CBSE Class 10 Mathematics Statistics Worksheet Set 07

Question. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of plants: 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14
Number of houses: 1, 2, 1, 5, 6, 2, 3
Which method did you use for finding the mean and why?
Answer: Calculation of mean number of plants per house.
Number of plants | Number of houses (\( f_i \)) | Class mark (\( x_i \)) | \( f_i x_i \)
0–2 | 1 | 1 | 1
2–4 | 2 | 3 | 6
4–6 | 1 | 5 | 5
6–8 | 5 | 7 | 35
8–10 | 6 | 9 | 54
10–12 | 2 | 11 | 22
12–14 | 3 | 13 | 39
Total | \( \sum f_i = 20 \) | | \( \sum f_i x_i = 162 \)
Hence, mean (\( \bar{X} \)) = \( \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1 \)
Here, we used direct method to find mean because numerical values of \( x_i \) and \( f_i \) are small.

Question. The following table shows the ages of the patients admitted in a hospital during a year.
Age (in years): 5 – 15, 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65
Number of patients: 6, 11, 21, 23, 14, 5
Find the mode of the data given above.
Answer: In the given data class interval 35 – 45 has maximum frequency.
\( \implies \) 35 – 45 is modal class.
Now, mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
Here, \( l = \) lower limit of modal class = 35
\( h = \) size of the class interval = 10
\( f_1 = \) frequency of the modal class = 23
\( f_0 = \) frequency of the class preceding the modal class = 21
\( f_2 = \) frequency of the class succeeding the modal class = 14
\( \therefore \) Mode = \( 35 + \left( \frac{23 - 21}{2 \times 23 - 21 - 14} \right) \times 10 \)
\( = 35 + \frac{2}{46 - 35} \times 10 = 35 + \frac{2}{11} \times 10 \)
\( = 35 + \frac{20}{11} = 35 + 1.8 = 36.8 \)

Question. The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure (in Rs.): 100–150, 150–200, 200–250, 250–300, 300–350
No. of households: 4, 5, 12, 2, 2
Find the mean daily expenditure on food by a suitable method.
Answer:
Daily expenditure | (\( x_i \)) | No. of houses (\( f_i \)) | \( u_i = \frac{x_i - 225}{50} \) | \( f_i u_i \)
100–150 | 125 | 4 | –2 | –8
150–200 | 175 | 5 | –1 | –5
200–250 | 225 | 12 | 0 | 0
250–300 | 275 | 2 | 1 | 2
300–350 | 325 | 2 | 2 | 4
Total | | \( \sum f_i = 25 \) | | \( \sum f_i u_i = -7 \)
Mean = \( 225 + 50 \times \left( \frac{-7}{25} \right) = 211 \)
Mean expenditure on food is Rs. 211. 

Question. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetime (in hours): 0–20, 20–40, 40–60, 60–80, 80–100, 100–120
Frequency: 10, 35, 52, 61, 38, 29
Determine the modal lifetimes of the components.
Answer: Here, the maximum class frequency is 61 and the class corresponding to this frequency is 60–80. So, the modal class is 60–80.
Here, \( l = 60, h = 20, f_1 = 61, f_0 = 52, f_2 = 38 \)
\( \therefore \) Mode = \( l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 60 + \frac{61 - 52}{2 \times 61 - 52 - 38} \times 20 = 60 + \frac{9}{122 - 90} \times 20 \)
\( = 60 + \frac{9}{32} \times 20 = 60 + \frac{45}{8} = 60 + 5.625 = 65.625 \)
Hence, modal lifetime of the components is 65.625 hours.

