CBSE Class 10 Mathematics Statistics MCQs Set 12

Answer: (c) 30

Explanation:
1. The number of families with income at least Rs. 15,000 is 115 and with income at least Rs. 20,000 is 85.
2. Families with income from Rs. 15,000 up to (but not including) Rs. 20,000 = 115 − 85 = 30.

Teacher's Note:
1. Subtract the two cumulative counts of the 'more than or equal to' table.
2. The difference of consecutive entries is the frequency of that class.

Answer: (c) 7

Explanation:
1. Number of students with marks of at least 30 is 45 and with marks of at least 40 is 38.
2. Marks from 30 up to 39 (that is, more than 29 but less than 40) = 45 − 38 = 7.

Teacher's Note:
1. Marks 30 to 39 fall in the class 30 – 40.
2. Frequency = difference of consecutive cumulative entries.

Answer: (d) 59

Explanation:
1. Students with height less than 165 cm are in the first three classes: 15 + 16 + 28 = 59.

Teacher's Note:
1. Add the frequencies of all classes below 165.
2. This is the 'less than' cumulative frequency at 165.

Answer: (c) 0 – 20,000

Explanation:
1. Frequencies from the 'below' table: 0 – 20,000: 16; 20,000 – 40,000: 25 − 16 = 9; 40,000 – 60,000: 32 − 25 = 7; 60,000 – 80,000: 38 − 32 = 6; 80,000 – 1,00,000: 45 − 38 = 7.
2. The highest frequency is 16, so the modal class is 0 – 20,000.

Teacher's Note:
1. Convert the cumulative table back to class frequencies first.
2. The modal class is the class with the greatest frequency.

Answer: (c) 18

Explanation:
1. Class mark = \( \frac{15.5 + 20.5}{2} = \frac{36}{2} = 18 \).

Teacher's Note:
1. Class mark is the mid-point of the class.
2. Average of the upper and lower limits.

Answer: (d) frequency of the class succeeding the modal class

Explanation:
1. In Mode = \( l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \), f₁ is the frequency of the modal class, f₀ is that of the class preceding it, and f₂ is that of the class succeeding it.

Teacher's Note:
1. Remember: f₀ before, f₁ modal, f₂ after.
2. h is the class size and l the lower limit of the modal class.

Answer: (a) 9

Explanation:
1. Less than 60: 47 and less than 50: 38.
2. Frequency of the class 50 – 60 = 47 − 38 = 9.

Teacher's Note:
1. Subtract consecutive cumulative frequencies.
2. Check that the class 50 – 60 lies between the two limits.

Answer: (c) 32

Explanation:
1. Total frequency = 12 + 26 + 10 + 9 + 15 = 72, so n/2 = 36.
2. Cumulative frequencies: 12, 38, 48, 57, 72. The first value ≥ 36 is 38, so the median class is 8 – 16.
3. The highest frequency is 26, so the modal class is also 8 – 16.
4. Sum of the upper limits = 16 + 16 = 32.

Teacher's Note:
1. Find n/2 and compare with the cumulative frequencies.
2. Both classes happen to be the same here.

Answer: (a) 80 – 100

Explanation:
1. Frequencies from the 'more than or equal to' table: 0 – 20: 53 − 51 = 2; 20 – 40: 51 − 45 = 6; 40 – 60: 45 − 37 = 8; 60 – 80: 37 − 25 = 12; 80 – 100: 25.
2. The highest frequency is 25, so the modal class is 80 – 100.

Teacher's Note:
1. The last entry of a 'more than' table is the frequency of the last class.
2. Differences give the other frequencies.

Answer: (b) 15

Explanation:
1. Total = 15 + 12 + 18 + 16 + 9 = 70, so n/2 = 35.
2. Cumulative frequencies: 15, 27, 45, 61, 70. The median class is 30 – 45, with upper limit 45.
3. The highest frequency is 18, so the modal class is 30 – 45, with lower limit 30.
4. Difference = 45 − 30 = 15.

