CBSE Class 10 Statistics Sure Shot Questions Set 05

CBSE Class 10 Statistics Sure Shot Questions Set E. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Statistics

Short Answer Type Questions 

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:

\( \sum f_i = 30 + f_1 + f_2 \) but \( \sum f_i = 50 \) (given)
So, \( 50 = 30 + f_1 + f_2 \Rightarrow f_1 + f_2 = 20 \) ...(i)
Now, mean = \( \frac{\sum f_ix_i}{\sum f_i} \Rightarrow 62.8 = \frac{2060 + 30f_1 + 70f_2}{50} \Rightarrow 3f_1 + 7f_2 = 108 \) ...(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:

\( \therefore \text{Mean } \bar{x} = \frac{\sum f_ix_i}{\sum f_i} \Rightarrow 53 = \frac{3340 + 70k}{72 + k} \Rightarrow 3340 + 70k = 53(72 + k) \Rightarrow 3340 + 70k = 3816 + 53k \Rightarrow 17k = 476 \Rightarrow k = 28 \)

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:

\( \therefore \bar{x} = a + \left[\frac{\sum f_id_i}{\sum f_i}\right] = 225 + \left[\frac{-350}{25}\right] = 225 - 14 = 211 \)
Thus, the mean daily expenditure of food is ₹ 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:

Since \( \bar{x} = 18, a = 18 \)
\( \therefore \bar{x} = a + \left[\frac{\sum f_id_i}{\sum f_i}\right] \Rightarrow 18 = 18 + \left[\frac{2f - 40}{f + 44}\right] \Rightarrow 0 = \left[\frac{2f - 40}{f + 44}\right] \Rightarrow 0 = 2f - 40 \Rightarrow 2f = 40 \Rightarrow f = 20 \)
Thus, the missing frequency is 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. 
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:

Let assumed mean a = 21 and class size h = 6
Mean, \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} = 21 + \frac{-276}{40} = 21 - 6.9 = 14.1 \)
Hence, mean number of days a student was absent is 14.1.

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? 
Answer: \( x_i \) = class-mark and a = assumed mean.

A = 13
\( \bar{x} = A + \frac{\sum f_id_i}{\sum f_i} = 13 + \frac{74}{50} = 13 + 1.48 = 14.48 \text{ km L}^{-1} \)
Hence, mean mileage of car is 14.48 km/litre. So, the manufacturer's statement is wrong that mileage is 16 km L⁻¹.

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:

Here, a = 110
\( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} = 110 + \frac{-8}{100} = 110 - 0.08 = 109.92 \)
but, seat cannot be in decimal. \( \Rightarrow \bar{x} = 110 \).
Hence, the mean number of seats occupied over the flights is 110.

Question. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency k. [CBSE Standard 2020, 2018]
Daily pocket allowance (in ₹): 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25
Number of children: 3, 6, 9, 13, k, 5, 4
Answer: k = 8

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:

Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{13760}{220} = 62.55 \) (approx)

Case Study Based Questions

I. 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 | Number of States/UTs 15-20 | 3 20-25 | 8 25-30 | 9 30-35 | 10 35-40 | 3 40-45 | 0 45-50 | 0 50-55 | 2

Question. In order to find the mean by direct method, we use the formula
(a) \( \frac{\sum_{i=1}^{n} f_ix_i}{n} \)
(b) \( \frac{n}{\sum_{i=1}^{n} f_ix_i} \)
(c) \( n \times \sum_{i=1}^{n} f_ix_i \)
(d) \( n + \sum_{i=1}^{n} f_ix_i \)
Answer: (a)

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

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

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

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

Case Study Based Questions

I. Females’ Literacy: The education of women helps to remove the social stigma that surrounds it. It is the key to eliminating social evils such as female infanticide, dowry, child marriage, harassment, etc. This will not just help the women of today but of the future generations who can live in a world where gender equality exists which ultimately raises the literacy rate. The following distribution shows the number of literate females in the age group 0 to 60 years of a particular area.
Age (in years): 0-10, 10-20, 20-30, 30-40, 40-50, 50-60
No. of literate females: 350, 1100, 900, 750, 600, 500

Question. The class marks of class 40-50 is
(a) 70
(b) 90
(c) 10
(d) 45
Answer: (d)

Question. The number of literate females whose ages are between 20 years and 50 years is
(a) 1350
(b) 1650
(c) 2000
(d) 2250
Answer: (d)

Question. The modal class of the above distribution is
(a) 0-10
(b) 10-20
(c) 20-30
(d) 30-40
Answer: (b)

Question. The number of literate females whose ages are less than 40 years is
(a) 1450
(b) 2350
(c) 3100
(d) 3700
Answer: (c)

Question. The upper limit of modal class is
(a) 10
(b) 20
(c) 30
(d) 60
Answer: (b)

II. 100 m Race: A stopwatch was used to find the time that it took a group of students to run 100 m.
Time (in sec.): 0-20, 20-40, 40-60, 60-80, 80-100
No. of students: 8, 10, 13, 6, 3

Question. The estimated mean time taken by a student to finish the race is
(a) 54
(b) 63
(c) 43
(d) 50
Answer: (c)

Question. What will be the upper limit of the modal class?
(a) 20
(b) 40
(c) 60
(d) 80
Answer: (c)

Question. The construction of cumulative frequency table is useful in determining the
(a) mean
(b) median
(c) mode
(d) All of the above
Answer: (b)

Question. The sum of lower limits of median class and modal class is
(a) 60
(b) 100
(c) 80
(d) 140
Answer: (c)

Question. How many students finished the race within 1 minute?
(a) 18
(b) 37
(c) 31
(d) 8
Answer: (c)

III. 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 among humans. The following tables 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 years
(b) 34.36 years
(c) 35.91 years
(d) 42.24 years
Answer: (c)

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

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)

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)

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)

Please click the link below to download CBSE Class 10 Statistics Sure Shot Questions Set E.