Question. The median of the following data is 525. Find the values of \( x \) and \( y \), if total frequency is 100 :
Class | 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: We have
Class | Frequency | Cumulative frequency
0–100 | 2 | 2
100–200 | 5 | 7
200–300 | \( x \) | \( 7 + x \)
300–400 | 12 | \( 19 + x \)
400–500 | 17 | \( 36 + x \)
500–600 | 20 | \( 56 + x \)
600–700 | \( y \) | \( 56 + x + y \)
700–800 | 9 | \( 65 + x + y \)
800–900 | 7 | \( 72 + x + y \)
900–1000 | 4 | \( 76 + x + y \)
Total | 100 |
\( \therefore 76 + x + y = 100 \)
\( \implies \) \( x + y = 24 \) ...(i)
Given median is 525.
500–600 is the median class.
\( l = 500, h = 100, cf = 36 + x, f = 20 \)
\( \therefore \) Median = \( l + \frac{\frac{N}{2} - cf}{f} \times h \)
\( \implies 525 = 500 + \frac{50 - 36 - x}{20} \times 100 \)
\( \implies 25 = (14 - x) \times 5 \)
\( \implies 14 - x = 5 \)
\( \implies x = 9 \)
Putting \( x = 9 \) in equation (i), we have
\( 9 + y = 24 \)
\( \implies y = 24 - 9 = 15 \)
\( \implies y = 15 \)

Question. If the median of the distribution given below is 28.5, find the values of \( x \) and \( y \).
Class interval: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Frequency: 5, \( x \), 20, 15, \( y \), 5
Answer: Here, median = 28.5 and \( n = 60 \)
Now, we have
Class interval | Frequency (\( f_i \)) | Cumulative frequency (\( cf \))
0–10 | 5 | 5
10–20 | \( x \) | \( 5 + x \)
20–30 | 20 | \( 25 + x \)
30–40 | 15 | \( 40 + x \)
40–50 | \( y \) | \( 40 + x + y \)
50–60 | 5 | \( 45 + x + y \)
Total | \( \sum f_i = 60 \) |
Since the median is given to be 28.5, thus the median class is 20 – 30.
\( \therefore \frac{n}{2} = 30, l = 20, h = 10, cf = 5 + x \) and \( f = 20 \)
\( \therefore \) Median = \( l + \left[ \frac{\frac{n}{2} - cf}{f} \right] \times h \)
\( \implies 28.5 = 20 + \left[ \frac{30 - (5 + x)}{20} \right] \times 10 \)
\( \implies 28.5 = 20 + \frac{25 - x}{20} \times 10 \)
\( \implies 28.5 = 20 + \frac{25 - x}{2} \)
\( \implies 57 = 40 + 25 - x \)
\( \implies 57 = 65 - x \)
\( \implies x = 65 - 57 = 8 \)
Also, \( n = \sum f_i = 60 \)
\( \implies 45 + x + y = 60 \)
\( \implies 45 + 8 + y = 60 \quad (\because x = 8) \)
\( \implies y = 60 - 53 \)
\( \implies y = 7 \)
Hence, \( x = 8 \) and \( y = 7 \).

Question. The lengths of 40 leaves of a plant are measured correctly to the nearest millimetre, and the data obtained is represented in the following table:
Length (in mm): 118–126, 127–135, 136–144, 145–153, 154–162, 163–171, 172–180
Number of leaves: 3, 5, 9, 12, 5, 4, 2
Find the median length of the leaves.
Answer: Here, the classes are not in inclusive form. So, we first convert them in inclusive form by subtracting \( \frac{h}{2} \) from the lower limit and adding \( \frac{h}{2} \) to the upper limit of each class, where \( h \) is the difference between the lower limit of a class and the upper limit of preceding class.
Now, we have
Class interval | Number of leaves | Cumulative frequency
117.5–126.5 | 3 | 3
126.5–135.5 | 5 | 8
135.5–144.5 | 9 | 17
144.5–153.5 | 12 | 29
153.5–162.5 | 5 | 34
162.5–171.5 | 4 | 38
171.5–180.5 | 2 | 40
Total | \( \sum f_i = 40 \) |
We have, \( n = 40 \)
\( \implies \frac{n}{2} = 20 \)
And, the cumulative frequency just greater than \( \frac{n}{2} \) is 29 and corresponding class is 144.5 – 153.5.
So median class is 144.5 – 153.5.
Here, we have \( \frac{n}{2} = 20, l = 144.5, h = 9, f = 12, cf = 17 \)
\( \therefore \) Median = \( l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h = 144.5 + \left( \frac{20 - 17}{12} \right) \times 9 \)
\( = 144.5 + \frac{3}{12} \times 9 = 144.5 + \frac{9}{4} \)
\( = 144.5 + 2.25 = 146.75 \text{ mm.} \)
Hence, the median length of the leaves is 146.75 mm.

