CBSE Class 10 Mathematics Statistics MCQs Set 11

Answer: (d) 17.5 and 45

Explanation:
1. Class mark = \( \frac{\text{lower limit}+\text{upper limit}}{2} \).
2. For 10 – 25: \( \frac{10+25}{2} = 17.5 \). For 35 – 55: \( \frac{35+55}{2} = 45 \).

Teacher's Note:
1. The class mark is the mid-point of the class.
2. Add the two limits and halve the sum.

Answer: (b) 30 – 40

Explanation:
1. Cumulative frequencies are 4, 8, 16, 26, 38, 46, 50 and N = 50.
2. \( \frac{N}{2} = 25 \). The first cumulative frequency greater than 25 is 26, which belongs to 30 – 40.
3. The median class is 30 – 40.

Teacher's Note:
1. Build the cumulative frequency column first.
2. The median class is the class whose cumulative frequency is just above N/2.

Answer: (a) 20

Explanation:
1. Sum of frequencies = 8 + 15 + p + 12 + 5 = 40 + p.
2. 40 + p = 60, so p = 20.

Teacher's Note:
1. Add all the known frequencies first.
2. Equate the total to the given sum of frequencies.

Answer: (b) 3

Explanation:
1. Mean = \( \frac{\sum f_i x_i}{\sum f_i} \), so \( 3 = \frac{3p+36}{15} \).
2. 45 = 3p + 36, so 3p = 9 and p = 3.

Teacher's Note:
1. Cross-multiply and solve the linear equation.

Answer: (a) 25

Explanation:
1. Empirical relation: Mode = 3 Median − 2 Mean.
2. 15 = 3 Median − 60, so 3 Median = 75 and Median = 25.

Teacher's Note:
1. Mode = 3 Median − 2 Mean is the key relation to remember.

Answer: (b) 5.5

Explanation:
1. The sum of the first 10 natural numbers is \( \frac{10 \times 11}{2} = 55 \).
2. Mean = \( \frac{55}{10} = 5.5 \).

Teacher's Note:
1. Mean of the first n natural numbers is (n + 1)/2.

Answer: (b) Mode = 3 Median − 2 Mean

Explanation:
1. The empirical relation between the three measures of central tendency is Mode = 3 Median − 2 Mean.

Teacher's Note:
1. Learn it as: 3 Median = Mode + 2 Mean.

Answer: (a) 6.5

Explanation:
1. Arrange x in ascending order with cumulative frequencies: 2 (13), 3 (20), 5 (28), 6 (37), 7 (49), 9 (63), 10 (74).
2. N = 74, so the median is the mean of the 37th and 38th observations. The 37th value is 6 and the 38th is 7.
3. Median = \( \frac{6+7}{2} = 6.5 \).

Teacher's Note:
1. When N is even, take the mean of the two middle observations.
2. Always sort the values of x before making the cumulative column.

Answer: (a) 14

Explanation:
1. \( 7 = \frac{4p+63}{17} \), so 4p + 63 = 119.
2. 4p = 56 and p = 14.

Teacher's Note:
1. Mean × Σf = Σfx.

Answer: (a) 78

Explanation:
1. Class marks are 55, 65, 75, 85, 95, so Σfx = 440 + 390 + 900 + 935 + 1235 = 3900.
2. Σf = 50, so mean = \( \frac{3900}{50} = 78 \).

Teacher's Note:
1. Use class marks as x in the direct method.

Answer: (c) \( a\bar{x} \)

Explanation:
1. Mean of the new data = \( \frac{ax_1 + ax_2 + \cdots + ax_n}{n} = a \cdot \frac{x_1 + x_2 + \cdots + x_n}{n} = a\bar{x} \).

Teacher's Note:
1. Multiplying every observation by a multiplies the mean by a.

Answer: (d) 29

Explanation:
1. Mean of the first n natural numbers = \( \frac{n+1}{2} \).
2. \( \frac{n+1}{2} = 15 \), so n + 1 = 30 and n = 29.

Teacher's Note:
1. Sum of first n natural numbers is n(n + 1)/2.

Answer: (b) 25

Explanation:
1. N = 66, so \( \frac{N}{2} = 33 \). Cumulative frequencies are 10, 25, 37, 57, 66, so the median class is 10 – 15.
2. The highest frequency is 20, so the modal class is 15 – 20.
3. Sum of lower limits = 10 + 15 = 25.

