STATISTICS - Questions & Answers

EXERCISE 17

1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) Which of the following variables are discrete?
(i) daily temperature of your city
(ii) sizes of shoes
(iii) distance travelled by a man
(iv) time
Answer: (ii) sizes of shoes. (Shoe sizes are exact numbers like 6, 7, 8 and do not take continuous decimal values like temperature or time.)

(b) The marks obtained by 15 students in a test (out of hundred) are given below :
81, 72, 90, 90, 80, 55, 72, 66, 69, 80, 36, 54, 62, 56 and 58
The range of data is :
(i) 46
(ii) 54
(iii) 90
(iv) 100
Answer: (ii) 54. (Range = Maximum value - Minimum value. Here, maximum marks = 90 and minimum marks = 36. Range = 90 - 36 = 54.)

(c) The class-mark of the class 35-45 is :
(i) 35
(ii) 45
(iii) 40
(iv) 42
Answer: (iii) 40. (Class mark is the mid-point of the class interval. Class mark = (Lower limit + Upper limit) / 2 = (35 + 45) / 2 = 80 / 2 = 40.)

(d) In the class-intervals 1-10, 11-20, 21-30; the class 11-20 after adjustment is :
(i) 0.5-10.5
(ii) 20.5-30.5
(iii) 10.5-30.5
(iv) 10.5-20.5
Answer: (iv) 10.5-20.5. (The adjustment factor is (11 - 10) / 2 = 0.5. We subtract this from the lower limit and add it to the upper limit. So, 11 - 0.5 = 10.5 and 20 + 0.5 = 20.5.)

(e) The class marks of a frequency distribution are 10, 15, 20, 25....
The class corresponding to the class mark 15 is :
(i) 12.5-17.5
(ii) 10-20
(iii) 10.25-17
(iv) 14-16
Answer: (i) 12.5-17.5. (The class size is the difference between two consecutive class marks: 15 - 10 = 5. The lower limit is 15 - (5/2) = 12.5 and upper limit is 15 + (5/2) = 17.5.)

2. State, which of the following variables are continuous and which are discrete :
(a) number of children in your class.
Answer: Discrete (Children can only be counted in whole numbers.)
(b) distance travelled by a car.
Answer: Continuous (Distance can be any value, including decimals like 12.5 km.)
(c) sizes of shoes.
Answer: Discrete (Shoe sizes jump by specific values, not all decimals are possible.)
(d) time.
Answer: Continuous (Time flows continuously and can be measured to exact fractions of a second.)
(e) number of patients in a hospital.
Answer: Discrete (People are counted as whole numbers.)

3. Given below are the marks obtained by 30 students in an examination :
08 17 33 41 47 23 20 34
09 18 42 14 30 19 29 11
36 48 40 24 22 02 16 21
15 32 47 44 33 01
Taking class intervals 1- 10, 11 - 20, ........... 41 - 50; make a frequency table for the above distribution.
Answer:
Class Interval (Marks) | Frequency (No. of Students)
1 - 10 | 4
11 - 20 | 8
21 - 30 | 6
31 - 40 | 6
41 - 50 | 6
Total | 30

4. The marks of 24 candidates in the subject mathematics are given below :
45 48 15 23 30 35 40 11
29 0 3 12 48 50 18 30
15 30 11 42 23 2 3 44
The maximum marks are 50. Make a frequency distribution taking class intervals 0 - 10, 10 - 20, ............ .
Answer:
(Note: In continuous exclusive intervals like 0-10, the upper limit is excluded. So 10 goes to the 10-20 class.)
Class Interval (Marks) | Frequency
0 - 10 | 4
10 - 20 | 6
20 - 30 | 3
30 - 40 | 4
40 - 50 | 6
50 - 60 | 1
Total | 24

5. Fill in the blanks :
(a) A quantity which can vary from one individual to another is called a variable.
(b) Sizes of shoes are discrete variables.
(c) Daily temperature is continuous variable.
(d) The range of the data 7, 13, 6, 25, 18, 20, 16 is 19. (Range = 25 - 6)
(e) In the class interval 35 - 46; the lower limit is 35 and upper limit is 46.
(f) The class mark of class interval 22 - 29 is 25.5. ((22 + 29) / 2)

