8500, 8500–9500
Number of tenants: 8, 10, 15, 25, 40, 20, 15, 7
Answer: Rs. 5,800

Question. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Age (in years): 0–10, 10–20, 20–30, 30–40, 40–50
Number of persons: 5, 25, \( f \), 18, 7
Answer: \( f = 25 \)

Question. Find the median for the following distribution.
Classes: 20–30, 30–40, 40–50, 50–60, 60–70, 70–80, 80–90
Frequency: 10, 8, 12, 24, 6, 25, 15
Answer: 58.33 (approx)

Question. 50 students enter for a school javelin throw competition. The distance (in metres) thrown are recorded below:
Distance (in m): 0–20, 20–40, 40–60, 60–80, 80–100
No. of students: 6, 11, 17, 12, 4
Calculate the median distance by using the formula for median.
Answer: 49.41 m

Question. The following distribution gives the daily income of 50 workers of a factory:
Daily income (in Rs.): 100–120, 120–140, 140–160, 160–180, 180–200
Number of workers: 12, 14, 8, 6, 10
Find the median.
Answer: 138

Question. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table:
Number of seats: 100–104, 104–108, 108–112, 112–116, 116–120
Frequency: 15, 20, 32, 18, 15
Determine the mean number of seats occupied over the flights.
Answer: 110 seats

Question. The marks obtained by 100 students of a class in an examination are given below.
Marks: 0-5, 5-10, 10-15, 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50
No. of students: 2, 5, 6, 8, 10, 25, 20, 18, 4, 2
Find median.
Answer: 29.5

Question. The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below:
Class: 13.8 – 14.0, 14.0 – 14.2, 14.2 – 14.4, 14.4 – 14.6, 14.6 – 14.8, 14.8 – 15.0
Frequency: 2, 4, 5, 71, 48, 20
Find the number of athletes who completed the race in less than 14.6 seconds.
(a) 80
(b) 81
(c) 82
(d) 83
Answer: (c) 82

Question. The mean of 11 numbers is 35. If the mean of first 6 numbers is 32 and that of last 6 numbers is 37, then the 6th number is equal to
(a) 28
(b) 29
(c) 31
(d) 25
Answer: (b) 29

Question. If the mean of the frequency distribution
\( x_i \): 2, 4, 6, 10, \( p + 5 \)
\( f_i \): 3, 2, 3, 1, 2
is 6, then the value of \( p \) is
(a) 7
(b) 8
(c) 9
(d) 4
Answer: (a) 7

Question. Construction of cumulative frequency table is useful in determining the
(a) mean
(b) median
(c) mode
(d) all, mean, mode and median
Answer: (b) median

Question. For the distribution:
Class: 0–5, 5–10, 10–15, 15–20, 20–25
Frequency: 10, 15, 12, 20, 9
The sum of upper limit of the median class and the modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Answer: (d) 35

Question. For the following distribution:
Marks: Below 10, Below 20, Below 30, Below 40, Below 50, Below 60
Number of Students: 3, 12, 27, 57, 75, 80
The modal class is
(a) 40–50
(b) 20–30
(c) 30–40
(d) 50–60
Answer: (c) 30–40

Question. Consider the following distribution,
Marks obtained: More than or equal to 0, More than or equal to 10, More than or equal to 20, More than or equal to 30, More than or equal to 40, More than or equal to 50
Number of Students: 63, 58, 55, 51, 48, 42
The frequency of the class 20–30 is
(a) 3
(b) 4
(c) 48
(d) 51
Answer: (b) 4

Question. The mode of the frequency distribution
Class: 10–15, 15–20, 20–25, 25–30, 30–35
Frequency: 4, 7, 20, 8, 1
is
(a) 22.6
(b) 24.4
(c) 23.4
(d) None of the options
Answer: (a) 22.6

Question. Consider the data:
Class: 65–85, 85–105, 105–125, 125–145, 145–165, 165 – 185, 185 – 205
Frequency: 4, 5, 13, 20, 14, 7, 4
The difference of the upper limit of the median class and the lower limit of the modal class is
(a) 0
(b) 19
(c) 20
(d) 38
Answer: (c) 20

Question. The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below:
Class: 13.8–14, 14–14.2, 14.2–14.4, 14.4–14.6, 14.6–14.8, 14.8 – 15
Frequency: 2, 4, 5, 71, 48, 20
The number of athletes who completed the race in less than 14.6 seconds is
(a) 11
(b) 71
(c) 82
(d) 130
Answer: (c) 82

Question. The middle most observation of a statistical data has value which is called
(a) mean of the data
(b) median of the data
(c) mode of the data
(d) None of the options
Answer: (b) median of the data

Question. Median of the observations 3, 6, 7, 11, 13, 17, 18, 25 is
(a) 11
(b) 13
(c) 4.5
(d) 12
Answer: (d) 12

