Chapter 13: Statistics — Online MCQ Test
MATHS · CLASS 10th · Karnataka State Board
Practice Chapter 13: Statistics with a free chapter-wise online MCQ test.
This chapter covers: Grouped data Frequency distribution Class interval Mean Median Mode Cumulative frequency Ogive Histogram.
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Chapter 13: Statistics — Important Questions & Answers
For a class interval 20-30, what is the class mark?
- A. 20
- B. 25
- C. 30
- D. 10
Answer: B. 25
Class mark = (lower limit + upper limit)/2 = (20 + 30)/2 = 25.
Class mark = (lower limit + upper limit)/2 = (20 + 30)/2 = 25.
Which formula is used to find the mean of grouped data by the direct method?
- A. Mean = Σfi/Σxi
- B. Mean = Σfixi/Σfi
- C. Mean = Σxi/Σfi
- D. Mean = Σfi/Σfixi
Answer: B. Mean = Σfixi/Σfi
For grouped data, the mean is calculated as Σfixi divided by Σfi, where xi is the class mark.
For grouped data, the mean is calculated as Σfixi divided by Σfi, where xi is the class mark.
The class with the highest frequency in a grouped frequency distribution is called the:
- A. Median class
- B. Modal class
- C. Mean class
- D. Cumulative class
Answer: B. Modal class
The modal class is the class interval having the maximum frequency.
The modal class is the class interval having the maximum frequency.
In the median formula, Median = l + [(n/2 - cf)/f] × h, what does cf represent?
- A. Frequency of the median class
- B. Cumulative frequency before the median class
- C. Total frequency
- D. Class size
Answer: B. Cumulative frequency before the median class
In the median formula, cf is the cumulative frequency of the class preceding the median class.
In the median formula, cf is the cumulative frequency of the class preceding the median class.
Find the mean of the grouped data: class intervals 0-10, 10-20, 20-30 with frequencies 4, 6, 10 respectively.
- A. 16
- B. 18
- C. 20
- D. 22
Answer: B. 18
Class marks are 5, 15 and 25. Mean = (4×5 + 6×15 + 10×25)/(4+6+10) = 360/20 = 18.
Class marks are 5, 15 and 25. Mean = (4×5 + 6×15 + 10×25)/(4+6+10) = 360/20 = 18.
For the distribution 0-10: 5, 10-20: 8, 20-30: 12, 30-40: 5, what is the median?
- A. 20.67
- B. 21.67
- C. 22.67
- D. 23.67
Answer: B. 21.67
Here n = 30 and n/2 = 15. Median class is 20-30, so Median = 20 + [(15 - 13)/12] × 10 = 21.67.
Here n = 30 and n/2 = 15. Median class is 20-30, so Median = 20 + [(15 - 13)/12] × 10 = 21.67.
If the classes 0-10, 10-20, 20-30, 30-40 have frequencies 5, x, 10, 5 and the mean is 20, find x.
- A. 5
- B. 8
- C. 10
- D. 12
Answer: C. 10
Using class marks 5, 15, 25, 35, mean = (450 + 15x)/(20 + x) = 20. Solving gives x = 10.
Using class marks 5, 15, 25, 35, mean = (450 + 15x)/(20 + x) = 20. Solving gives x = 10.
For the distribution 0-10: 5, 10-20: x, 20-30: 20, 30-40: 10, the median is 22. Find x.
- A. 15
- B. 16
- C. 17
- D. 18
Answer: C. 17
Here n = 35 + x and median class is 20-30. Using 22 = 20 + [((35 + x)/2 - (5 + x))/20] × 10 gives x = 17.
Here n = 35 + x and median class is 20-30. Using 22 = 20 + [((35 + x)/2 - (5 + x))/20] × 10 gives x = 17.
For the distribution 0-10: 4, 10-20: 8, 20-30: x, 30-40: 10, 40-50: 6, the median is 27. Find x.
- A. 8
- B. 10
- C. 12
- D. 14
Answer: B. 10
Here n = 28 + x and median class is 20-30. Using 27 = 20 + [((28 + x)/2 - 12)/x] × 10 gives x = 10.
Here n = 28 + x and median class is 20-30. Using 27 = 20 + [((28 + x)/2 - 12)/x] × 10 gives x = 10.
For the distribution 0-10: 12, 10-20: 18, 20-30: 18, 30-40: 6, if 10-20 is taken as the modal class for applying the NCERT mode formula, what value is obtained?
- A. 18
- B. 20
- C. 22
- D. 24
Answer: B. 20
Using l = 10, f1 = 18, f0 = 12, f2 = 18, h = 10, Mode = 10 + [(18 - 12)/(36 - 12 - 18)] × 10 = 20.
Using l = 10, f1 = 18, f0 = 12, f2 = 18, h = 10, Mode = 10 + [(18 - 12)/(36 - 12 - 18)] × 10 = 20.