6. Find the actual lower class limits, upper class limits and the mid-values of the classes :
10 - 19, 20 - 29, 30 - 39 and 40 - 49.
Answer:
Step 1: Find the adjustment factor. (20 - 19) / 2 = 0.5. Subtract 0.5 from lower limits and add 0.5 to upper limits.
Class 10 - 19: Actual Lower Limit = 9.5, Actual Upper Limit = 19.5, Mid-value = (9.5 + 19.5)/2 = 14.5
Class 20 - 29: Actual Lower Limit = 19.5, Actual Upper Limit = 29.5, Mid-value = (19.5 + 29.5)/2 = 24.5
Class 30 - 39: Actual Lower Limit = 29.5, Actual Upper Limit = 39.5, Mid-value = (29.5 + 39.5)/2 = 34.5
Class 40 - 49: Actual Lower Limit = 39.5, Actual Upper Limit = 49.5, Mid-value = (39.5 + 49.5)/2 = 44.5

7. Find the actual lower and upper class limits and also the class marks of the classes :
1.1 - 2.0, 2.1 - 3.0 and 3.1 - 4.0.
Answer:
Step 1: Find the adjustment factor. (2.1 - 2.0) / 2 = 0.1 / 2 = 0.05.
Class 1.1 - 2.0: Actual Limits = 1.05 to 2.05, Class Mark = (1.05 + 2.05) / 2 = 1.55
Class 2.1 - 3.0: Actual Limits = 2.05 to 3.05, Class Mark = (2.05 + 3.05) / 2 = 2.55
Class 3.1 - 4.0: Actual Limits = 3.05 to 4.05, Class Mark = (3.05 + 4.05) / 2 = 3.55

8. Use the table given below to find :
Class Interval: 30-34 (f=7), 35-39 (f=10), 40-44 (f=12), 45-49 (f=13), 50-54 (f=8), 55-59 (f=4)
(a) The actual class limits of the fourth class.
Answer: The fourth class is 45 - 49. Subtract 0.5 from lower and add 0.5 to upper. Actual limits are 44.5 and 49.5.
(b) The class boundaries of the sixth class.
Answer: The sixth class is 55 - 59. Class boundaries (actual limits) are 54.5 and 59.5.
(c) The class mark of the third class.
Answer: The third class is 40 - 44. Class mark = (40 + 44) / 2 = 84 / 2 = 42.
(d) The upper and lower limits of the fifth class.
Answer: The fifth class is 50 - 54. Lower limit is 50, and Upper limit is 54.
(e) The size of the third class.
Answer: The actual limits of the third class are 39.5 to 44.5. Size = True Upper Limit - True Lower Limit = 44.5 - 39.5 = 5.

9. Construct a cumulative frequency distribution table from the frequency table given below :
(i)
Class Interval: 0-8 (f=9), 8-16 (f=13), 16-24 (f=12), 24-32 (f=7), 32-40 (f=15)
Answer (i):
Class Interval | Frequency | Cumulative Frequency
0 - 8 | 9 | 9
8 - 16 | 13 | 22 (9 + 13)
16 - 24 | 12 | 34 (22 + 12)
24 - 32 | 7 | 41 (34 + 7)
32 - 40 | 15 | 56 (41 + 15)

(ii)
Class Interval: 1-10 (f=12), 11-20 (f=18), 21-30 (f=23), 31-40 (f=15), 41-50 (f=10)
Answer (ii):
Class Interval | Frequency | Cumulative Frequency
1 - 10 | 12 | 12
11 - 20 | 18 | 30 (12 + 18)
21 - 30 | 23 | 53 (30 + 23)
31 - 40 | 15 | 68 (53 + 15)
41 - 50 | 10 | 78 (68 + 10)

10. Construct a frequency distribution table from the following cumulative frequency distribution:
(i)
Class Interval: 10-19 (CF=8), 20-29 (CF=19), 30-39 (CF=23), 40-49 (CF=30)
Answer (i):
Step 1: Subtract the previous cumulative frequency to get the simple frequency.
Class Interval | Frequency
10 - 19 | 8
20 - 29 | 11 (19 - 8)
30 - 39 | 4 (23 - 19)
40 - 49 | 7 (30 - 23)