Question. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg): 40–45, 45–50, 50–55, 55–60, 60–65, 65–70, 70–75
Number of students: 2, 3, 8, 6, 6, 3, 2
Answer: Calculation of median
Weight (in kg) | Number of students (\( f \)) | Cumulative frequency (\( cf \))
40–45 | 2 | 2
45–50 | 3 | 5
50–55 | 8 | 13
55–60 | 6 | 19
60–65 | 6 | 25
65–70 | 3 | 28
70–75 | 2 | 30
Total | \( \sum f_i = 30 \) |
We have, \( \sum f_i = n = 30 \)
\( \implies \frac{n}{2} = 15 \)
The cumulative frequency just greater than \( \frac{n}{2} = 15 \) is 19, and the corresponding class is 55 – 60.
\( \therefore \) 55 – 60 is the median class.
Now, we have \( \frac{n}{2} = 15, l = 55, cf = 13, f = 6, h = 5 \)
\( \therefore \) Median = \( l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( = 55 + \left( \frac{15 - 13}{6} \right) \times 5 = 55 + \frac{2}{6} \times 5 = 55 + 1.67 = 56.67 \)
Hence, median weight is 56.67 kg.

Multiple Choice Questions

Choose and write the correct option in the following questions.

Question. The table below shows the time taken by a group of students to complete 100 m race.
Time taken (in sec): 18–20, 20–22, 22–24, 24–26, 26–28, 28–30
No. of students: 3, 18, 26, 19, 9, 5
Which of these is the mean time taken, in sec, by the group of students to complete the 100 m race when calculated using direct method?
(a) 18.16
(b) 18.96
(c) 23.7
(d) 33.7
Answer: (c) 23.7

Question. Vatsal grows cucumbers in his farm. He collects some of them and measures their lengths and represents his data as shown below.
Height (in cm): 110–120, 120–130, 130–140, 140–150, 150–160, 160–170, 170–180
No. of cucumbers: 10, 18, 12, \( n \), 26, 11, 19
When calculated using assumed mean method, Vatsal gets the mean length of the cucumber as 147.25 cm. Which of the following statement is true? 
(a) There are a smaller number of cucumbers of length (140 – 150) mm than of length (120 – 130) mm.
(b) There are a greater number of cucumbers of length (140 – 150) mm than of length (130 – 140) mm.
(c) There are a greater number of cucumbers of length (140 – 150) mm than of length (150 – 160) mm.
(d) There are a smaller number of cucumbers of length (140 – 150) mm than of length (170 – 180) mm.
Answer: (b) There are a greater number of cucumbers of length (140 – 150) mm than of length (130 – 140) mm.

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

Question. A grouped data is shown below:
Class Interval | Frequency | Cumulative frequency
0–10 | 4 | 4
10–20 | \( a \) | \( 4 + a \)
20–30 | 4 | \( 8 + a \)
30–40 | \( b \) | \( 8 + a + b \)
40–50 | 5 | \( 13 + a + b \)
Total | \( 13 + a + b \) |
If the median of the grouped data is 22.50 and the total frequency is 20, then what is the value of \( a \)?
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (c) 5

Question. Consider the following frequency distribution of the height of 60 students of a class:
Height (in cm): 150–155, 155–160, 160–165, 165–170, 170–175, 175–180
No. of students: 15, 13, 10, 8, 9, 5
The sum of the lower limit of the modal class and upper limit of the median class is 
(a) 310
(b) 315
(c) 320
(d) 330
Answer: (b) 315

Question. Consider the following frequency distribution:
Class: 0–5, 6–11, 12–17, 18–23, 24–29
Frequency: 13, 10, 15, 8, 11
The upper limit of the median class is [NCERT Exemplar]
(a) 17
(b) 17.5
(c) 18
(d) 18.5
Answer: (b) 17.5

