CBSE Class 10 Mathematics Statistics VBQs Set H

The following data gives the information on the observed life-times (in hours) of 225 electrical components.

Question. Determine the mean of the above data. 
Life-time (in hours): 0–20, 20–40, 40–60, 60–80, 80–100, 100–120
Frequency: 10, 35, 52, 61, 38, 29

Answer:
Solution.
\[ \begin{array}{|c|c|c|c|c|} \hline \text{Class interval} & f_i & x_i & d_i = x_i - A & f_i d_i \\ \hline 0-20 & 10 & 10 & -60 & -600 \\ \hline 20-40 & 35 & 30 & -40 & -1400 \\ \hline 40-60 & 52 & 50 & -20 & -1040 \\ \hline 60-80 & 61 & 70 = A & 0 & 0 \\ \hline 80-100 & 38 & 90 & 20 & 760 \\ \hline 100-120 & 29 & 110 & 40 & 1160 \\ \hline & \sum f_i = 225 & & & \sum f_i d_i = -1120 \\ \hline \end{array} \] \( \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} \)
\( = 70 + \frac{-1120}{225} \)
\( = 70 - 4.98 = 65.02 \)
\( \bar{x} = 65.02 \) approximately.

Exercise 

I. Very Short Answer Type Questions 

Question. While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes
(b) centered at the classmarks of the classes
(c) centered at the upper limits of the classes
(d) centered at the lower limits of the classes

Answer: (b) centered at the classmarks of the classes

Question. If \( x_i \) are the mid-points of the class intervals of grouped data, \( f_i \)'s are the corresponding frequencies and \( \bar{x} \) is the mean, then \( \sum (f_i x_i - \bar{x}) \) is equal to
(a) 0
(b) -1
(c) 1
(d) 2

Answer: (a) 0

Question. In the following \( \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} \), for finding the mean of grouped frequency distribution, \( d_i = \)
(a) \( x_i + A \)
(b) \( A - x_i \)
(c) \( x_i - A \)
(d) \( \frac{A - x_i}{f_i} \)

Answer: (c) \( x_i - A \)

Question. If the arithmetic mean of \( n \) numbers of a series is \( \bar{x} \) and the sum of first \( (n - 1) \) numbers is \( k \), the value of the last number is
(a) \( n\bar{x} - k \)
(b) \( n\bar{x} + k \)
(c) \( \frac{\bar{x} + k}{n} \)
(d) \( n(\bar{x} + k) \)

Answer: (a) \( n\bar{x} - k \)

Question. Arithmetic mean of all factors of 20 is
(a) 5
(b) 6
(c) 7
(d) 8

Answer: (c) 7

Question. The mean of 5 numbers is 27. If one number is excluded their mean is 25. The excluded number is
(a) 30
(b) 35
(c) 32
(d) 36

Answer: (b) 35

Question. Assertion (A): The arithmetic mean of the following given frequency distribution table is 13.81.
\( x \): 4, 7, 10, 13, 16, 19
\( f \): 7, 10, 15, 20, 25, 30
Reason (R): \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \)
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Answer: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Assertion (A): To find mean of a grouped data, we use \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \) where \( a \) is the assumed mean and \( d_i \) the deviation.
Reason (R): To find deviation, we use \( d_i = a - x_i \) where \( a \) is the assumed mean and \( x_i \) is the class mark.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Answer: (c) Assertion (A) is true but reason (R) is false.

Question. Find the class-mark of class 25–35.

Answer: Class-mark \( = \frac{25 + 35}{2} = 30 \)

Question. Find the mean of first ten odd natural numbers.

Answer: First ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. Mean \( = \frac{100}{10} = 10 \)

II. Short Answer Type Questions-I

Question. If the mean of the following data is 20.6, find the value of \( p \).
\( x \): 10, 15, \( p \), 25, 35
\( f \): 3, 10, 25, 7, 5

Answer:
Using formula \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \sum f_i = 50 \)
\( \sum f_i x_i = (10 \times 3) + (15 \times 10) + (p \times 25) + (25 \times 7) + (35 \times 5) = 530 + 25p \)
\( 20.6 = \frac{530 + 25p}{50} \Rightarrow 1030 = 530 + 25p \Rightarrow 25p = 500 \Rightarrow p = 20 \)

Question. Find the mean of the following distribution:
Class: 5–15, 15–25, 25–35, 35–45
Frequency: 2, 4, 3, 1

Answer:
\( x_i \): 10, 20, 30, 40
\( \sum f_i = 10 \)
\( \sum f_i x_i = 20 + 80 + 90 + 40 = 230 \)
Mean \( = \frac{230}{10} = 23 \)

III. Short Answer Type Questions-II 

Question. The mean of the following frequency distribution is 62.8 and 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

Answer:
\( 5 + f_1 + 10 + f_2 + 7 + 8 = 50 \Rightarrow f_1 + f_2 = 20 \)...(i)
\( \sum f_i x_i = 5(10) + f_1(30) + 10(50) + f_2(70) + 7(90) + 8(110) = 2060 + 30f_1 + 70f_2 \)
\( 62.8 = \frac{2060 + 30f_1 + 70f_2}{50} \Rightarrow 3140 = 2060 + 30f_1 + 70f_2 \Rightarrow 3f_1 + 7f_2 = 108 \)...(ii)
Solving (i) and (ii), we get \( f_1 = 8 \) and \( f_2 = 12 \)