Question. Consider the following frequency distribution
Class: 0 – 9, 10 – 19, 20 – 29, 30 – 39, 40 – 49
Frequency: 13, 10, 15, 8, 11
The upper limit of the median class is
(a) 29
(b) 19.5
(c) 29.5
(d) 30
Answer: (c) 29.5

Question. For the following distribution:
Class: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25
Frequency: 10, 15, 12, 20, 9
The sum of the lower limits of the median class and modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Answer: (b) 25

Question. In statistics, an outlier is a data point that differs significantly from other observations of a data set. If an outlier is in included in the following data set, which measure(s) of central tendency would change? 12, 15, 22, 44, 44, 48, 50, 51 
(a) only mean
(b) only mean and median
(c) all-mean, median, mode
(d) Cannot be said without knowing the outlier.
Answer: (a) only mean

Question. Find the mean of first 10 composite numbers.
Answer: The first 10 composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18.
Sum \( = 4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 112 \)
Mean \( = \frac{112}{10} = 11.2 \)

Question. The mean monthly salary of the 12 employees of a firm is Rs. 1450. If one more person joins the firm who gets Rs. 1600 per month, what will be the mean monthly salary now?
Answer: Total monthly salary of 12 employees \( = 12 \times 1450 = 17400 \)
Salary of 13 employees \( = 17400 + 1600 = 19000 \)
Mean monthly salary now \( = \frac{19000}{13} \approx Rs. 1461.54 \)

Question. Nine persons went to a hotel for taking their meals. Eight of them spent Rs. 12 each on their meals and the ninth spent Rs. 8 more than the average expenditure of all the nine. What was the total money spent by them? 
Answer: Let the average expenditure of all nine persons be \( x \).
Total expenditure \( = 9x \)
Expenditure of 8 persons \( = 8 \times 12 = 96 \)
Expenditure of 9th person \( = x + 8 \)
According to the question: \( 9x = 96 + (x + 8) \)

\( \implies 8x = 104 \)

\( \implies x = 13 \)
Total money spent \( = 9 \times 13 = Rs. 117 \)

Question. There were 35 students in a hostel. Due to the admission of 7 new students, the expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs. 1. What was the original expenditure of the mess? 
Answer: Let the original average expenditure per head be \( x \).
Original total expenditure \( = 35x \)
New total number of students \( = 35 + 7 = 42 \)
New average expenditure per head \( = x - 1 \)
New total expenditure \( = 42(x - 1) \)
According to the question: \( 42(x - 1) - 35x = 42 \)

\( \implies 42x - 42 - 35x = 42 \)

\( \implies 7x = 84 \)

\( \implies x = 12 \)
Original expenditure of the mess \( = 35 \times 12 = Rs. 420 \)

Question. Find the mode of the following distribution:
Class Interval: 10 – 15, 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40
Frequency: 30, 45, 75, 35, 25, 15
Answer: The modal class is 20 – 25 as it has the maximum frequency (75).
\( l = 20, f_1 = 75, f_0 = 45, f_2 = 35, h = 5 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
Mode \( = 20 + \left( \frac{75 - 45}{2(75) - 45 - 35} \right) \times 5 \)
Mode \( = 20 + \left( \frac{30}{150 - 80} \right) \times 5 \)
Mode \( = 20 + \frac{30}{70} \times 5 = 20 + \frac{15}{7} \approx 20 + 2.14 = 22.14 \)

Question. Complete mean of the grouped data:
Monthly salary: 325.5 – 350.5, 350.5 – 375.5, 375.5 – 400.5, 400.5 – 425.5, 425.5 – 450.5, 450.5 – 475.5, 475.5 – 500.5
No. of persons: 20, 10, 10, 5, 1, 2, 2
Answer: To find the mean, we calculate the class marks (\( x_i \)): 338, 363, 388, 413, 438, 463, 488.
\( \sum f_i = 50 \)
\( \sum f_i x_i = (20 \times 338) + (10 \times 363) + (10 \times 388) + (5 \times 413) + (1 \times 438) + (2 \times 463) + (2 \times 488) \)
\( \sum f_i x_i = 6760 + 3630 + 3880 + 2065 + 438 + 926 + 976 = 18675 \)
Mean salary \( = \frac{18675}{50} = Rs. 373.5 \)

