Class 10 Maths Chapter 13 Statistics MCQ / Objective Questions with Answers in English

Contents hide
1 Direct Questions
2 Assertion-Reason
3 Match the Following
4 Statement-Based Questions
5 Arrange in the Correct Order
6 Fill in the Blanks
7 Odd One Out
8 Passage-Based Questions
9 True / False

21 multiple-choice (MCQ / objective) questions from this chapter, in the common exam-style shapes. Every question comes with an explanation along with the answer, so you remember why the option is correct, not just which one it is.

1 Direct Questions

Direct Questions
  1. 1 markIf ∑fi=40\sum f_i = 40 and ∑fixi=1320\sum f_i x_i = 1320, the mean of the data is:
    (A) 3.3(B) 30(C) 33(D) 330
  2. 1 markIn the step-deviation method, a=25a = 25, h=10h = 10, ∑fiui=15\sum f_i u_i = 15 and ∑fi=50\sum f_i = 50. The mean is:
    (A) 28(B) 25.3(C) 22(D) 40
  3. 1 markThe values 2, 4, 6, 8 occur with frequencies 3, 2, 4, 1 respectively. Their mean is:
    (A) 4.6(B) 5(C) 4.5(D) 46
  4. 1 markFor a modal class 20 – 30 with f1=18f_1 = 18, f0=12f_0 = 12, f2=9f_2 = 9, the mode is:
    (A) 26(B) 22(C) 24(D) 25
  5. 1 markIf the modal class is the very first class of the distribution, then in the mode formula:
    (A) f0=0f_0 = 0(B) f2=0f_2 = 0(C) f1=0f_1 = 0(D) l=0l = 0
  6. 1 markFor the classes 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50 with frequencies 3, 7, 12, 8, 5, the median class is:
    (A) 10 – 20(B) 20 – 30(C) 30 – 40(D) 40 – 50
  7. 1 markFor a distribution, mean = 30 and median = 32. By the empirical relationship, the mode is:
    (A) 34(B) 28(C) 31(D) 36
  8. 1 markIn a less than type distribution, 12 students scored less than 20 and 30 scored less than 30. The frequency of the class 20 – 30 is:
    (A) 18(B) 30(C) 42(D) 12
  9. 1 markBefore applying the median formula, the class 21 – 30 (in data with classes 11 – 20, 21 – 30, 31 – 40) should be written as:
    (A) 21 – 30.5(B) 20 – 31(C) 20.5 – 29.5(D) 20.5 – 30.5
  10. 1 markA shopkeeper wants to know the shoe size demanded by the largest number of customers. The best measure to use is the:
    (A) mean(B) median(C) class mark(D) mode
  11. 1 markFor 100 observations arranged in ascending order, the median of the ungrouped data is:
    (A) the 50th observation(B) the 51st observation(C) the average of the 50th and 51st observations(D) the 50.5th observation
  12. 1 markThe exact mean of 30 students’ marks is 59.3, but after grouping the direct method gives 62. The difference is because:
    (A) the wrong method was used(B) each class is represented by its mid-point (class mark)(C) the assumed mean was chosen badly(D) the frequencies were added wrongly

2 Assertion-Reason

Assertion-Reason
  1. 1 markAssertion (A): The mean of grouped data found by the direct method is an approximate value.
    Reason (R): In grouped data, the frequency of each class is assumed to be centred at its class mark.
    (A) Both A and R are true, and R is the correct explanation of A(B) Both A and R are true, but R is not the correct explanation of A(C) A is true, but R is false(D) A is false, but R is true

3 Match the Following

Match the Following
  1. 1 markMatch Column A with Column B.
    Column A: (i) Direct method (ii) Assumed mean method (iii) Step-deviation method (iv) Class mark
    Column B: (a) a+∑fidi∑fia + \frac{\sum f_i d_i}{\sum f_i} (b) upper limit+lower limit2\frac{\text{upper limit} + \text{lower limit}}{2} (c) ∑fixi∑fi\frac{\sum f_i x_i}{\sum f_i} (d) a+(∑fiui∑fi)ha + \left(\frac{\sum f_i u_i}{\sum f_i}\right)h
    (A) i-c, ii-a, iii-d, iv-b(B) i-a, ii-c, iii-d, iv-b(C) i-c, ii-d, iii-a, iv-b(D) i-b, ii-a, iii-d, iv-c

4 Statement-Based Questions

Statement-Based Questions
  1. 1 markStatement I: In the mode formula, f0f_0 is the frequency of the class succeeding the modal class.
    Statement II: The mode of grouped data is a value inside the modal class.
    (A) Both Statement I and Statement II are true(B) Statement I is true, but Statement II is false(C) Statement I is false, but Statement II is true(D) Both Statement I and Statement II are false

5 Arrange in the Correct Order

Arrange in the Correct Order
  1. 1 markArrange the steps for finding the median of grouped data in the correct order: (1) find n2\frac{n}{2} (2) make the cumulative frequency column (3) substitute ll, cfcf, ff, hh in the formula (4) locate the median class.
    (A) 1, 2, 3, 4(B) 2, 1, 4, 3(C) 2, 4, 1, 3(D) 1, 4, 2, 3

6 Fill in the Blanks

Fill in the Blanks
  1. 1 markThe class interval with the maximum frequency is called the ______ class.
    (A) median(B) cumulative(C) upper(D) modal

7 Odd One Out

Odd One Out
  1. 1 markWhich of the following does NOT belong with the other three?
    (A) Direct method(B) Assumed mean method(C) Median formula(D) Step-deviation method