Question. The arithmetic mean of the following frequency distribution is 53. Find the value of \( k \).  
Class: 0–20, 20–40, 40–60, 60–80, 80–100
Frequency: 12, 15, 32, \( k \), 13

Answer:
\( \sum f_i = 72 + k \)
\( \sum f_i x_i = 120 + 450 + 1600 + 70k + 1170 = 3340 + 70k \)
\( 53 = \frac{3340 + 70k}{72 + k} \Rightarrow 3816 + 53k = 3340 + 70k \Rightarrow 17k = 476 \Rightarrow k = 28 \)

Question. The table below shows the daily expenditure on grocery of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.  
Daily expenditure (in ₹): 100–150, 150–200, 200–250, 250–300, 300–350
No. of households: 4, 5, 12, 2, 2

Answer:
Using Assumed Mean Method (A = 225):
\( \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} = 225 + \frac{-350}{25} = 225 - 14 = Rs 211 \)

Question. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency \( f \).
Daily pocket allowance (in ₹): 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25
Number of children: 7, 6, 9, 13, \( f \), 5, 4

Answer:
\( \sum f_i = 44 + f \)
\( \sum f_i d_i = 2f - 40 \) (taking A = 18)
\( 18 = 18 + \frac{2f - 40}{44 + f} \Rightarrow 2f - 40 = 0 \Rightarrow f = 20 \)

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. [AI 2019]
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:
\( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 21 + \frac{-276}{40} = 21 - 6.9 = 14.1 \) days.

IV. Long Answer Type Questions 

Question. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are as follows:
Mileage (km / l): 10–12, 12–14, 14–16, 16–18
No. of Cars: 7, 12, 18, 13
Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km/l. Do you agree with this claim? [NCERT Exemplar] [Imp]

Answer:
\( \sum f_i = 50 \); taking A = 13, \( \sum f_i d_i = 74 \).
Mean \( \bar{x} = 13 + \frac{74}{50} = 13 + 1.48 = 14.48 \text{ km/l} \).
The manufacturer's claim of 16 km/l is wrong.

Question. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given as follows:
No. 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:
Taking A = 110, \( \sum f_i = 100 \), \( \sum f_i d_i = -8 \).
Mean \( = 110 + \frac{-8}{100} = 109.92 \).
Since seats cannot be in decimal, mean number of seats \( \approx 110 \).

Question. Find the mean of the following data:
Classes: 0–20, 20–40, 40–60, 60–80, 80–100, 100–120
Frequency: 20, 35, 52, 44, 38, 31

Answer:
\( \sum f_i = 220 \), \( \sum f_i x_i = 13760 \)
Mean \( = \frac{13760}{220} = 62.55 \) (approx).

Case Study Based Questions

Student-Teacher Ratio: Student-teacher ratio expresses the relationship between the number of students enrolled in a school and the number of teachers in that school. It is important for a number of reasons. For example, it can be an indicator of the amount of individual attention any child is likely to receive, keeping in mind that not all class size are going to be the same. The following distribution gives the state-wise student-teacher ratio in higher secondary schools of India (28 states and 7 UTs only).
Number of students per teacher: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55
Number of States/UTs: 3, 8, 9, 10, 3, 0, 0, 2

Question. In order to find the mean by direct method, we use the formula
(a) \( \frac{\sum f_i x_i}{n} \)
(b) \( \frac{n}{\sum f_i x_i} \)
(c) \( n \times \sum f_i x_i \)
(d) \( n + \sum f_i x_i \)

Answer: (a) \( \frac{\sum f_i x_i}{n} \)

Question. The mean of the above data is
(a) 29.2
(b) 30.5
(c) 38.3
(d) 40.1

Answer: (a) 29.2

Question. The formula for assumed mean method to find the mean is
(a) \( A - \frac{\sum f_i d_i}{\sum f_i} \)
(b) \( A + \frac{\sum f_i}{\sum f_i d_i} \)
(c) \( A \times \frac{\sum f_i d_i}{\sum f_i} \)
(d) \( A + \frac{\sum f_i d_i}{\sum f_i} \)

Answer: (d) \( A + \frac{\sum f_i d_i}{\sum f_i} \)

Question. The sum of class marks of 25-30 and 45-50 is
(a) 62
(b) 70
(c) 75
(d) 85

Answer: (c) 75

Question. The sum of the upper and lower limits of modal class is
(a) 55
(b) 65
(c) 85
(d) 75

Answer: (b) 65

2. MODE AND MEDIAN OF GROUPED DATA

Mode

The mode or modal value is that value of the variate which occurs most frequently. To find the mode of a grouped data, we proceed as follows:
(i) Obtain the grouped data.
(ii) Locate the class having maximum frequency. This class is called modal class.
(iii) Mode of a grouped data is given by the formula:
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]
where, \( l \) = lower limit of the modal class
\( f_1 \) = frequency of the modal class
\( f_0 \) = frequency of the class preceding the modal class
\( f_2 \) = frequency of the class succeeding the modal class
\( h \) = size of the modal class

Question. Example 1. The marks distribution of 30 students in a science examination are as follows. Find the mode of this data. [Imp.]
Marks obtained (\( x_i \)): 10, 20, 36, 40, 50, 56, 60, 70, 72, 80, 88, 92, 95
Number of students (\( f_i \)): 1, 1, 3, 4, 3, 2, 4, 4, 1, 1, 2, 3, 1

Answer:
First, we will make the class interval with class size of 15.