Question. The table below shows the time spent by 100 students on exercising every day.
Time (in minutes): 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Number of students: \( x \), 24, 32, 21, \( y \), 5
If the median time spent by the students on exercising is 26.25 minutes which statement correctly compares the frequencies of the class intervals 0 – 10 and 40 – 50?
(a) There are twice as many students who exercise 40 – 50 minutes each day as the number of students who exercise 0 – 10 minutes each day.
(b) There are thrice as many students who exercise 40 – 50 minutes each day as the number of students who exercise 0 – 10 minutes each day.
(c) There are twice as many students who exercise 0 – 10 minutes each day as the number of students who exercise 40 – 50 minutes each day.
(d) There are thrice as many students who exercise 0 – 10 minutes each day as the number of students who exercise 40 – 50 minutes each day.
Answer: (a) There are twice as many students who exercise 40 – 50 minutes each day as the number of students who exercise 0 – 10 minutes each day.

Question. The table below shows the age of people attending a musical concert.
Age (in years): 0–10, 10–20, 20–30, 30–40, 40–50
Number of people: 2, 19, 32, 23, 19
If five more people of age 21 yrs., 32 yrs., 35 yrs., 44 yrs., and 11 yrs. attend the concert, which statement describes the central tendency of the new data?
(a) The central tendency of the data increases by 0.1 as the mean increases by 0.1.
(b) The central tendency of the data increases by 0.2 as the median increases by 0.2.
(c) The central tendency of the data increases by 0.1 as the median increases by 0.1.
(d) The central tendency of the data increases by 0.2 as the mean increases by 0.2.
Answer: (b) The central tendency of the data increases by 0.2 as the median increases by 0.2.

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

Very Short Answer Questions

Question. Find the class-marks of the classes 10–25 and 35–55. 
Answer: We have,
Class mark of the class 10–25 = \( \frac{25 + 10}{2} \)
\( = \frac{35}{2} = 17.5 \)
and, class mark of the class 35–55 = \( \frac{55 + 35}{2} = \frac{90}{2} = 45 \)

Question. If the mean of the first \( n \) natural numbers is 15, then find \( n \).
Answer: We have, mean of first \( n \) natural numbers = \( \frac{n(n + 1)}{2n} \)
\( \implies 15 = \frac{n + 1}{2} \)
\( \implies n + 1 = 30 \)
\( \implies n = 30 - 1 = 29 \)
\( \therefore n = 29 \)

Question. Using the empirical formula, find the mode of a distribution whose mean is 8.32 and the median is 8.05. 
Answer: Sol. Given, Mean = 8.32 and Median = 8.05
\( \therefore \) Mode = 3 Median – 2 Mean
= \( 3 \times 8.05 - 2 \times 8.32 \)
= 24.15 – 16.64 = 7.51

Question. A data has 13 observations arranged in descending order. Which observation represents the median of data?
Answer: Sol. Total no. of observations = 13, which is odd.
\( \therefore \) The median will be \( \left( \frac{n+1}{2} \right)^{\text{th}} \text{ term} = \left( \frac{13+1}{2} \right)^{\text{th}} = \left( \frac{14}{2} \right)^{\text{th}} = 7^{\text{th}} \)
i.e., 7th term will be the median.

Question. In an arranged series of an even number of \( 2n \) terms which term is median?
Answer: Sol. No. of terms = \( 2n \) which are even.
\( \therefore \) The median term will be \( \frac{\left[ \left( \frac{n}{2} \right)^{\text{th}} + \left( \frac{n}{2} + 1 \right)^{\text{th}} \right]}{2} \)
\( = \frac{\left[ \left( \frac{2n}{2} \right)^{\text{th}} + \left( \frac{2n}{2} + 1 \right)^{\text{th}} \right]}{2} = \left[ \frac{n^{\text{th}} + (n+1)^{\text{th}}}{2} \right] \) [Put \( n = 2n \)]
i.e., the mean of \( n^{\text{th}} \) and \( (n + 1)^{\text{th}} \) term will be the median.

