CBSE Class 10 Statistics Sure Shot Questions Set D

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. The marks distribution of 30 students in a science examination are as follows. Find the mode of this data. 
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.
Class interval | Number of students
10–25 | 2
25–40 | 3 (\( f_0 \))
40–55 | 7 (Modal class, \( f_1 = 7 \))
55–70 | 6 (\( f_2 \))
70–85 | 6
85–100 | 6
Total \( \sum f_i = 30 \)
Since the maximum number of students (7) have got marks in the interval 40–55, the model class is 40–55.
So lower limit of the model class, \( l = 40 \)
class size, \( h = 15 \)
frequency, \( f_1 \) of the model class = 7
frequency, \( f_0 \) of the class preceding the model class = 3
frequency, \( f_2 \) of the class succeeding the model class = 6
Now, using the formula
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 40 + \left( \frac{7 - 3}{14 - 3 - 6} \right) \times 15 = 52 \)
\( \therefore \text{Mode} = 52 \)

Median

The median is the middle value of a distribution i.e., median of a distribution is the value of the observation which divides it into two equal parts.

• Median of ungrouped data: (i) Arrange the data in ascending order. (ii) If \( n \) (number of observations) is odd, then \( \text{median} = \left( \frac{n + 1}{2} \right)^{\text{th}} \) observation. (iii) If \( n \) (number of observations) is even, then \( \text{median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right)^{\text{th}} \text{observation} + \left( \frac{n}{2} + 1 \right)^{\text{th}} \text{observation} \right] \)
• Median of grouped data: Median of a grouped data or continuous frequency distribution can be found by using the formula: \[ \text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \] where, \( l \) = lower limit of the median class \( n \) = number of observations \( f \) = frequency of the median class \( h \) = size of the median class (assuming all class sizes to be equal) \( cf \) = cumulative frequency of the class preceding the median class

Question. Find the median of the following data:
\( x_i \): 22, 24.50, 28, 31.50, 34, 36.50
\( f_i \): 10, 23, 32, 28, 12, 05

Answer:

Since \( \sum f_i = n = 110 \), which is even.
\( \therefore \text{Median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right)^{\text{th}} \text{observation} + \left( \frac{n}{2} + 1 \right)^{\text{th}} \text{observation} \right] \)
\( = \frac{1}{2} \left[ 55^{\text{th}} \text{observation} + 56^{\text{th}} \text{observation} \right] = \frac{1}{2} (28 + 28) = 28 \)
(\( \because \) Both 55 and 56 lies in \( cf \) 65, therefore \( 55^{\text{th}} \text{observation} = 56^{\text{th}} \text{observation} = 28 \))

Question. The distribution below gives the marks of 30 students of a class in mathematics. Find the median marks of the students.
Marks: 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:

\( \sum f_i = n = 30 \), \( \frac{n}{2} = 15 \)
Since \( cf \) just greater than \( \frac{n}{2} = 15 \) is 19.
\( \therefore \) The corresponding class is 55–60 which is the median class.
Now, we have \( \frac{n}{2} = 15, l = 55, cf = 13, f = 6, h = 5 \)
\( \therefore \text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( = 55 + \left( \frac{15 - 13}{6} \right) \times 5 = 55 + 1.67 = 56.67 \)

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, Total
Frequency: 5, \( x \), 20, 15, \( y \), 5, 60

Answer: Here, median = 28.5, \( n = 60 \)

Since the median = 28.5, therefore, median class is 20–30
\( \therefore \frac{n}{2} = 30, l = 20, h = 10, cf = 5 + x, f = 20 \)
\( \therefore \text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( \Rightarrow 28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10 \)
\( \Rightarrow 28.5 = 20 + \frac{25 - x}{20} \times 10 \)
\( \Rightarrow 28.5 = 20 + \frac{25 - x}{2} \)
\( \Rightarrow 57 = 40 + 25 - x \)
\( \Rightarrow 57 = 65 - x \Rightarrow x = 65 - 57 = 8 \)
Also, 45 + \( x \) + \( y \) = 60 \( \Rightarrow y = 7 \)

I. Very Short Answer Type Questions 

Multiple Choice Questions 

Question. Consider the following frequency distribution of the heights 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 upper limit of the median class in the given data is
(a) 165
(b) 155
(c) 160
(d) 170
Answer: (a)

Question. For the following distribution:
Marks | Number of students
Below 10 | 3
Below 20 | 12
Below 30 | 28
Below 40 | 57
Below 50 | 75
Below 60 | 80
The modal class is
(a) 0-20
(b) 20-30
(c) 30-40
(d) 50-60
Answer: (c)

Question. The cumulative frequency of a given class is obtained by adding the frequencies of all the classes
(a) preceding it
(b) succeeding it
(c) Both (a) and (b)
(d) None of these
Answer: (a)

