CBSE Class 10 Statistics Sure Shot Questions Set 02

CBSE Class 10 Statistics Sure Shot Questions Set B. 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.

Topics Covered

• 1. Mean of Grouped Data
• 2. Mode and Median of Grouped Data

1. Mean of Grouped Data

We know that mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations. The mean of the values \( x_1, x_2, x_3, ..., x_n \) of the variable \( x \) is denoted by \( \bar{x} \) and is given by \[ \bar{x} = \frac{x_1 + x_2 + x_3 + ... + x_n}{n} = \frac{\sum_{i=1}^{n} x_i}{n} \] The mean of grouped data can be found by the following methods: (i) Direct method: The mid-points \( x_1, x_2, x_3, ..., x_n \) of the class intervals are taken as values of the variate whose respective frequencies are \( f_1, f_2, f_3, ..., f_n \). Then mean of the data is given by \[ \text{Mean } \bar{x} = \frac{f_1x_1 + f_2x_2 + ... + f_nx_n}{f_1 + f_2 + ... + f_n} \Rightarrow \bar{x} = \frac{\sum f_ix_i}{\sum f_i} \] (ii) Assumed mean method: In this method, first of all obtain the frequency distribution and prepare the frequency table in such a way that its first column consists of class intervals of the variable and the second column consists of the corresponding frequencies. For each class, find the mid-point i.e., class-mark \( x_i = \frac{1}{2} (\text{lower limit + upper limit}) \). Now choose an arbitrary number preferably among the middle values of \( x_i \), in the third column. We call it assumed mean and denote it by \( A \). Find the deviations \( d_i = x_i - A \). Thus, Mean \( \bar{x} = A + \frac{\sum f_id_i}{\sum f_i} \) where, \( A \) is the assumed mean and \( d_i = x_i - A \).

Question. Find the arithmetic mean of 1, 2, 3, ..., n
Answer: We know that, \( \bar{x} = \frac{x_1 + x_2 + ... + x_n}{n} = \frac{\sum x_i}{n} = \frac{1 + 2 + 3 + ... + n}{n} = \frac{1}{n} \left[ \frac{n}{2} (2 + n - 1) \right] \) [\(\because 1 + 2 + 3 + ... + n\) is an A.P. and sum of this A.P. = \( \frac{n}{2} (2 + n - 1) \)] \( = \frac{1}{n} \left[ \frac{n(n + 1)}{2} \right] = \frac{n + 1}{2} \)

Question. If the mean of the following distribution is 6.4, then find the value of ‘p’. 
x: 2, 4, 6, 8, 10, 12
f: 3, p, 5, 3, 2, 1
Answer: \( \bar{x} = \frac{\sum f_ix_i}{\sum f} \Rightarrow 6.4 = \frac{6 \times 3 + 4 \times p + 6 \times 5 + 8 \times 3 + 10 \times 2 + 12 \times 1}{3 + p + 5 + 3 + 2 + 1} = \frac{6 + 4p + 30 + 24 + 20 + 12}{14 + p} = \frac{92 + 4p}{14 + p} \)
\( \Rightarrow 6.4 = \frac{92 + 4p}{14 + p} \Rightarrow 64 = \frac{(92 + 4p) \times 10}{14 + p} \Rightarrow 920 + 40p = 896 + 64p \)
\( \Rightarrow 24p = 24 \Rightarrow p = \frac{24}{24} = 1 \)

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

\( \therefore \bar{x} = A + \frac{\sum f_id_i}{\sum f_i} = 70 - \frac{1120}{225} = 70 - 4.98 = 65.02 \)
\( \Rightarrow \bar{x} = 65.02 \) approximately.

Very Short Answer Type Questions

Choose the correct answer from the given options:
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)

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 \( \sum(f_ix_i - \bar{x}) \) is equal to
(a) 0
(b) –1
(c) 1
(d) 2
Answer: (a)

Question. In the following \( \bar{x} = A + \left(\frac{\sum f_id_i}{\sum f_i}\right) \), 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)

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)

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

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)

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): 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): \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} \)
Answer: (a)

Question. Assertion (A): To find mean of a grouped data, we use \( \bar{x} = a + \frac{\sum f_id_i}{\sum f_i} \) 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)

Question. Answers the following: Find the class-mark of class 25–35.
Answer: Class-mark = \( \frac{25 + 35}{2} = \frac{60}{2} = 30 \)

Question. Find the mean of first ten odd natural numbers.
Answer: 10

Question. If the mean of the first n natural number is 15, then find n. 
Answer: \( \bar{x} = \frac{n + 1}{2} \Rightarrow 15 = \frac{n + 1}{2} \Rightarrow n + 1 = 30 \Rightarrow n = 29 \)

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

Short Answer Type Questions 

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:

Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} \Rightarrow 20.6 = \frac{530 + 25p}{50} \Rightarrow 530 + 25p = 1030 \Rightarrow 25p = 500 \Rightarrow p = 20 \)

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:

Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{326}{40} = 8.15 \)

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

Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{230}{10} = 23 \)

Case Based MCQs

Case I : Read the following passage and answer the questions.
Toll Tax Collection
On a particular day, National Highway Authority of India (NHAI) checked the toll tax collection of a particular toll plaza in Rajasthan.
The following table shows the toll tax paid by drivers and the number of vehicles on that particular day.
Toll tax (in ₹): 30-40, 40-50, 50-60, 60-70, 70-80
Number of vehicles: 80, 110, 120, 70, 40