Teacher's Note:
1. Find the median class from cumulative frequency and the modal class from the highest frequency.

Answer: (b) 17.5

Explanation:
1. The classes are inclusive, so convert to exclusive classes: 0.5 – 5.5, 5.5 – 11.5, 11.5 – 17.5, 17.5 – 23.5, 23.5 – 29.5.
2. N = 57 and \( \frac{N}{2} = 28.5 \). Cumulative frequencies are 13, 23, 38, 46, 57, so the median class is 11.5 – 17.5.
3. Its upper limit is 17.5.

Teacher's Note:
1. Gaps between classes are removed by subtracting 0.5 from lower limits and adding 0.5 to upper limits.

Answer: (b) centred at the class marks of the classes

Explanation:
1. In the mean of grouped data every observation of a class is taken to be equal to the class mark, so the frequency is centred at the class mark.

Teacher's Note:
1. This is the basic assumption behind the direct, assumed mean and step deviation methods.

Answer: (c) class marks of the classes

Explanation:
1. Here \( x_i \) denotes the class mark (mid-point) of the i-th class and \( d_i = x_i - a \) is its deviation from the assumed mean a.

Teacher's Note:
1. Deviation is always taken for the class marks.

Answer: (c) Mean = Mode = Median

Explanation:
1. In a symmetrical distribution the three measures of central tendency coincide: Mean = Median = Mode.

Teacher's Note:
1. Substituting Mean = Median = Mode in Mode = 3 Median − 2 Mean confirms this.

Answer: (c) 0

Explanation:
1. \( \sum f_i(x_i - \bar{x}) = \sum f_i x_i - \bar{x}\sum f_i = n\bar{x} - n\bar{x} = 0 \).

Teacher's Note:
1. The mean is the balancing point of the data.

Answer: (b) 8

Explanation:
1. Mode − Mean = 12. Since Mode = 3 Median − 2 Mean, Mode − Mean = 3(Median − Mean), so Median − Mean = 4.
2. Mode − Median = (Mode − Mean) − (Median − Mean) = 12 − 4 = 8.

Teacher's Note:
1. Rewrite Mode = 3 Median − 2 Mean as Mode − Mean = 3 (Median − Mean).

Answer: (a) 3

Explanation:
1. Number of students with marks 30 or more = 51 and with marks 40 or more = 48.
2. Frequency of 30 – 40 = 51 − 48 = 3.

Teacher's Note:
1. For a more-than table, subtract the next cumulative value from the present one.

Answer: (b) 8

Explanation:
1. Σfx = 4 + 10 + 3y + 4 + 10 = 28 + 3y and Σf = 12 + y.
2. \( 2.6 = \frac{28+3y}{12+y} \), so 31.2 + 2.6y = 28 + 3y.
3. 0.4y = 3.2, so y = 8.

Teacher's Note:
1. Form Σfx and Σf in terms of y and cross-multiply.

Answer: (c) 82

Explanation:
1. Athletes finishing in less than 14.6 seconds are those in the first four classes.
2. 2 + 4 + 5 + 71 = 82.

Teacher's Note:
1. Less than an upper limit means all classes up to that limit.

Answer: (b) 315

Explanation:
1. The highest frequency is 16, so the modal class is 150 – 155 and its lower limit is 150.
2. \( \frac{N}{2} = 30 \). The first cumulative frequency above 30 is 37, so the median class is 160 – 165 with upper limit 165.
3. Required sum = 150 + 165 = 315.

Teacher's Note:
1. Modal class: highest frequency. Median class: cumulative frequency just above N/2.

Answer: (c) 13

Explanation:
1. Families with income more than Rs. 20000 = 50 and more than Rs. 25000 = 37.
2. Families with income from Rs. 20000 to Rs. 25000 = 50 − 37 = 13.

Teacher's Note:
1. Subtract the larger limit's cumulative value from the smaller limit's.

Answer: (d) 15

Explanation:
1. Employees who received less than 70% are in the classes 50 – 60 and 60 – 70.
2. 9 + 6 = 15.

Teacher's Note:
1. A larger cut means a lower percentage received.

Answer: (c) 16

Explanation:
1. These are the classes 60 – 70, 70 – 80 and 80 – 90.
2. 6 + 8 + 2 = 16.

Teacher's Note:
1. Add the frequencies of the required classes.

Answer: (a) 50 – 60

Explanation:
1. The maximum frequency is 9, so the modal class is 50 – 60.

