RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
Statistics is the art of turning a messy pile of numbers into a single sentence someone can act on. You will learn to organise raw data into frequency tables, picture it with bar graphs, histograms and frequency polygons, and summarise it with the mean, median and mode. Each summary tells a slightly different story, so choosing the right one matters as much as calculating it.
The range of the data 18, 27, 5, 33, 12, 41 is:
Answer
The range is 36. Range = highest observation - lowest observation. The highest value is 41 and the lowest is 5. Range = 41 - 5 = 36.
The class mark of the class interval 25-35 is:
Answer
The class mark is 30. Class mark = (lower limit + upper limit)/2. = (25 + 35)/2 = 60/2 = 30. (The value 10 is the class size, not the class mark.)
The mean of 5 numbers is 24. If the number 18 among them is replaced by 38, the new mean is:
Answer
The new mean is 28. Original sum = mean × number of observations = 24 × 5 = 120. Replacing 18 by 38 increases the sum by 38 - 18 = 20. New sum = 120 + 20 = 140, and the count is still 5. New mean = 140/5 = 28.
The median of the data 7, 11, 4, 9, 15, 6 is:
Answer
The median is 8. First arrange the data in ascending order: 4, 6, 7, 9, 11, 15. Here n = 6, which is even, so the median is the mean of the 3rd and 4th observations. The 3rd is 7 and the 4th is 9. Median = (7 + 9)/2 = 16/2 = 8.
Assertion (A): For the data 3, 5, 5, 7, 9, 9, 9, 12 the mode is 9. Reason (R): The mode of a data set is the observation that occurs with the greatest frequency.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states the definition of mode correctly. Counting frequencies in the data: 3 occurs once, 5 occurs twice, 7 occurs once, 9 occurs three times and 12 occurs once. The greatest frequency is 3, belonging to the observation 9, so the mode is 9 and Assertion A is true. A follows directly from the definition in R, so R explains A.
The mean of 8 observations is 15. A ninth observation of value 24 is added to the set. Find the new mean.
Answer
Sum of the 8 observations = mean × number = 15 × 8 = 120. After including 24, the new sum = 120 + 24 = 144 and the number of observations is 9. New mean = 144/9 = 16.
Find the mean of the following distribution of marks scored in a quiz: Marks (x): 10, 20, 30, 40 Number of students (f): 3, 5, 8, 4
Answer
Compute each product fx: 10 × 3 = 30 20 × 5 = 100 30 × 8 = 240 40 × 4 = 160 Σfx = 30 + 100 + 240 + 160 = 530. Σf = 3 + 5 + 8 + 4 = 20. Mean = Σfx/Σf = 530/20 = 26.5 marks.
The data 12, 15, x, 20, 25 is written in ascending order and its median is 18. Find x and then the mean of the data.
Answer
There are n = 5 observations, which is odd, so the median is the ((5 + 1)/2)th = 3rd observation. The 3rd observation is x, so x = 18. (This is consistent with the ascending order, since 15 < 18 < 20.) Mean = (12 + 15 + 18 + 20 + 25)/5 = 90/5 = 18.
The number of hours 30 students of a class spent on outdoor games in one week is recorded below: 3, 12, 7, 18, 22, 9, 14, 27, 5, 11, 16, 2, 24, 8, 13, 19, 6, 21, 10, 15, 28, 4, 17, 23, 1, 26, 12, 9, 20, 14 (a) Find the range of the data. (b) Construct a grouped frequency distribution with classes 0-5, 5-10, 10-15, 15-20, 20-25 and 25-30, taking the lower limit as included and the upper limit as excluded. (c) Write the class marks of all six classes. (d) State the class with the highest frequency and how many students spent 15 hours or more.
Answer
(a) The highest value is 28 and the lowest is 1, so range = 28 - 1 = 27 hours. (b) Sorting each value into its class: 0-5 contains 3, 2, 4, 1 → frequency 4. 5-10 contains 7, 9, 5, 8, 6, 9 → frequency 6. 10-15 contains 12, 14, 11, 13, 10, 12, 14 → frequency 7. 15-20 contains 18, 16, 19, 15, 17 → frequency 5. 20-25 contains 22, 24, 21, 23, 20 → frequency 5. 25-30 contains 27, 28, 26 → frequency 3. Check: 4 + 6 + 7 + 5 + 5 + 3 = 30, which matches the number of students. (c) Class mark = (lower limit + upper limit)/2, so the class marks are 2.5, 7.5, 12.5, 17.5, 22.5 and 27.5. (d) The highest frequency is 7, belonging to the class 10-15. Students spending 15 hours or more = 5 + 5 + 3 = 13.
(a) The mean of 10 observations was calculated as 42, but later it was found that one observation had been misread as 26 instead of the correct value 56. Find the correct mean. (b) For the data 14, 18, 14, 20, 25, 14, 18, 22, 30, 25, find the mean, the median and the mode.
Answer
(a) Incorrect sum = incorrect mean × number of observations = 42 × 10 = 420. The misread value was 26 while the true value is 56, so the sum was short by 56 - 26 = 30. Correct sum = 420 + 30 = 450. Correct mean = 450/10 = 45. (b) Sum = 14 + 18 + 14 + 20 + 25 + 14 + 18 + 22 + 30 + 25. Adding step by step: 14, 32, 46, 66, 91, 105, 123, 145, 175, 200. Mean = 200/10 = 20. Arranging in ascending order: 14, 14, 14, 18, 18, 20, 22, 25, 25, 30. Here n = 10 is even, so the median is the mean of the 5th and 6th observations, which are 18 and 20. Median = (18 + 20)/2 = 19. Frequencies: 14 occurs 3 times, 18 twice, 20 once, 22 once, 25 twice and 30 once. The greatest frequency is 3, so the mode is 14.
Read the following and answer the questions that follow: A school library recorded the number of books issued each day over 40 working days. The librarian grouped the results as shown: Books issued: 0-10, 10-20, 20-30, 30-40, 40-50 Number of days: 5, 9, 12, 8, 6 (a) Write the class size and the class mark of the class 20-30. (b) On how many days were fewer than 30 books issued? (c) What percentage of the days saw 30 or more books issued? (d) Which class has the highest frequency, and what type of graph should the librarian draw for this data and why?
Answer
(a) Class size = upper limit - lower limit = 30 - 20 = 10. Class mark of 20-30 = (20 + 30)/2 = 25. (b) Days with fewer than 30 books = 5 + 9 + 12 = 26 days. (c) Days with 30 or more books = 8 + 6 = 14. Percentage = (14/40) × 100 = 35%. (Check: 26 + 14 = 40, the total number of days.) (d) The highest frequency is 12, belonging to the class 20-30. The librarian should draw a histogram, because the data is continuous grouped data with classes that run into one another, so the bars must be drawn adjacent with no gaps and with areas proportional to the frequencies. Here all class widths are equal, so the bar heights may be taken as the frequencies themselves.
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