Question. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (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. Assertion (A): If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, then median is 30.
Reason (R): \( \text{Median} = \left( \frac{n + 1}{2} \right)^{\text{th}} \) value, if \( n \) is odd.
Answer: (d)

Question. Assertion (A): If the value of mode and mean is 60 and 66 respectively, then the value of median is 64.
Reason (R): Median = (Mode + 2 Mean) / 3
Answer: (c)

Question. Answer the following. Write the modal class of 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: 30-40

Question. Write the median of the following data: 3, 5, 2, 9, 7, 11
Answer: 6

II. Short Answer Type Questions

Question. Find the median class of following data:
Class interval: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Frequency: 8, 10, 12, 22, 30, 18

Answer: 30-40

Question. Find the mode of the following data: 
Class: 0–20, 20–40, 40–60, 60–80, 80–100, 100–120, 120–140
Frequency: 6, 8, 10, 12, 6, 5, 3

Answer: 65

Question. The following distribution shows the transport expenditure of 100 employees:
Expenditure (in ₹): 200–400, 400–600, 600–800, 800–1000, 1000–1200
No. of employees: 21, 25, 19, 23, 12
Find the mode of the distribution.

Answer: 480

III. Short Answer Type Questions

Question. The weight of tea in 70 packets are as follows:
Weight (in kg): 200–201, 201–202, 202–203, 203–204, 204–205, 205–206
No. of packets: 12, 26, 20, 4, 2, 1
Determine the modal weight. 

Answer: 201.7 kg

Question. The annual rainfall record of a city of 100 days is given in the following table:
Rainfall (in cm): 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
No. of days: 8, 8, 14, 22, 30, 8, 10
Calculate the median rainfall. 

Answer: 39.09 cm

Question. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in year): 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
No. of patients: 22, 10, 8, 15, 5, 6
Find the median of the data given above. 

Answer: 21.25 years

Question. The table below shows the salaries of 280 persons:
Salary (In thousand ₹) | No. of Persons
5 – 10 | 49
10 – 15 | 133
15 – 20 | 63
20 – 25 | 15
25 – 30 | 6
30 – 35 | 7
35 – 40 | 4
40 – 45 | 2
45 – 50 | 1
Calculate the median salary of the data.

Answer: ₹13420

Question. The following data gives the information on the observed life-times (in hours) of 25 electrical components. Determine the model life-time of the components. 
Life-time (in hrs): 0–50, 50–100, 100–150, 150–200, 200–250, 250–300, 300–350
No. of components: 2, 3, 5, 6, 5, 3, 1

Answer: 175 hrs.

Question. The table shows the daily expenditure on grocery of 25 households in a locality. Find the modal daily expenditure on grocery 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: ₹220.59

Question. The median of the following data is 16. Find the missing frequencies \( a \) and \( b \), if the total of the frequencies is 70.
Class: 0-5, 5-10, 10-15, 15-20, 20-25, 25-30, 30-35, 35-40
Frequency: 12, \( a \), 12, 15, \( b \), 6, 6, 4

Answer: \( a = 8, b = 7 \)

IV. Long Answer Type Questions 

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: \( f_1 = 3, f_2 = 6 \)

Question. Compare the modal age of two groups of students A and B appearing for an entrance test. 
Class interval: 16–18, 18–20, 20–22, 22–24, 24–26
Frequency Group A: 50, 78, 46, 28, 23
Frequency Group B: 54, 89, 40, 25, 17

Answer: Modal age of group A = 18.93 years. Modal age of group B = 18.83 years. Modal age of students of group A > modal age of students of group B.

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

Answer: \( x = 9, y = 15 \)

Question. Daily wages of 110 workers, obtained in a survey, are tabulated below: 
Daily Wages (in ₹): 100–120, 120–140, 140–160, 160–180, 180–200, 200–220, 220–240
Number of Workers: 10, 15, 20, 22, 18, 12, 13
Compute the mean daily wages and modal daily wages of these workers.

Answer: Mean daily wages = ₹170.19 (approx.). Modal daily wages = ₹166.67 (approx.)

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
No. of bowlers: 7, 5, 16, 12, 2, 3

Answer: Mean = 125.33. Median = 126.25.

Question. The mode of the following data is 67. Find the missing frequency \( x \). 
Class: 40-50, 50-60, 60-70, 70-80, 80-90
Frequency: 5, \( x \), 15, 12, 7

Answer: \( x = 8 \)

Question. A survey regarding the heights (in cm) of 51 girls of class X of a school was conducted and the following data was obtained. Find the median height and the mean using the formulae.
Height (in cm): Less than 140, Less than 145, Less than 150, Less than 155, Less than 160, Less than 165
Number of Girls: 4, 11, 29, 40, 46, 51

Answer: Median height = 149.03. Mean height = 149.75.