Question. Write the modal class for the following frequency distribution:
Class Interval: 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 33, 38, 65, 52, 19, 48
Answer: Sol. Maximum frequency, i.e., 65 corresponds to the class 30–40.
\( \therefore \) Modal class is 30 – 40.

Short Answer Questions-I

Question. Find the mean of the following distribution:
Class: 3–5, 5–7, 7–9, 9–11, 11–13
Frequency: 5, 10, 10, 7, 8
Answer: Sol. We have
Class | Frequency (\( f_i \)) | Mid-value (\( x_i \)) | \( f_i \cdot x_i \)
3 – 5 | 5 | 4 | 20
5 – 7 | 10 | 6 | 60
7 – 9 | 10 | 8 | 80
9 – 11 | 7 | 10 | 70
11 – 13 | 8 | 12 | 96
Total | \( \sum f_i = 40 \) | | \( \sum f_i x_i = 326 \)
We have, \( \sum f_i = 40, \sum f_i x_i = 326 \)
\( \therefore \) Mean = \( \frac{\sum f_i x_i}{\sum f_i} = \frac{326}{40} = 8.15 \)

Question. Find the mode of the following distribution: 
Marks: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60
Number of Students: 4, 6, 7, 12, 5, 6
Answer: Sol. We have,
Marks | No. of Students
0 – 10 | 4
10 – 20 | 6
20 – 30 | 7
30 – 40 | 12
40 – 50 | 5
50 – 60 | 6
Here, modal class is 30 – 40
\( \therefore \) \( l = 30, h = 10, f_1 = 12, f_0 = 7, f_2 = 5 \)
\( \therefore \) Mode = \( l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
\( = 30 + \frac{12 - 7}{2 \times 12 - 7 - 5} \times 10 \)
\( = 30 + \frac{50}{12} \)
Mode = 30 + 4.16 = 34.16

Question. Find the median class of the following distribution:
Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 4, 4, 8, 10, 12, 8, 4
Answer: Sol. First we find the cumulative frequency
Classes | Frequency | Cumulative frequency
0–10 | 4 | 4
10–20 | 4 | 8
20–30 | 8 | 16
30–40 | 10 | 26
40–50 | 12 | 38
50–60 | 8 | 46
60–70 | 4 | 50
Total | 50 |
Here, \( \frac{n}{2} = \frac{50}{2} = 25 \)
\( \therefore \) Median class = 30 – 40.

Question. Median of a data is 52.5 and its Mean is 54, use empirical relationship between three measures of central tendency to find its mode.
Answer: Sol. We have,
Median = 52.5
and, Mean = 54
We know that 3 Median = Mode + 2 Mean
\( \implies \) \( 3 \times 52.5 = \text{Mode} + 2 \times 54 \)
\( \implies \) Mode = 157.5 – 108 = 49.5

Question. Consider the following frequency distribution.
Class: 0–5, 6–11, 12–17, 18–23, 24–29
Frequency: 13, 10, 15, 8, 11
Find the upper limit of median class.
Answer: Sol. Classes are not continuous, hence make them continuous by adding 0.5 to the upper limits and subtracting 0.5 from the lower limits.
Class Interval | Frequency | Cumulative frequency
0–5.5 | 13 | 13
5.5–11.5 | 10 | 23
11.5–17.5 | 15 | 38
17.5–23.5 | 08 | 46
23.5–29.5 | 11 | 57
Total | \( \sum f = 57 \) |
Note: Class interval can't be negative hence the first CI is starting from 0.
Now to find median class we calculate \( \frac{\sum f}{2} = \frac{57}{2} = 28.5 \)
\( \therefore \) Median class = 11.5 – 17.5.
So, the upper limit is 17.5.

