CBSE Class 10 Mathematics Statistics Worksheet Set 08

Case Study-based Questions

COVID-19 Pandemic
The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans.
The following table shows the age distribution of case admitted during a day in two different hospitals
Table 1:
Age (in years): 5 – 15, 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65
No. of cases: 6, 11, 21, 23, 14, 5
Table 2:
Age (in years): 5 – 15, 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65
No. of cases: 8, 16, 10, 42, 24, 12

Question. Refer to table 1: The average age for which maximum cases occurred is
(a) 32.24
(b) 34.36
(c) 36.82
(d) 42.24
Answer: (c) 36.82

Question. Refer to table 1: The upper limit of modal class is
(a) 15
(b) 25
(c) 35
(d) 45
Answer: (d) 45

Question. Refer to table 1: The mean of the given data is
(a) 26.2
(b) 32.4
(c) 33.5
(d) 35.4
Answer: (d) 35.4

Question. Refer to table 2: The mode of the given data is
(a) 41.4
(b) 48.2
(c) 55.3
(d) 64.6
Answer: (a) 41.4

Question. Refer to table 2: The median of the given data is
(a) 32.7
(b) 40.2
(c) 42.3
(d) 48.6
Answer: (b) 40.2

Electricity Energy Consumption
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).
Tariff : LT - Residential
Bill Number : 384756
Type of Supply : Single Phase
Connected Load : 3 KW
Meter Reading Date : 31-11-13
Meter Reading : 65789
Previous Reading Date : 31-10-13
Previous Meter Reading : 65500
Units Consumed : 289
A survey is conducted for 56 families of a colony. The following table gives the weekly consumption of electricity of these families.
Weekly consumption (in units): 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
No. of families: 16, 12, 18, 6, 4, 0

Question. Refer to data received from colony: What is the median weekly consumption?
Answer: From the cumulative frequency table, \( N = 56 \implies \frac{N}{2} = 28 \). The cumulative frequency just greater than or equal to 28 is 28, which corresponds to class 10 – 20. However, the next cumulative frequency 46 corresponds to 20 – 30. Using the formula:
\( \therefore \) 20 – 30 is the median class.
Median \( = l + \frac{\frac{N}{2} - cf}{f} \times h \)
\( = 20 + \frac{28 - 28}{18} \times 10 = 20 \)

Question. Refer to data received from colony: What is the mean weekly consumption?
Answer: Mean \( = \frac{\sum f_i x_i}{\sum f_i} = \frac{1100}{56} = 19.64 \)

Objective Type Questions: 

Question. When calculated using direct method, the mean of the data set shown in the table below is 31.
Class Interval | Frequency
0–10 | 22
10–20 | 24
20–30 | 35
30–40 | 30
40–50 | 27
50–60 | \( m \)
60–70 | 6
70–80 | 4
What is the frequency for the class interval 50–60?
(a) 8
(b) 10
(c) 12
(d) 20
Answer: (c) 12

Question. The table below summarizes the data about the heights of students in a class.
Height (in cm): 130–140, 140–150, 150–160, 160–170, 170–180
Number of students: 6, 14, 20, 12, 8
When calculated using assumed mean method what is the mean height of students in the class?
(a) 154.33 cm
(b) 155.33 cm
(c) 156.33 cm
(d) 157.33 cm
Answer: (b) 155.33 cm

Question. If the mean of data is 27 and mode is 45 then median is
(a) 30
(b) 27
(c) 32
(d) 33
Answer: (d) 33

Question. The runs scored by a batsman in 35 different matches are given below:
Runs Scored: 0–15, 15–30, 30–45, 45–60, 60–75, 75–90
Frequency: 5, 7, 4, 8, 8, 3
Number of matches in which the batsman scored less than 60 runs are
(a) 16
(b) 24
(c) 8
(d) 19
Answer: (b) 24

Question. \( \sum f_i = 18, \sum f_i x_i = 2p + 24 \) and mean of any distribution is 2, then \( p \) is equal to
(a) 3
(b) 4
(c) 8
(d) 6
Answer: (d) 6

Very Short Answer Questions: 

Question. Consider the following frequency distribution.
Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Frequency: 3, 9, 15, 30, 18, 5
Determine the modal class.
Answer: 30–40

Question. In the formula \( \bar{X} = A + h \left( \frac{\sum f_i u_i}{\sum f_i} \right) \) for finding the mean of grouped frequency distribution, what is the value of \( u_i \)? 
Answer: \( \frac{x_i - a}{h} \)

Question. The time, 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
Find the number of athletes who completed the race in less then 14.6 seconds. 
Answer: 82

Question. In the following distribution:
Monthly income range (in Rs.): More than 10000, More than 13000, More than 16000, More than 19000, More than 22000, More than 25000
Number of families: 100, 85, 69, 50, 33, 15
Find the number of families having income range Rs. 16000 – Rs. 19000.
Answer: 19