Question. The table below gives the frequency distribution of the number of teachers in Higher Secondary Schools in 20XX in India. Find the average number of teachers per Higher Secondary School in India for 20XX.
Number of Teachers: 6 – 10, 11 – 15, 16 – 20, 21 – 25, 26 – 30, 31 – 35, 36 – 40, 41 – 45, 46 – 50
Number of H.S. Schools: 955, 1067, 1663, 1492, 1220, 1129, 745, 637, 442
Answer: Total number of schools (\( \sum f_i \)) \( = 955 + 1067 + 1663 + 1492 + 1220 + 1129 + 745 + 637 + 442 = 9350 \)
Class marks (\( x_i \)): 8, 13, 18, 23, 28, 33, 38, 43, 48
\( \sum f_i x_i = (955 \times 8) + (1067 \times 13) + (1663 \times 18) + (1492 \times 23) + (1220 \times 28) + (1129 \times 33) + (745 \times 38) + (637 \times 43) + (442 \times 48) \)
\( \sum f_i x_i = 7640 + 13871 + 29934 + 34316 + 34160 + 37257 + 28310 + 27391 + 21216 = 234095 \)
Average number of teachers \( = \frac{234095}{9350} \approx 25.04 \)

Question. Following is the distribution of marks obtained by 60 students in Economics test:
Marks: More than 0, More than 10, More than 20, More than 30, More than 40, More than 50
No. of students: 60, 56, 40, 20, 10, 3
Calculate the arithmetic mean.
Answer: Convert to class intervals:
0–10: \( 60 - 56 = 4 \)
10–20: \( 56 - 40 = 16 \)
20–30: \( 40 - 20 = 20 \)
30–40: \( 20 - 10 = 10 \)
40–50: \( 10 - 3 = 7 \)
50–60: 3
Class marks (\( x_i \)): 5, 15, 25, 35, 45, 55
\( \sum f_i = 60 \)
\( \sum f_i x_i = (4 \times 5) + (16 \times 15) + (20 \times 25) + (10 \times 35) + (7 \times 45) + (3 \times 55) \)
\( \sum f_i x_i = 20 + 240 + 500 + 350 + 315 + 165 = 1590 \)
Arithmetic Mean \( = \frac{1590}{60} = 26.5 \)

Question. Heights of 50 students of class X of a school are recorded and following data is obtained:
Height (in cm): 130 – 135, 135 – 140, 140 – 145, 145 – 150, 150 – 155, 155 – 160
Number of students: 4, 11, 12, 7, 10, 6
Find the median height of the students. 
Answer: Total frequency \( n = 50 \). Cumulative frequencies are: 4, 15, 27, 34, 44, 50.
\( \frac{n}{2} = 25 \). The cumulative frequency just greater than 25 is 27, which belongs to the class 140 – 145.
Median class = 140 – 145.
\( l = 140, f = 12, cf = 15, h = 5 \)
Median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
Median \( = 140 + \left( \frac{25 - 15}{12} \right) \times 5 = 140 + \frac{10 \times 5}{12} = 140 + \frac{50}{12} \approx 140 + 4.17 = 144.17 \text{ cm} \)

Question. The mean of the following frequency table is 50. But the frequencies \( f_1 \) and \( f_2 \) in classes 20 — 40 and 60 — 80 are missing. Find the missing frequencies.
Class interval: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100
Frequency: 17, \( f_1 \), 32, \( f_2 \), 19
Total: 120
Answer: Total frequency: \( 17 + f_1 + 32 + f_2 + 19 = 120 \implies f_1 + f_2 = 52 \) ...(i)
Class marks (\( x_i \)): 10, 30, 50, 70, 90
Mean \( = \frac{\sum f_i x_i}{\sum f_i} = 50 \)

\( \implies 50 = \frac{17(10) + f_1(30) + 32(50) + f_2(70) + 19(90)}{120} \)

\( \implies 6000 = 170 + 30f_1 + 1600 + 70f_2 + 1710 \)

\( \implies 30f_1 + 70f_2 = 2520 \implies 3f_1 + 7f_2 = 252 \) ...(ii)
Multiplying (i) by 3: \( 3f_1 + 3f_2 = 156 \)
Subtracting from (ii): \( 4f_2 = 96 \implies f_2 = 24 \)
From (i): \( f_1 + 24 = 52 \implies f_1 = 28 \)
The missing frequencies are \( f_1 = 28 \) and \( f_2 = 24 \).

Question. Find median for the following data:
Class Interval: 10 – 19, 20 – 29, 30 – 39, 40 – 49, 50 – 59, 60 – 69, 70 – 79
Frequency: 2, 4, 8, 9, 4, 2, 1
Answer: Convert to exclusive classes: 9.5 – 19.5, 19.5 – 29.5, 29.5 – 39.5, 39.5 – 49.5, 49.5 – 59.5, 59.5 – 69.5, 69.5 – 79.5.
Frequencies: 2, 4, 8, 9, 4, 2, 1. Cumulative frequencies: 2, 6, 14, 23, 27, 29, 30.
\( n = 30, \frac{n}{2} = 15 \). The median class is 39.5 – 49.5.
\( l = 39.5, f = 9, cf = 14, h = 10 \)
Median \( = 39.5 + \left( \frac{15 - 14}{9} \right) \times 10 = 39.5 + \frac{10}{9} \approx 39.5 + 1.11 = 40.61 \)