Question. The data regarding the heights of 50 girls of class X of a school is given below:
Heights (in cm): 120–130, 130–140, 140–150, 150–160, 160–170, Total
No. of girls: 2, 8, 12, 20, 8, 50
Change the above distribution to 'More than type' distribution.
Answer: Sol. We have,
Heights (in cm) | No. of girls
More than 120 | 50
More than 130 | 48
More than 140 | 40
More than 150 | 28
More than 160 | 8

Short Answer Questions-II

Question. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. 
Number of days: 0–6, 6–12, 12–18, 18–24, 24–30, 30–36, 36–42
Number of students: 10, 11, 7, 4, 4, 3, 1
Answer: Sol. Calculating mean using step deviation method.
No. of days (class interval) | No. of students (\( f_i \)) | \( x_i \) | \( u_i = \frac{x_i - A}{h} \) | \( f_i u_i \)
0 – 6 | 10 | 3 | –3 | –30
6 – 12 | 11 | 9 | –2 | –22
12 – 18 | 7 | 15 | –1 | –7
18 – 24 | 4 | 21 (A) | 0 | 0
24 – 30 | 4 | 27 | 1 | 4
30 – 36 | 3 | 33 | 2 | 6
36 – 42 | 1 | 39 | 3 | 3
Total | \( \sum f_i = 40 \) | | | \( \sum f_i u_i = -46 \)
Class size (\( h \)) = 6 - 0 = 6
Assumed mean (\( A \)) = 21
Mean = \( A + \frac{\sum f_i u_i}{\sum f_i} \times h \)
\( \implies \) Mean = \( 21 + \left( \frac{-46}{40} \right) \times 6 \)
= 21 – \( \frac{23 \times 3}{10} \)
= 21 – 6.9 = 14.1 days
Mean of no. of days student remains absent = 14.1 days.

Question. If the mean of the following distribution is 6, find the value of \( p \).
\( x \): 2, 4, 6, 10, \( p + 5 \)
\( f \): 3, 2, 3, 1, 2
Answer: Sol. Calculation of mean
\( x_i \) | \( f_i \) | \( f_i x_i \)
2 | 3 | 6
4 | 2 | 8
6 | 3 | 18
10 | 1 | 10
\( p + 5 \) | 2 | \( 2p + 10 \)
Total | \( \sum f_i = 11 \) | \( \sum f_i x_i = 2p + 52 \)
We have, \( \sum f_i = 11, \sum f_i x_i = 2p + 52, \bar{X} = 6 \)
\( \therefore \) Mean \( (X) = \frac{\sum f_i x_i}{\sum f_i} \)
\( \implies \) \( 6 = \frac{2p + 52}{11} \)
\( \implies \) \( 66 = 2p + 52 \)
\( \implies \) \( 2p = 14 \)
\( \implies \) \( p = 7 \)

Question. Compute the mode for the following frequency distribution
Size of items (in cm): 0–4, 4–8, 8–12, 12–16, 16–20, 20–24, 24–28
Frequency: 5, 7, 9, 17, 12, 10, 6
Answer: Sol.
Size of items (in cm) | Frequency
0 – 4 | 5
4 – 8 | 7
8 – 12 | 9
12 – 16 | 17
16 – 20 | 12
20 – 24 | 10
24 – 28 | 6
Here, maximum frequency is 17.
\( \therefore \) 12 – 16 is the modal class.
Now, we have
\( l = 12, h = 4, f_1 = 17 \) and \( f_0 = 9, f_2 = 12 \)
\( \therefore \) Mode = \( l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
\( = 12 + \frac{17 - 9}{2 \times 17 - 9 - 12} \times 4 \)
\( \implies \) Mode = \( 12 + \frac{32}{13} = 12 + 2.46 = 14.46 \)
\( \therefore \) Mode = 14.46

Question. Find the value of \( p \), if the mean of the following distribution is 7.5. 
Classes: 2–4, 4–6, 6–8, 8–10, 10–12, 12–14
Frequency (\( f_i \)): 6, 8, 15, \( p \), 8, 4
Answer: Sol. We have,
Class | Frequency (\( f \)) | \( x \) | \( fx \)
2 – 4 | 6 | 3 | 18
4 – 6 | 8 | 5 | 40
6 – 8 | 15 | 7 | 105
8 – 10 | \( p \) | 9 | \( 9p \)
10 – 12 | 8 | 11 | 88
12 – 14 | 4 | 13 | 52
| \( 41 + p \) | | \( 303 + 9p \)
Mean = \( 7.5 = \frac{303 + 9p}{p + 41} \)
\( \implies \) \( p = 3 \)

