CBSE Class 10 Mathematics Statistics MCQs Set 08

Practice MCQs for Class 10 Mathematics Chapter 13 Statistics

Access targeted multiple-choice questions for Chapter 13 Statistics designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Chapter 13 Statistics Questions and Solutions

Access the complete set of multiple-choice questions for Chapter 13 Statistics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question: The middle most observation of every data arranged in order is called
(a) mode
(b) median
(c) mean
(d) deviation
Show Answer & Explanation

Answer: (b) median

Explanation:
1. After arranging the observations in order, the middle observation is the median.

Question: For some data \(x_1, x_2, \ldots, x_n\) with respective frequencies \(f_1, f_2, \ldots, f_n\), the value of \(\sum_{i=1}^{n} f_i (x_i - \bar{x})\) is equal to
(a) \(n\bar{x}\)
(b) \(1\)
(c) \(\sum f_i\)
(d) \(0\)
Show Answer & Explanation

Answer: (d) \(0\)

Explanation:
1. \(\sum f_i (x_i - \bar{x}) = \sum f_i x_i - \bar{x}\sum f_i = \sum f_i x_i - n\bar{x}\).
2. Since \(\bar{x} = \frac{\sum f_i x_i}{n}\), this is \(0\).

Question: The median and mode respectively of a frequency distribution are \(26\) and \(29\). Then its mean is
(a) 27.5
(b) 24.5
(c) 28.4
(d) 25.8
Show Answer & Explanation

Answer: (b) 24.5

Explanation:
1. Empirical relation: \(\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}\).
2. \(29 = 3(26) - 2M\), so \(2M = 49\) and \(M = 24.5\).

Question: The cumulative frequency table is useful in determining
(a) Mean
(b) Median
(c) Mode
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Cumulative frequencies help in finding the mean, median and mode of grouped data.

Question: In a frequency distribution, the mid value of a class is \(10\) and the width of the class is \(6\). The lower limit of the class is
(a) 6
(b) 7
(c) 8
(d) 12
Show Answer & Explanation

Answer: (b) 7

Explanation:
1. Let the limits be \(x\) (upper) and \(y\) (lower). Then \(\frac{x + y}{2} = 10\), so \(x + y = 20\).
2. \(x - y = 6\). Solving: \(y = 7\).

Question: Consider the following frequency distribution of the heights of \(60\) students of a class: \(150\)–\(155\): \(15\), \(155\)–\(160\): \(13\), \(160\)–\(165\): \(10\), \(165\)–\(170\): \(8\), \(170\)–\(175\): \(9\), \(175\)–\(180\): \(5\). The upper limit of the median class is
(a) 165
(b) 155
(c) 160
(d) 170
Show Answer & Explanation

Answer: (a) 165

Explanation:
1. Cumulative frequencies are \(15, 28, 38, 46, 55, 60\). \(\frac{N}{2} = 30\).
2. The first cumulative frequency greater than \(30\) is \(38\), so the median class is \(160\)–\(165\). Its upper limit is \(165\).

Question: A set of numbers consists of three \(4\)'s, five \(5\)'s, six \(6\)'s, eight \(8\)'s and seven \(10\)'s. The mode of this set of numbers is
(a) 6
(b) 7
(c) 8
(d) 10
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. The value that occurs most often is \(8\) (eight times).

Question: If the mean of the observations \(x, x+3, x+5, x+7\) and \(x+10\) is \(9\), the mean of the last three observations is
(a) \(10\frac{1}{3}\)
(b) \(10\frac{2}{3}\)
(c) \(11\frac{1}{3}\)
(d) \(11\frac{2}{3}\)
Show Answer & Explanation

Answer: (c) \(11\frac{1}{3}\)

Explanation:
1. \(\frac{5x + 25}{5} = 9\), so \(x = 4\).
2. Last three observations: \(x+5, x+7, x+10\), with sum \(3x + 22 = 34\). Mean \(= \frac{34}{3} = 11\frac{1}{3}\).

Question: The mean and median of the data \(a, b\) and \(c\) are \(50\) and \(35\) respectively, where \(a < b < c\). If \(c - a = 55\), then \(b - a\) is
(a) 8
(b) 7
(c) 3
(d) 5
Show Answer & Explanation

Answer: (d) 5

Explanation:
1. Median \(b = 35\). Mean: \(a + b + c = 150\), so \(a + c = 115\).
2. With \(c - a = 55\): \(a = 30\). Therefore \(b - a = 35 - 30 = 5\).

Question: The frequency distribution of marks is: \(0\)–\(10\): \(3\), \(10\)–\(20\): \(9\), \(20\)–\(30\): \(16\), \(30\)–\(40\): \(29\), \(40\)–\(50\): \(18\), \(50\)–\(60\): \(5\). The modal class is
(a) 10–20
(b) 20–30
(c) 30–40
(d) 50–60
Show Answer & Explanation

Answer: (c) 30–40

Explanation:
1. The maximum frequency is \(29\), and it belongs to the class \(30\)–\(40\).
2. So the modal class is \(30\)–\(40\).

Question: If \(\bar{X}, M\) and \(Z\) denote mean, median and mode of a data, and \(\bar{X} : M = 9 : 8\), then the ratio \(M : Z\) is
(a) 3 : 4
(b) 4 : 9
(c) 4 : 3
(d) 2 : 5
Show Answer & Explanation

Answer: (c) 4 : 3

Explanation:
1. \(Z = 3M - 2\bar{X}\), and \(\bar{X} = \frac{9}{8}M\).
2. \(Z = 3M - \frac{9}{4}M = \frac{3}{4}M\). So \(M : Z = 4 : 3\).

Question: A student noted the number of cars passing through a spot on a road for \(100\) periods of \(3\) min each. The frequencies are: \(0\)–\(10\): \(7\), \(10\)–\(20\): \(14\), \(20\)–\(30\): \(13\), \(30\)–\(40\): \(12\), \(40\)–\(50\): \(20\), \(50\)–\(60\): \(11\), \(60\)–\(70\): \(15\), \(70\)–\(80\): \(8\). The mode of the number of cars is
(a) 34.7
(b) 44.7
(c) 54.7
(d) 64.7
Show Answer & Explanation

Answer: (b) 44.7

Explanation:
1. Modal class \(40\)–\(50\): \(l = 40\), \(f_1 = 20\), \(f_0 = 12\), \(f_2 = 11\), \(h = 10\).
2. \(M_o = 40 + \frac{20 - 12}{2(20) - 12 - 11} \times 10 = 40 + \frac{8}{17} \times 10 \approx 44.7\).

Question: If the mean of \(a, b, c\) is \(M\) and \(ab + bc + ca = 0\), the mean of \(a^2, b^2\) and \(c^2\) is \(KM^2\). Then \(K\) is equal to
(a) 3
(b) 9
(c) 6
(d) 4
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\), so \((3M)^2 = 3KM^2\).
2. \(9M^2 = 3KM^2\), giving \(K = 3\).

Question: In the formula \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\) for finding the mean of grouped data, \(d_i\) are deviations of \(x_i\) from \(a\), where \(x_i\) are the
(a) lower limits of the classes
(b) upper limits of the classes
(c) mid-points of the classes
(d) frequencies of the class marks
Show Answer & Explanation

Answer: (c) mid-points of the classes

Explanation:
1. Here \(d_i = x_i - a\), and \(x_i\) is the mid-point of each class.

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Download Chapter MCQs: Class 10 Mathematics Chapter 13 Statistics

Class 10 Mathematics Chapter 13 Statistics Objective Test Questions

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FAQs

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