(ii)
C.I.: 5-10 (C.F.=18), 10-15 (C.F.=30), 15-20 (C.F.=46), 20-25 (C.F.=73), 25-30 (C.F.=90)
Answer (ii):
C.I. | Frequency
5 - 10 | 18
10 - 15 | 12 (30 - 18)
15 - 20 | 16 (46 - 30)
20 - 25 | 27 (73 - 46)
25 - 30 | 17 (90 - 73)

11. Construct a frequency polygon for the following distribution :
Class-intervals: 0-4 (f=4), 4-8 (f=7), 8-12 (f=10), 12-16 (f=15), 16-20 (f=11), 20-24 (f=6)
Answer:
Step 1: Find the class marks (mid-values) for each interval.
Class marks are 2, 6, 10, 14, 18, and 22.
Step 2: Add an imaginary class at the beginning (-4 to 0, class mark = -2, frequency = 0) and at the end (24 to 28, class mark = 26, frequency = 0).
Step 3: Plot these points on a graph: (-2, 0), (2, 4), (6, 7), (10, 10), (14, 15), (18, 11), (22, 6), and (26, 0).
Step 4: Connect all these points consecutively using a ruler to form the frequency polygon.

12. Construct a combined histogram and frequency polygon for the following frequency distribution :
Class-intervals: 10-20 (f=3), 20-30 (f=5), 30-40 (f=6), 40-50 (f=4), 50-60 (f=2)
Answer:
Step 1: Draw a histogram by making rectangles on the X-axis for class intervals 10-20, 20-30 etc., with heights corresponding to their frequencies (3, 5, 6, 4, 2) on the Y-axis.
Step 2: Mark the center points at the top of each rectangle. These points are (15, 3), (25, 5), (35, 6), (45, 4), (55, 2).
Step 3: Mark the mid-points of an imaginary class before the first and after the last class on the X-axis: (5, 0) and (65, 0).
Step 4: Join all these mid-points together with straight lines to draw the frequency polygon over the histogram.

13. Construct a frequency polygon for the following data :
Class-intervals: 10-14 (f=5), 15-19 (f=8), 20-24 (f=12), 25-29 (f=9), 30-34 (f=4)
Answer:
Step 1: These are inclusive intervals, so first find the class marks. Class mark = (Lower limit + Upper limit) / 2.
Class marks are 12, 17, 22, 27, and 32.
Step 2: Add an imaginary class at the beginning (class mark = 7, frequency = 0) and at the end (class mark = 37, frequency = 0).
Step 3: Plot these points on a graph paper: (7, 0), (12, 5), (17, 8), (22, 12), (27, 9), (32, 4), and (37, 0).
Step 4: Join them successively with straight lines using a ruler. Put a kink (zig-zag line) on the X-axis near zero to show a break in the scale since we start plotting from 7.

14. The daily wages in a factory are distributed as follows :
Daily wages (in ₹): 125-175 (f=4), 175-225 (f=20), 225-275 (f=22), 275-325 (f=10), 325-375 (f=6)
Draw a frequency polygon for this distribution.
Answer:
Step 1: Find the class marks (mid-points) for the daily wages.
Class marks are 150, 200, 250, 300, and 350.
Step 2: Include imaginary classes at the beginning (class mark = 100, frequency = 0) and at the end (class mark = 400, frequency = 0).
Step 3: Plot the points (100, 0), (150, 4), (200, 20), (250, 22), (300, 10), (350, 6), and (400, 0) on a graph.
Step 4: Join the points with straight lines. Ensure a kink is drawn on the X-axis from zero to indicate a jump in the scale.

1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) The class-mark of class 19-30 is :
(i) (30+19)/2
(ii) (29.5-19.5)/2
(iii) (30.5-19)/2
(iv) (29.5-18.5)/2
Answer: (i) (30+19)/2. (Class mark is calculated by adding the upper limit and lower limit and dividing by 2.)