Question. The table below shows the salaries of 280 persons:
Salary (In thousand Rs.): 5–10, 10–15, 15–20, 20–25, 25–30, 30–35, 35–40, 40–45, 45–50
No. of Persons: 49, 133, 63, 15, 6, 7, 4, 2, 1
Calculate the median salary of the data. 
Answer: Sol. Distribution of frequencies:
Salary in thousand Rs. | No. of persons | Cumulative frequency
5–10 | 49 | 49
10–15 | 133 | 182
15–20 | 63 | 245
20–25 | 15 | 260
25–30 | 6 | 266
30–35 | 7 | 273
35–40 | 4 | 277
40–45 | 2 | 279
45–50 | 1 | 280
To find median:
No. of people = 280.
\( \implies \frac{n}{2} = 140 \), the 140th term lies in class interval 10–15.
\( \implies \) median class = 10–15.
\( l = 10, h = 5, f = 133, \frac{n}{2} = 140, cf = 49. \)
We know, \( \text{median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h. \)
\( \implies \text{median} = 10 + \frac{140 - 49}{133} \times 5. \)
\( = 10 + \frac{91}{133} \times 5. \)
\( = 10 + \frac{65}{19} \)
= 10 + 3.421
= 13.421.
The median salary is 13.421 thousand rupees. 

Long Answer Questions

Question. The mean of the following distribution is 18. Find the frequency \( f \) of the class 19 – 21. 
Class | 11–13 | 13–15 | 15–17 | 17–19 | 19–21 | 21–23 | 23–25
Frequency | 3 | 6 | 9 | 13 | \( f \) | 5 | 4
Answer: Frequency distribution.
Class | Frequency | \( x_i \) | \( f_i x_i \)
11–13 | 3 | 12 | \( 3 \times 12 = 36 \)
13–15 | 6 | 14 | \( 6 \times 14 = 84 \)
15–17 | 9 | 16 | \( 9 \times 16 = 144 \)
17–19 | 13 | 18 | \( 13 \times 18 = 234 \)
19–21 | \( f \) | 20 | \( f \times 20 = 20f \)
21–23 | 5 | 22 | \( 5 \times 22 = 110 \)
23–25 | 4 | 24 | \( 4 \times 24 = 96 \)
Total | \( \sum f_i = 40 + f \) | | \( \sum f_i x_i = 704 + 20f \)
Given, mean = 18. To find: value of \( f \).
We know,
mean \( (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \)
\( \implies \) \( 18 = \frac{704 + 20f}{40 + f} \)
\( 720 + 18f = 704 + 20f \)
\( 720 - 704 = 20f - 18f \)
\( 16 = 2f \)
\( \implies \) \( f = 8 \)
The value of \( f \) is 8. 

Question. The mean of the following frequency table is 50. But the frequencies \( f_1 \) and \( f_2 \) in class 20–40 and 60–80 respectively are missing. Find the missing frequencies. 
Classes | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 | Total
Frequency | 17 | \( f_1 \) | 32 | \( f_2 \) | 19 | 120
Answer: Let the assumed mean \( A = 50 \) and \( h = 20 \).
Calculation of mean
Class interval | Mid-values (\( x_i \)) | Frequency (\( f_i \)) | \( u_i = \frac{x_i - 50}{20} \) | \( f_i u_i \)
0–20 | 10 | 17 | –2 | –34
20–40 | 30 | \( f_1 \) | –1 | \( -f_1 \)
40–60 | 50 | 32 | 0 | 0
60–80 | 70 | \( f_2 \) | 1 | \( f_2 \)
80–100 | 90 | 19 | 2 | 38
Total | | \( \sum f_i = 68 + f_1 + f_2 \) | | \( \sum f_i u_i = 4 - f_1 + f_2 \)
We have, \( \sum f_i = 120 \) (Given)
\( \implies \) \( 68 + f_1 + f_2 = 120 \)
\( \implies \) \( f_1 + f_2 = 52 \) ...(i)
Now, mean = 50
\( \implies \) \( \bar{X} = A + h \left( \frac{\sum f_i u_i}{\sum f_i} \right) \)
\( \implies \) \( 50 = 50 + 20 \times \left\{ \frac{4 - f_1 + f_2}{120} \right\} \)
\( \implies \) \( 50 = 50 + \frac{4 - f_1 + f_2}{6} \)
\( \implies \) \( 0 = \frac{4 - f_1 + f_2}{6} \)
\( f_1 - f_2 = 4 \) ...(ii)
From equations (i) and (ii), we get
\( f_1 + f_2 = 52 \)
\( f_1 - f_2 = 4 \)
\( 2f_1 = 56 \)
\( \implies \) \( f_1 = 28 \)
Putting the value of \( f_1 \) in equation (i), we get
\( 28 + f_2 = 52 \)
\( \implies \) \( f_2 = 24 \)
Hence, the missing frequencies \( f_1 \) is 28 and \( f_2 \) is 24.