Question. Consider the following distribution:
Marks Obtained: Less than 10, Less than 20, Less than 30, Less than 40, Less than 50
No. of students: 02, 10, 15, 30, 40
Find the number of students having marks in range 30–40.
Answer: 15

Question. What is the empirical relation between mean, median and mode?
Answer: 3 Median = Mode + 2 Mean

Short Answer Questions-I: 

Question. Find the mode of the following frequency distribution:
Class Interval: 25–30, 30–35, 35–40, 40–45, 45–50, 50–55
Frequency: 25, 34, 50, 42, 38, 14
Answer: 38.33

Question. Write the median class of the following distribution: 
Classes: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 14, 6, 8, 20, 15, 12, 9
Answer: 30–40

Question. The mean, median and mode of grouped data are always different. State True or False and justify your answer.
Answer: False, it depends on the data

Question. Find the class marks of classes 15–35 and 20–40.
Answer: 25, 30

Question. If the mean of the following distribution is 2.6, then find the value of \( k \).
\( x_i \): 1, 2, 3, 4, 5
\( f_i \): \( k \), 5, 8, 1, 2
Answer: \( k = 4 \)

Short Answer Questions-II: 

Question. Obtain the median for the following frequency distribution.
\( x_i \): 10, 20, 30, 40, 50
\( f_i \): 5, 3, 6, 8, 8
Calculate the median salary of the data. 
Answer: 40

Question. Find the mode of the following frequency distribution.
Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 8, 10, 10, 16, 12, 6, 7
Answer: Mode = 36

Question. The weights of tea in 70 packets is given in the following table:
Weight (In gms.): 200–201, 201–202, 202–203, 203–204, 204–205, 205–206
Number of Packets: 12, 26, 20, 9, 2, 1
Find the modal weight. 
Answer: Mode = 201.7

Question. Find the value of \( p \), if the mean of the following distribution is 20.
\( x \): 15, 17, 19, \( 20+p \), 23
\( f \): 2, 3, 4, 5, 6
Answer: \( p = 1 \)

Question. For the following distribution, calculate mean:
Classes: 25–29, 30–34, 35–39, 40–44, 45–49, 50–54, 55–59
Frequency: 14, 22, 16, 6, 5, 3, 4
Answer: 36.36

Question. Find the mean age of 100 residents of a town from the following data:
Age equal and above (in years): 0, 10, 20, 30, 40, 50, 60, 70
No. of Persons: 100, 90, 75, 50, 25, 15, 5, 0
Answer: 31 years

Question. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below:
Mileage (km/L): 10–12, 12–14, 14–16, 16–18
Number of cars: 7, 12, 18, 13
Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km/L.
Answer: 14.48 km/L

Long Answer Questions: 

Question. Thirty women were examined in a hospital by a doctor and the number of heart beat per minutes were recorded and summarised as follows. Find the mean heart beats per minute for the women, choosing a suitable method.
Number of heart beat per minute: 65–68, 68–71, 71–74, 74–77, 77–80, 80–83, 83–86
Number of women: 2, 4, 3, 8, 7, 4, 2
Answer: 75.9

Question. Find the mean and mode for the following data:
Classes: 10–20, 20–30, 30–40, 40–50, 50–60, 60–70, 70–80
Frequency: 4, 8, 10, 12, 10, 4, 2
[CBSE 2018 (C) (30/1)]
Answer: Mean = 42.2, Mode = 45

Question. Calculate the mean of the following frequency distribution:
Class: 10–30, 30–50, 50–70, 70–90, 90–110, 110–130
Frequency: 5, 8, 12, 20, 3, 2
[CBSE 2019(30/4/2)]
Answer: 65.6

Question. If the mean of the following frequency distribution is 62.8, then find the missing frequency \( x \):
Class: 0–20, 20–40, 40–60, 60–80, 80–100, 100–120
Frequency: 5, 8, \( x \), 12, 7, 8
[CBSE 2019 (C) (30/1/1)]
Answer: \( x = 10 \)

Question. The annual rainfall record of a city of 66 days is given in the following table:
Rainfall (in cm): 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Number of days: 22, 10, 8, 15, 5, 6
Calculate the median rainfall.
Answer: 21.25 cm

Question. Find the median of the following.
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

Question. The following is the frequency distribution of duration for 100 calls made on a mobile phone:
Duration (in seconds): 95–125, 125–155, 155–185, 185–215, 215–245
Number of calls: 14, 22, 28, 21, 15
Calculate the average duration (in sec.) of a call. Find the median.
Answer: Average = 170.3 sec, Median = 170 sec

Question. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency \( f_1 \) and \( f_2 \).
Classes: 0–20, 20–40, 40–60, 60–80, 80–100, 100–120
Frequency: 5, \( f_1 \), 10, \( f_2 \), 7, 8
Answer: \( f_1 = 8, f_2 = 12 \)