Question. Find the missing frequencies \( f_1, f_2 \) and \( f_3 \) in the following frequency distribution, when it is given that \( f_2 : f_3 = 4 : 3 \), and mean = 50 
Class interval: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100
Frequency: 17, \( f_1 \), \( f_2 \), \( f_3 \), 19
Total: 120
Answer: Total frequency \( = 17 + f_1 + f_2 + f_3 + 19 = 120 \implies f_1 + f_2 + f_3 = 84 \).
Given \( f_2 : f_3 = 4 : 3 \), let \( f_2 = 4k \) and \( f_3 = 3k \).
So, \( f_1 + 4k + 3k = 84 \implies f_1 + 7k = 84 \implies f_1 = 84 - 7k \).
Mean \( = 50 = \frac{17(10) + f_1(30) + f_2(50) + f_3(70) + 19(90)}{120} \)

\( \implies 6000 = 170 + 30(84 - 7k) + 50(4k) + 70(3k) + 1710 \)

\( \implies 6000 = 170 + 2520 - 210k + 200k + 210k + 1710 \)

\( \implies 6000 = 4400 + 200k \implies 200k = 1600 \implies k = 8 \).
Thus, \( f_2 = 4 \times 8 = 32 \), \( f_3 = 3 \times 8 = 24 \), and \( f_1 = 84 - 7(8) = 84 - 56 = 28 \).
The frequencies are \( f_1 = 28, f_2 = 32, f_3 = 24 \).

Question. Find the value of \( f_1 \) from the following data, if its mode is 65:
Class: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100, 100 – 120
Frequency: 6, 8, \( f_1 \), 12, 6, 5
Where frequency 6, 8, \( f_1 \) and 12 are in ascending order. 
Answer: Given mode = 65, which lies in the class 60 – 80.
Modal class = 60 – 80.
\( l = 60, f_1 (\text{modal frequency}) = 12, f_0 = f_1, f_2 = 6, h = 20 \).
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)

\( \implies 65 = 60 + \left( \frac{12 - f_1}{2(12) - f_1 - 6} \right) \times 20 \)

\( \implies 5 = \left( \frac{12 - f_1}{18 - f_1} \right) \times 20 \)

\( \implies 1 = \frac{4(12 - f_1)}{18 - f_1} \)

\( \implies 18 - f_1 = 48 - 4f_1 \)

\( \implies 3f_1 = 30 \implies f_1 = 10 \).

Question. An incomplete distribution is given below:
Variable: 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70, 70 – 80
Frequency: 12, 30, \( x \), 65, \( y \), 25, 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) Total frequency \( = 12 + 30 + x + 65 + y + 25 + 18 = 230 \implies x + y = 80 \).
Median \( = 46 \), lies in class 40 – 50. Modal class = 40 – 50.
\( l = 40, f = 65, h = 10, cf = 12 + 30 + x = 42 + x, n/2 = 115 \).
\( 46 = 40 + \left( \frac{115 - (42 + x)}{65} \right) \times 10 \)

\( \implies 6 = \frac{73 - x}{6.5} \implies 39 = 73 - x \implies x = 34 \).
From \( x + y = 80 \), \( y = 80 - 34 = 46 \).
Frequencies are \( x = 34 \) and \( y = 46 \).
(ii) Calculate A.M. using class marks (\( x_i \)): 15, 25, 35, 45, 55, 65, 75.
\( \sum f_i x_i = (12 \times 15) + (30 \times 25) + (34 \times 35) + (65 \times 45) + (46 \times 55) + (25 \times 65) + (18 \times 75) \)
\( \sum f_i x_i = 180 + 750 + 1190 + 2925 + 2530 + 1625 + 1350 = 10550 \)
A.M. \( = \frac{10550}{230} \approx 45.87 \).

ASSERTION AND REASON QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct answer out of the following choices.
(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.

Question. Assertion (A): Consider the following frequency distribution
Class Interval: 0–4, 4–8, 8–12, 12–16, 16–20
Frequency: 6, 3, 5, 20, 10
The median class is 12–16.
Reason (R): Let \( n = \sum f_i \). Then, the class whose cumulative frequency is just lesser than \( \frac{n}{2} \) is the median.
Answer: (c) A is true but R is false.

Question. Assertion (A): The arithmetic mean of the following frequency distribution is 25.
Class Interval: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 5, 18, 15, 16, 6
Reason (R): Mean \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \), where \( x_i = \frac{1}{2} (\text{lower limit} + \text{upper limit}) \) of \( i \)-th class interval and \( f_i \) is its frequency.
Answer: (a) Both A and R are true and R is the correct explanation of A.