Question. The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean and the median of the number of wickets taken.
Number of wickets | 20 – 60 | 60 – 100 | 100 – 140 | 140 – 180 | 180 – 220 | 220 – 260
Number of bowlers | 7 | 5 | 16 | 12 | 2 | 3
Answer: We have,
For Mean
No. of wickets | \( (f_i) \) No. of bowlers | \( (x_i) \) Mid values | \( u_i = \frac{x_i - A}{h} \) | \( f_i \cdot u_i \)
20 – 60 | 7 | 40 | –3 | –21
60 – 100 | 5 | 80 | –2 | –10
100 – 140 | 16 | 120 | –1 | –16
140 – 180 | 12 | 160 = A | 0 | 0
180 – 220 | 2 | 200 | 1 | 2
220 – 260 | 3 | 240 | 2 | 6
Total | 45 | | | –39
We have,
\( A = 160, h = 40, \sum f_i = 45, \sum f_i u_i = -39 \)
Mean \( = A + \frac{\sum f_i u_i}{\sum f_i} \times h \)
\( = 160 + \frac{-39 \times 40}{45} = 160 - 34.67 = 125.33 \)

For Median
Number of wickets | Number of bowlers | Cumulative Frequency
20 – 60 | 7 | 7
60 – 100 | 5 | 12
100 – 140 | 16 | 28
140 – 180 | 12 | 40
180 – 220 | 2 | 42
220 – 260 | 3 | 45
Total | 45 |
We have,
\( n = 45 \implies \frac{n}{2} = 22.5 \)
\( \therefore \) 100 – 140 is the median class.
\( \therefore \) \( l = 100, cf = 12, f = 16, h = 40 \)
\( \therefore \) Median \( = l + \frac{\frac{n}{2} - cf}{f} \times h \)
\( = 100 + \frac{22.5 - 12}{16} \times 40 = 100 + \frac{10.5 \times 40}{16} = 100 + 26.25 = 126.25 \)
\( \therefore \) Median \( = 126.25 \)

Question. If the median of the following frequency distribution is 32.5. Find the values of \( f_1 \) and \( f_2 \). 
Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 | Total
Frequency | \( f_1 \) | 5 | 9 | 12 | \( f_2 \) | 3 | 2 | 40
Answer:
Class | Frequency | Cumulative Frequency
0–10 | \( f_1 \) | \( f_1 \)
10–20 | 5 | \( 5 + f_1 \)
20–30 | 9 | \( 14 + f_1 \)
30–40 | 12 | \( 26 + f_1 \)
40–50 | \( f_2 \) | \( 26 + f_1 + f_2 \)
50–60 | 3 | \( 29 + f_1 + f_2 \)
60–70 | 2 | \( 31 + f_1 + f_2 \)
Total | \( \sum f = 40 \) |
Median = 32.5 \( \implies \) Median class is 30–40.
Now \( 32.5 = 30 + \frac{10}{12} (20 - 14 - f_1) \)
\( \implies \) \( f_1 = 3 \)
Also \( 31 + f_1 + f_2 = 40 \)
\( \implies \) \( f_2 = 6 \)