(b) In a frequency distribution, the mid-value of a class is 10 and width of the class is 6, the lower limit of the class is :
(i) 6
(ii) 7
(iii) 8
(iv) 9
Answer: (ii) 7. (Lower limit = Mid-value - (Width / 2) = 10 - (6 / 2) = 10 - 3 = 7.)

(c) Statement (1) : Let m be the mid-value and x be the upper limit of a class in a continuous frequency distribution, then lower limit of this class is 2m - x.
Statement (2) : For a given class : (lower limit + upper limit) / 2 = mid-value of the class
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Answer: (i) Both the statements are true. (Statement 2 is the exact formula for mid-value. Using that formula: (Lower Limit + x) / 2 = m. Multiply by 2: Lower Limit + x = 2m. Therefore, Lower Limit = 2m - x. So, statement 1 is also mathematically correct.)

(d) Assertion (A) : 30 children were asked about the number of hours they watched TV programme every day. The results are recorded as under.
Number of hours: 0-5 (f=8), 5-10 (f=16), 10-15 (f=4), 15-20 (f=2)
Then the number of children who watched TV for 10 or more hours a day is 22.
Reason (R) : The assertion is not correct as the required number is 4 + 2 = 6
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Answer: (ii) A is false, R is true. (Children watching 10 or more hours belong to the 10-15 and 15-20 categories. The frequency is 4 + 2 = 6, not 22. So the Assertion is false, and the Reason correctly explains it.)

2. Construct a frequency table from the following data :
Marks | No. of students
less than 10 | 6
less than 20 | 15
less than 30 | 30
less than 40 | 39
less than 50 | 53
less than 60 | 70
Answer:
Step 1: Convert 'less than' cumulative frequencies to simple class frequencies by subtracting previous values.
Marks (Class Interval) | No. of students (Frequency)
0 - 10 | 6
10 - 20 | 9 (15 - 6)
20 - 30 | 15 (30 - 15)
30 - 40 | 9 (39 - 30)
40 - 50 | 14 (53 - 39)
50 - 60 | 17 (70 - 53)

3. Construct the frequency distribution table from the following cumulative frequency table:
Ages: Below 4 (0), Below 7 (85), Below 10 (140), Below 13 (243), Below 16 (300)
(i) State the number of students in the age group 10 - 13.
(ii) State the age group which has the least number of students.
Answer:
First, we make the frequency distribution table:
Ages | No. of Students (Frequency)
0 - 4 | 0
4 - 7 | 85 (85 - 0)
7 - 10 | 55 (140 - 85)
10 - 13 | 103 (243 - 140)
13 - 16 | 57 (300 - 243)

(i) The number of students in the age group 10 - 13 is 103.
(ii) The age group which has the least number of students (with a non-zero count) is 7 - 10 (with 55 students).

4. Fill in the blanks in the following table:
Class Interval: 25-34, 35-44, 45-54, 55-64, 65-74, 75-84
Frequency: ........., ........., 21, 16, ........., 12
Cumulative Frequency: 15, 28, ........., ........., 73, .........
Answer:
Class Interval | Frequency | Cumulative Frequency
25 - 34 | 15 | 15
35 - 44 | 13 (28 - 15) | 28
45 - 54 | 21 | 49 (28 + 21)
55 - 64 | 16 | 65 (49 + 16)
65 - 74 | 8 (73 - 65) | 73
75 - 84 | 12 | 85 (73 + 12)

5. The value of π upto 50 decimal places is :
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution table of the digits from 0 to 9 after the decimal place.
(ii) Which are the most and the least occurring digits ?
Answer:
(i) By carefully counting each digit appearing after the decimal point:
Digit | Frequency
0 | 2
1 | 5
2 | 5
3 | 8
4 | 4
5 | 5
6 | 4
7 | 4
8 | 5
9 | 8

(ii) The most occurring digits are 3 and 9 (they both occur 8 times). The least occurring digit is 0 (it occurs only 2 times).

