Statistics CBSE Class 9 Mathematics Worksheets Set 03

Exercise I

Question 1. The marks out of 50 of 30 students are: 40, 12, 46, 37, 17, 27, 30, 6, 2, 23, 19, 39, 25, 5, 33, 19, 21, 12, 41, 17, 19, 17, 12, 8, 10, 1, 9, 21, 13, 48. Arrange them in ascending order and present them as grouped data (i) in exclusive form (ii) in inclusive form.
Answer:
Ascending: 1, 2, 5, 6, 8, 9, 10, 12, 12, 12, 13, 17, 17, 17, 19, 19, 19, 21, 21, 23, 25, 27, 30, 33, 37, 39, 40, 41, 46, 48.
(i) Exclusive: 0-10 : 6, 10-20 : 11, 20-30 : 5, 30-40 : 4, 40-50 : 4 (a value equal to the upper limit goes to the next class).
(ii) Inclusive: 0-10 : 7, 11-20 : 10, 21-30 : 6, 31-40 : 4, 41-50 : 3. Both totals are 30.
In simple words: Sort the marks, then tally each class.

Teacher's Note:
1. Mind the boundary values 10, 20, 30, 40.

Question 2. A cumulative frequency table shows: below 10 : 17, below 20 : 22, below 30 : 29, below 40 : 37, below 50 : 50, below 60 : 60. Construct the frequency table.
Answer:
Each class frequency is the difference of consecutive cumulative frequencies.
0-10 : 17, 10-20 : 5, 20-30 : 7, 30-40 : 8, 40-50 : 13, 50-60 : 10. Total = 60.
In simple words: Subtract each cumulative value from the next.

Teacher's Note:
1. The first class equals its cumulative frequency.

Question 3. Make a frequency table from: more than 60 : 0, more than 50 : 16, more than 40 : 40, more than 30 : 75, more than 20 : 87, more than 10 : 92, more than 0 : 100.
Answer:
Frequency of a class = (more than its lower limit) − (more than its upper limit).
0-10 : 100 − 92 = 8, 10-20 : 5, 20-30 : 12, 30-40 : 35, 40-50 : 24, 50-60 : 16 and 60 or more : 0. Total = 100.
In simple words: Subtract each 'more than' count from the one before it.

Teacher's Note:
1. Check that the frequencies add up to 100.

Question 4. The number of literate females in the age groups 10-17, 18-25, 26-33, 34-41, 42-49 and 50-57 is 300, 980, 740, 580, 260 and 140. Describe how to draw the histogram.
Answer:
The gap between classes is 1, so subtract 0.5 from each lower limit and add 0.5 to each upper limit.
Classes: 9.5-17.5, 17.5-25.5, 25.5-33.5, 33.5-41.5, 41.5-49.5, 49.5-57.5.
Mark these boundaries on the x-axis and draw touching rectangles of heights 300, 980, 740, 580, 260 and 140 (the class width, 8, is equal for all).
In simple words: Make the classes continuous, then draw equal-width rectangles.

Teacher's Note:
1. Use a scale such as 1 unit = 100 females.

Question 5. Describe the histogram for: class interval 10-15, 15-20, 20-30, 30-50, 50-80 with frequency 6, 10, 10, 8, 18.
Answer:
Widths are 5, 5, 10, 20, 30, so adjust to the minimum width 5.
Adjusted heights: 6, 10, 10 × 5/10 = 5, 8 × 5/20 = 2, 18 × 5/30 = 3.
Draw touching rectangles on 10-15, 15-20, 20-30, 30-50, 50-80 with heights 6, 10, 5, 2 and 3.
In simple words: Divide each frequency by its width in units of 5.

Teacher's Note:
1. Heights are not the raw frequencies for unequal classes.

Question 6. The mean of 8 numbers is 25. If 5 is subtracted from each number, what is the new mean?
Answer:
Subtracting 5 from every number lowers the mean by 5.
New mean = 25 − 5 = 20.
In simple words: The mean shifts by the same constant.

Teacher's Note:
1. Sum falls by 40, from 200 to 160, and 160 ÷ 8 = 20.

Question 7. The mean of the following data is 20.6. Find p: x = 10, 15, p, 25, 35 with f = 3, 10, 25, 7, 5.
Answer:
Σf = 50, so Σfx = 20.6 × 50 = 1030.
Σfx = 30 + 150 + 25p + 175 + 175 = 530 + 25p.
530 + 25p = 1030, so p = 20.
In simple words: Use mean = Σfx ÷ Σf.

Teacher's Note:
1. Check: 530 + 500 = 1030.

Question 8. Find the missing frequencies f1 and f2 if the mean of the distribution is 1.46: accidents 0, 1, 2, 3, 4, 5 with frequencies 46, f1, f2, 25, 10, 5 and total 200.
Answer:
46 + f1 + f2 + 25 + 10 + 5 = 200, so f1 + f2 = 114.
Σfx = f1 + 2f2 + 75 + 40 + 25 = 1.46 × 200 = 292, so f1 + 2f2 = 152.
Subtracting: f2 = 38 and f1 = 76.
In simple words: Two equations from the total and from the mean.

Teacher's Note:
1. Answer: f1 = 76, f2 = 38.

Question 9. The daily minimum temperatures (in degrees F) in a week were 35.5, 30.8, 27.3, 32.1, 23.8 and 29.9. Find the mean temperature.
Answer:
Sum = 35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9 = 179.4.
Mean = 179.4 ÷ 6 = 29.9 degrees F.
In simple words: Add the six values and divide by 6.

Teacher's Note:
1. The source lists six days.

Question 10. The pocket expenses (in rupees) of 206 students are: 0-5 : 10, 5-10 : 16, 10-15 : 30, 15-20 : 42, 20-25 : 50, 25-30 : 30, 30-35 : 16, 35-40 : 12. Describe how to construct the frequency polygon.
Answer:
Class marks: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5, 32.5, 37.5.
Add the end class marks −2.5 and 42.5 with frequency 0.
Plot the points (class mark, frequency) and join them by line segments, from (−2.5, 0) to (42.5, 0). The peak is at (22.5, 50).
In simple words: Plot class mark against frequency and close the ends at zero.

Teacher's Note:
1. The total frequency 206 is a check.