Question. If A is taken as assumed mean, then the possible value of A is
(a) 32
(b) 42
(c) 85
(d) 55
Answer: (d)

Question. If \( x_i \)’s denote the class marks and \( d_i \)’s denote the deviation of assumed mean (A) from \( x_i \)’s, then the minimum value of \( |d_i| \) is
(a) –200
(b) –100
(c) 0
(d) 100
Answer: (c)

Question. The mean of toll tax received by NHAI by assumed mean method is
(a) ₹ 52
(b) ₹ 52.14
(c) ₹ 52.50
(d) ₹ 53.50
Answer: (b)

Question. The mean of toll tax received by NHAI by direct method is
(a) equal to the mean of toll tax received by NHAI by assumed mean method
(b) greater than the mean of toll tax received by NHAI by assumed mean method
(c) less than the mean of toll tax received by NHAI by assumed mean method
(d) can't say
Answer: (a)

Question. The average toll tax received by NHAI in a day, from that particular toll plaza, is
(a) ₹ 21000.60
(b) ₹ 21900.80
(c) ₹ 30000.90
(d) ₹ 21898.80
Answer: (d)

Case II : Read the following passage and answer the questions.
Distance Analysis of Public Transport Buses
Transport department of a city wants to buy some Electric buses for the city. For which they want to analyse the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 60 existing public transport buses in a day.
Daily distance travelled (in km): 200-209, 210-219, 220-229, 230-239, 240-249
Number of buses: 4, 14, 26, 10, 6

Question. The upper limit of a class and lower limit of its succeeding class is differ by
(a) 9
(b) 1
(c) 10
(d) none of these
Answer: (b)

Question. The median class is
(a) 229.5-239.5
(b) 230-239
(c) 220-229
(d) 219.5-229.5
Answer: (d)

Question. The cumulative frequency of the class preceding the median class is
(a) 14
(b) 18
(c) 26
(d) 10
Answer: (b)

Question. The median of the distance travelled is
(a) 222 km
(b) 225 km
(c) 223 km
(d) none of these
Answer: (d)

Question. If the mode of the distance travelled is 223.78 km, then mean of the distance travelled by the bus is
(a) 225 km
(b) 220 km
(c) 230.29 km
(d) 224.29 km
Answer: (d)

Case III : Read the following passage and answer the questions.
Electric Scooter Manufacturing Company
An electric scooter manufacturing company wants to declare the mileage of their electric scooters. For this, they recorded the mileage (km/charge) of 50 scooters of the same model. Details of which are given in the following table.
Mileage (km/charge): 100-120, 120-140, 140-160, 160-180
Number of scooters: 7, 12, 18, 13

Question. The average mileage is
(a) 140 km/charge
(b) 150 km/charge
(c) 130 km/charge
(d) 144.8 km/charge
Answer: (d)

Question. The modal value of the given data is
(a) 150
(b) 150.91
(c) 145.6
(d) 140.9
Answer: (b)

Question. The median value of the given data is
(a) 140
(b) 146.67
(c) 130
(d) 136.6
Answer: (b)

Question. Assumed mean method is useful in determining the
(a) Mean
(b) Median
(c) Mode
(d) All of these
Answer: (a)

Question. The manufacturer can claim that the mileage for his scooter is
(a) 144 km/charge
(b) 155 km/charge
(c) 165 km/charge
(d) 175 km/charge
Answer: (d)

Assertion & Reasoning Based MCQs

Question. Assertion : If the arithmetic mean of 5, 7, \( x \), 10, 15 is \( x \), then \( x = 9.25 \).
Reason : If \( x_1, x_2, x_3, ..., x_n \) are \( n \) values of a variable \( X \), then the arithmetic mean of these values is given by \( \frac{x_1 + x_2 + x_3 + ... + x_n}{n} \).
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Answer: (a)

Question. Assertion : Consider the following frequency distribution:
Class interval: 10-15, 15-20, 20-25, 25-30, 30-35
Frequency: 5, 9, 12, 6, 8
The modal class is 10-15.
Reason : The class having maximum frequency is called the modal class.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Answer: (d)

Question. Assertion : If for a certain frequency distribution, \( l = 24.5, h = 4, f_0 = 14, f_1 = 14, f_2 = 15 \), then the value of mode is 25.
Reason : Mode of a frequency distribution is given by :
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Answer: (d)

Question. Assertion : Consider the following frequency distribution:
Class interval: 0-4, 4-8, 8-12, 12-16, 16-20
Frequency: 6, 3, 5, 20, 10
The median class is 12-16.
Reason : Let \( n = \sum f_i \). Then, the class whose cumulative frequency is just lesser than \( \frac{n}{2} \) is the median class.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Answer: (c)

Question. Assertion : If the median and mode of a frequency distribution are 8.9 and 9.2 respectively, then its mean is 9.
Reason : Mean, median and mode of a frequency distribution are related as:
mode = 3 median – 2 mean
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Answer: (d)

Question. Assertion : The arithmetic mean of the following frequency distribution is 25.
Class interval: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 5, 18, 15, 16, 6
Reason : \( \text{Mean} (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \), where \( x_i = \frac{1}{2} \) (lower limit + upper limit) of \( i^{th} \) class interval and \( f_i \) is its frequency.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Answer: (a)

Question. Assertion : Consider the following frequency distribution:
Class interval: 3-6, 6-9, 9-12, 12-15, 15-18, 18-21
Frequency: 2, 5, 21, 23, 10, 12
The mode of the above data is 12.4.
Reason : The value of the variable which occurs most often is the mode.
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.
Answer: (a)

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