6. Draw frequency polygons for each of the following frequency distributions :
(a) using histogram (b) without using histogram.
(i) C.I.: 10-30 (f=4), 30-50 (f=7), 50-70 (f=5), 70-90 (f=9), 90-110 (f=5), 110-130 (f=6), 130-150 (f=4)
Answer (i):
(a) Using Histogram: Draw adjacent rectangles for the classes 10-30 to 130-150 with respective heights. Find the center of the top of each rectangle. Add points for imaginary classes at (-10 to 10) on the X-axis (0,0) and at (150 to 170) on the X-axis (160,0). Connect these points successively with straight lines.
(b) Without Histogram: Calculate class marks (20, 40, 60, 80, 100, 120, 140). Plot the coordinates: (0,0), (20,4), (40,7), (60,5), (80,9), (100,5), (120,6), (140,4), and (160,0). Connect them with straight lines.

(ii) C.I.: 5-15 (f=8), 15-25 (f=16), 25-35 (f=18), 35-45 (f=14), 45-55 (f=8), 55-65 (f=2)
Answer (ii):
(a) Using Histogram: Draw rectangles for the data and mark the mid-points on the top boundaries. Join these points to the X-axis at (0,0) and (70,0) with straight lines.
(b) Without Histogram: Calculate class marks (10, 20, 30, 40, 50, 60). Plot points: (0,0), (10,8), (20,16), (30,18), (40,14), (50,8), (60,2), and (70,0). Connect with straight lines.

7. Using the class intervals 0-9, 10-19, 20-29,..., construct the frequency distribution for :
15, 8, 12, 7, 13, 16, 22, 29, 35, 49, 37 and 48.
Answer:
Step 1: Group the given numbers into their respective class intervals.
Class Interval | Tally Marks (Frequency)
0 - 9 | 2 (Contains: 8, 7)
10 - 19 | 4 (Contains: 15, 12, 13, 16)
20 - 29 | 2 (Contains: 22, 29)
30 - 39 | 2 (Contains: 35, 37)
40 - 49 | 2 (Contains: 49, 48)

8. Construct the cumulative frequency table for :
Marks: 20-29 (18), 30-39 (23), 40-49 (36), 50-59 (42)
Answer:
Marks | No. of students (Freq) | Cumulative Frequency
20 - 29 | 18 | 18
30 - 39 | 23 | 41 (18 + 23)
40 - 49 | 36 | 77 (41 + 36)
50 - 59 | 42 | 119 (77 + 42)

9. Construct a combined histogram and frequency polygon for the following frequency distribution :
Class-intervals: 20-30 (f=8), 30-40 (f=12), 40-50 (f=15), 50-60 (f=10), 60-70 (f=3)
Answer:
Step 1: On a graph paper, draw a kink near the origin on the X-axis because we are starting from 20 directly.
Step 2: Draw the rectangles of the histogram for classes 20-30 up to 60-70 taking their respective frequencies as heights.
Step 3: Mark the mid-points at the top of each rectangle: (25, 8), (35, 12), (45, 15), (55, 10), and (65, 3).
Step 4: Add points for imaginary intervals on the X-axis: (15, 0) and (75, 0).
Step 5: Join all the points consecutively with straight line segments to complete the polygon over the histogram.

Case-Study Based Question
Srikanth is a class IX Mathematics teacher in Chennai. After a class test, he asks a student to record the marks that all the students obtained. Narayanan scored the least marks 6 in the class and Keerthana scored the highest marks 59 in the class out of 60 marks. He prepares a frequency distribution table using the collected data and draws a histogram as shown below :
(i) What is the width of each class ?
Answer: 10. (The class intervals on the graph are 0-10, 10-20, etc. Width = 10 - 0 = 10.)
(ii) What is the total number of students in class IX ?
Answer: 60 students. (By reading heights of rectangles on the given histogram and adding them: 2 + 10 + 21 + 17 + 9 + 1 = 60.)
(iii) How many students scored 50% and above marks ?
Answer: 27 students. (50% of the total 60 marks is 30 marks. Look at the bars from 30 onwards on the X-axis. Their frequencies are 17, 9, and 1. Sum = 17 + 9 + 1 = 27.)
(iv) What is the range of the collected data ?
Answer: 53. (Range is calculated by subtracting the least mark from the highest mark. Range = 59 - 6 = 53.)