Teacher's Note:
1. The median class and modal class are the same here.
2. Difference of the limits of the same class equals the class size.

Answer: (b) 24

Explanation:
1. Matches with fewer than 60 runs are in the classes 0–15, 15–30, 30–45 and 45–60.
2. 5 + 7 + 4 + 8 = 24.

Teacher's Note:
1. Add the frequencies of the classes below 60.
2. Do not include the class 60 – 75.

Answer: (d) 80

Explanation:
1. Frequencies: 0 – 20: 7; 20 – 40: 18 − 7 = 11; 40 – 60: 33 − 18 = 15; 60 – 80: 47 − 33 = 14; 80 – 100: 60 − 47 = 13.
2. n = 60, n/2 = 30, and cumulative frequency 33 is the first value ≥ 30, so the median class is 40 – 60.
3. The highest frequency is 15, so the modal class is also 40 – 60.
4. Sum of the lower limits = 40 + 40 = 80.

Teacher's Note:
1. Convert the 'below' table into class frequencies.
2. Both classes are 40 – 60.

Answer: (c) 145

Explanation:
1. Total = 3 + 9 + 22 + 15 + 8 + 5 + 2 = 64, so n/2 = 32.
2. Cumulative frequencies: 3, 12, 34, ... The first value ≥ 32 is 34, so the median class is 145 – 150.
3. Its lower limit is 145.

Teacher's Note:
1. The median class is where the cumulative frequency first reaches n/2.
2. Write the cumulative column step by step.

Answer: (b) 150

Explanation:
1. The highest frequency is 22, for the class 145 – 150.
2. The upper limit of the modal class is 150.

Teacher's Note:
1. Pick the class with the greatest frequency.
2. Read the upper limit, not the lower.

Answer: (b) 57

Explanation:
1. Students with height less than 160 cm: 3 + 9 + 22 + 15 + 8 = 57.

Teacher's Note:
1. Add up the frequencies of all classes below 160.
2. This is the 'less than 160' cumulative frequency.

Answer: (b) 30

Explanation:
1. Students with height ≥ 150 cm: 15 + 8 + 5 + 2 = 30.

Teacher's Note:
1. Add the frequencies of classes starting at 150 or more.
2. Check: 64 − 34 = 30.

Answer: (c) 12

Explanation:
1. The modal class is 145 – 150. The class preceding it is 140 – 145.
2. Cumulative frequency of 140 – 145 = 3 + 9 = 12.

Teacher's Note:
1. Cumulative frequency adds all frequencies up to and including that class.
2. The preceding class is the one just before the modal class.

Answer: (b) 15

Explanation:
1. The modal class is 145 – 150. The class succeeding it is 150 – 155, with frequency 15.

Teacher's Note:
1. Succeeding means the class that comes next.
2. Read its frequency directly from the table.

Answer: (b) The modal class and the median class of a data may be different.

Explanation:
1. The modal class is the class of highest frequency, while the median class is where the cumulative frequency reaches n/2. They need not be the same class, so (b) is true.
2. For symmetric data the mean, median and mode coincide, so (a) is false.
3. A cumulative frequency table is used to find the median, not the mode, so (c) is false.
4. The mode can be smaller than the mean (for example in a negatively skewed data), so (d) is false.

Teacher's Note:
1. Use the empirical relation Mode = 3 Median − 2 Mean to see how they can differ.
2. Cumulative frequency gives the median.

Answer: (c) The median of ungrouped data and the median calculated when the same data is grouped are always the same.

Explanation:
1. When data is grouped, the exact values are lost and every observation in a class is treated as lying at the class mark, so the grouped median is only an estimate and usually differs from the ungrouped median. So (c) is false.
2. (a) is the assumption used for the mean of grouped data, so it is true.
3. (b) is true as explained by the definition of the two classes.
4. (d) is the definition of the class mark, so it is true.

Teacher's Note:
1. Grouping loses information, so values calculated from grouped data are estimates.
2. The same reason applies to the mean.