8 Passage-Based Questions

Passage-Based Questions
Passage
A school recorded the time (in minutes) that 40 students take to reach school. Time 0 – 10: 6 students; 10 – 20: 11 students; 20 – 30: 14 students; 30 – 40: 5 students; 40 – 50: 4 students.
  1. 1 markUsing the passage, the mode of the time taken is:
    (A) 25 minutes(B) 21.5 minutes(C) 22.5 minutes(D) 20 minutes
  2. 1 markUsing the passage, the median time is about:
    (A) 22.14 minutes(B) 22.5 minutes(C) 25 minutes(D) 20.71 minutes

9 True / False

True / False
  1. 1 markConsider: P: The class mark of 5 – 15 is 10. Q: The median class is the class with the maximum frequency. R: 3 Median = Mode + 2 Mean. Mark each True (T) or False (F), in the order P, Q, R.
    (A) T T T(B) F F T(C) T F T(D) T F F
Answer Key and Explanations
Direct Questions
  1. (C) xˉ=∑fixi∑fi=132040=33\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{1320}{40} = 33.
  2. (A) xˉ=a+(∑fiui∑fi)×h=25+1550×10=25+3=28\bar{x} = a + \left(\dfrac{\sum f_i u_i}{\sum f_i}\right)\times h = 25 + \dfrac{15}{50}\times 10 = 25 + 3 = 28.
  3. (A) ∑fi=10\sum f_i = 10, ∑fixi=6+8+24+8=46\sum f_i x_i = 6 + 8 + 24 + 8 = 46, so xˉ=4610=4.6\bar{x} = \dfrac{46}{10} = 4.6.
  4. (C) Mode =20+(18−1236−12−9)×10=20+615×10=20+4=24= 20 + \left(\dfrac{18 – 12}{36 – 12 – 9}\right)\times 10 = 20 + \dfrac{6}{15}\times 10 = 20 + 4 = 24.
  5. (A) f0f_0 is the frequency of the class preceding the modal class. The first class has no class before it, so f0=0f_0 = 0.
  6. (B) n=35n = 35, n2=17.5\frac{n}{2} = 17.5. Cumulative frequencies: 3, 10, 22, … The first cf greater than 17.5 is 22, for 20 – 30.
  7. (D) 3 Median=Mode+2 Mean⇒Mode=96−60=363\,\text{Median} = \text{Mode} + 2\,\text{Mean} \Rightarrow \text{Mode} = 96 – 60 = 36.
  8. (A) Frequency of 20 – 30 =30−12=18= 30 – 12 = 18 (subtract consecutive cumulative frequencies).
  9. (D) The gap between 20 and 21 is 1, so subtract 0.5 from the lower limit and add 0.5 to the upper limit: 20.5 – 30.5.
  10. (D) The mode gives the most frequent value or most popular item, such as the consumer item in greatest demand.
  11. (C) n=100n = 100 is even, so the median is the average of the (n2)\left(\frac{n}{2}\right)th and (n2+1)\left(\frac{n}{2} + 1\right)th observations, the 50th and 51st.
  12. (B) Grouped mean uses the mid-point assumption: every value in a class is taken to be the class mark. So 59.3 is exact and 62 is an approximate mean.
Assertion-Reason
  1. (A) Both are true. A is true because the exact values are lost when data is grouped; R gives the reason: every value is replaced by the class mark (mid-point assumption).
Match the Following
  1. (A) Direct =∑fixi∑fi= \frac{\sum f_i x_i}{\sum f_i} (c); assumed mean =a+∑fidi∑fi= a + \frac{\sum f_i d_i}{\sum f_i} (a); step-deviation =a+(∑fiui∑fi)h= a + \left(\frac{\sum f_i u_i}{\sum f_i}\right)h (d); class mark = (upper + lower)/2 (b).
Statement-Based Questions
  1. (C) Statement I is false: f0f_0 is the frequency of the class preceding the modal class (f2f_2 is the succeeding one). Statement II is true.
Arrange in the Correct Order
  1. (B) First build the cumulative frequencies (2), then find n2\frac{n}{2} (1), then the class whose cf is greater than and nearest to n2\frac{n}{2} (4), then use the formula (3). Steps (1) and (2) do not depend on each other and may be done in either order; B is the only option in which both come before locating the median class (4), and the formula (3) comes last.
Fill in the Blanks
  1. (D) The class with the maximum frequency is the modal class; the mode lies inside it.
Odd One Out
  1. (C) Direct, assumed mean and step-deviation are three methods for the mean; the median formula finds a different measure.
Passage-Based Questions
  1. (C) Modal class 20 – 30: l=20l = 20, f1=14f_1 = 14, f0=11f_0 = 11, f2=5f_2 = 5, h=10h = 10. Mode =20+14−1128−11−5×10=20+2.5=22.5= 20 + \frac{14 – 11}{28 – 11 – 5}\times 10 = 20 + 2.5 = 22.5 minutes.
  2. (A) n=40n = 40, n2=20\frac{n}{2} = 20; cf: 6, 17, 31, … so the median class is 20 – 30 with cf=17cf = 17, f=14f = 14. Median =20+20−1714×10=22.14= 20 + \frac{20 – 17}{14}\times 10 = 22.14 minutes.
True / False
  1. (C) P is true: 5+152=10\frac{5 + 15}{2} = 10. Q is false: that describes the modal class. R is true.
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