Teacher's Note:
1. Highest frequency gives the modal class.

Answer: (b) 60 – 70

Explanation:
1. Cumulative frequencies are 9, 15, 23, 25 and \( \frac{N}{2} = 12.5 \).
2. The first cumulative frequency above 12.5 is 15, so the median class is 60 – 70.

Teacher's Note:
1. Use cumulative frequency.

Answer: (a) 3 Median = Mode + 2 Mean

Explanation:
1. From Mode = 3 Median − 2 Mean we get 3 Median = Mode + 2 Mean.

Teacher's Note:
1. Check by substituting Mean = Median = Mode.

Answer: (b) 1

Explanation:
1. For example, 209 and 210 differ by 1. The classes are inclusive and must be made continuous before using the formulae.

Teacher's Note:
1. Subtract half of this gap (0.5) from lower limits and add it to upper limits.

Answer: (d) 219.5 – 229.5

Explanation:
1. Exclusive classes: 199.5 – 209.5, 209.5 – 219.5, 219.5 – 229.5, 229.5 – 239.5, 239.5 – 249.5, with cumulative frequencies 4, 18, 44, 54, 60.
2. \( \frac{N}{2} = 30 \), so the median class is 219.5 – 229.5.

Teacher's Note:
1. Convert inclusive classes to exclusive classes first.

Answer: (b) 18

Explanation:
1. The class before 219.5 – 229.5 is 209.5 – 219.5 with cumulative frequency 4 + 14 = 18.

Teacher's Note:
1. This is c.f. in the median formula.

Answer: (d) 224.12 km

Explanation:
1. Median = \( 219.5 + \frac{30-18}{26} \times 10 = 219.5 + 4.615 = 224.12 \) km.

Teacher's Note:
1. Use l = 219.5, f = 26, c.f. = 18, h = 10.

Answer: (d) 224.29 km

Explanation:
1. Mode = 3 Median − 2 Mean, so Mean = \( \frac{3 \text{ Median} - \text{Mode}}{2} \).
2. Mean = \( \frac{672.36 - 223.78}{2} = 224.29 \) km.

Teacher's Note:
1. Rearrange Mode = 3 Median − 2 Mean for the mean.

Answer: (c) 24

Explanation:
1. The known frequencies add to 76, so 76 + x + y = 100 and x + y = 24.

Teacher's Note:
1. Total frequency is 100.

Answer: (c) 9

Explanation:
1. The median 525 lies in 500 – 600, so l = 500, f = 20, h = 100, N/2 = 50 and c.f. = 2 + 5 + x + 12 + 17 = 36 + x.
2. \( 525 = 500 + \frac{50-36-x}{20} \times 100 \), so 25 = 5(14 − x).
3. 14 − x = 5, so x = 9.

Teacher's Note:
1. Write c.f. in terms of x and solve.

Answer: (b) 600

Explanation:
1. With x = 9 and y = 15, the maximum frequency is 20, so the modal class is 500 – 600 with upper limit 600.

Teacher's Note:
1. Find x and y before comparing frequencies.

Answer: (b) 522 kWh

Explanation:
1. Σfx with class marks 50, 150, ..., 950 and frequencies 2, 5, 9, 12, 17, 20, 15, 9, 7, 4 is 100 + 750 + 2250 + 4200 + 7650 + 11000 + 9750 + 6750 + 5950 + 3800 = 52200.
2. Mean = \( \frac{52200}{100} = 522 \) kWh.

Teacher's Note:
1. Find the missing frequencies first, then use the direct method.

Answer: (d) not necessarily greater, less or equal to the actual mean; it can be any value

Explanation:
1. The assumed mean is only a convenient guess, usually a class mark near the middle, so it may be greater than, less than or equal to the actual mean.

Teacher's Note:
1. The final formula corrects for the difference between A and the actual mean.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Reason: the mode of a data set is the value that occurs most often, so R is true.
2. The highest frequency is 23, so the modal class is 12 – 15 with l = 12, f₁ = 23, f₀ = 21, f₂ = 10, h = 3.
3. Mode = \( 12 + \frac{23-21}{46-21-10} \times 3 = 12 + \frac{6}{15} = 12.4 \), so A is true and R gives the meaning of mode: option (a).

Teacher's Note:
1. Recompute the mode with the formula before judging the Assertion.
2. If both are true and the Reason explains the idea, choose (a).