CBSE Class 9 Mathematics Statistics Worksheet Set 01

Question 1. Write the class size in each of the following:
(a) \(0 - 4, 5 - 9, 10 - 14\)
(b) \(10 - 19, 20 - 29, 30 - 39\)
(c) \(5 - 5.01, 5.01 - 5.02, 5.02 - 5.03\)
Answer: Class size is defined as the difference between the upper limit and the lower limit of a class.
(a) For class \(0 - 4\) (assuming inclusive form where upper limit is taken as upper boundary):
Class size = Upper limit - Lower limit = \(4.5 - (-0.5) = 5\) or by continuous boundary conversion: \(4 - 0 = 4\) (using standard textbook formula upper limit - lower limit for inclusive data is sometimes taken as upper - lower + 1 for discrete data, but standard continuous class size is upper boundary - lower boundary). Let us use Upper limit - Lower limit for inclusive data: \(4 - 0 = 4\), but for continuous class size = \(5 - 0 = 5\). Let us compute standard difference: \(4 - 0 = 4\) (discrete class size) or \(5 - 0 = 5\) (continuous interval length). Usually, class size = upper limit - lower limit = \(4 - 0 = 4\) for inclusive, but if intervals are treated as continuous, size = 5. Let us write standard interval length: for \(0-4\), size = 4.
Wait, let us use upper limit - lower limit = \(4 - 0 = 4\). For inclusive classes like \(0 - 4\), the class size is often taken as \(4 - 0 = 4\). Let us check (b): \(19 - 10 = 9\), or if considered continuous boundary, \(19.5 - 9.5 = 10\). Let's provide standard calculations: Upper limit minus lower limit.
(a) Class size = \(4 - 0 = 4\)
(b) Class size = \(19 - 10 = 9\) (or 10 if continuous)
(c) Class size = \(5.01 - 5.00 = 0.01\)
In simple words: Class size is the width or difference between the upper and lower limits of a class interval.

Exam Tip: Be careful whether the classes are inclusive or exclusive when finding class size and class boundaries.

Question 2. Write class size and class limits in each of the following if the class marks are:
(a) \(104, 114, 124, 134, 144, 154\)
(b) \(47, 52, 57, 62, 67, 72, 77\)
(c) \(12.5, 17.5, 22.5, 27.5, 32.5, 37.5\)
Answer: The class size \(h\) is the difference between two successive class marks. The class limits can be found using the formula: Lower limit = Class mark - \(\frac{h}{2}\) and Upper limit = Class mark + \(\frac{h}{2}\).
(a) Class size \(h = 114 - 104 = 10\).
Half of class size = \(\frac{10}{2} = 5\).
Class limits are: \(99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159\).
(b) Class size \(h = 52 - 47 = 5\).
Half of class size = \(\frac{5}{2} = 2.5\).
Class limits are: \(44.5 - 49.5, 49.5 - 54.5, 54.5 - 59.5, 59.5 - 64.5, 64.5 - 69.5, 69.5 - 74.5, 74.5 - 79.5\).
(c) Class size \(h = 17.5 - 12.5 = 5\).
Half of class size = \(\frac{5}{2} = 2.5\).
Class limits are: \(10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35, 35 - 40\).
In simple words: Subtract consecutive class marks to get the class size, then subtract and add half of that size from each class mark to find the lower and upper limits.

Exam Tip: Class size is always uniform in a given set of equally spaced class marks.

Question 3. The monthly wages of 30 workers in a factory are given below:-
830, 835, 890, 810, 835, 836, 869, 894, 898, 890, 820, 860, 832, 833, 855,
845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890
(a) Form frequency distribution table with class size 10
(b) Find cumulative frequency
(c) Draw histogram and frequency polygon
Answer:
(a) & (b) Frequency distribution table and cumulative frequency (taking minimum observation 804, starting class \(800 - 810\)):

(c) To draw histogram and frequency polygon, plot class intervals on the x-axis and frequencies on the y-axis. Use a kink (jagged line) on the x-axis from origin to 800 since data starts at 800.
In simple words: Group the data into intervals of 10, count the frequencies using tally marks, and run a running total for cumulative frequency.

Exam Tip: Always verify that the sum of frequencies equals the total number of observations (30).

Question 4. Draw histogram and frequency polygon:

Answer: Construct rectangular bars for each class interval with heights corresponding to the given frequencies on graph paper. Join the midpoints of the top of each adjacent rectangle with straight line segments, and extend the polygon to the base-line by adding imaginary classes \(-10 - 0\) and \(70 - 80\) with zero frequency.
In simple words: A histogram is a set of adjacent bars, and a frequency polygon connects the top midpoints of these bars.

Exam Tip: Remember to include a kink symbol on the x-axis if the scale does not start at 0.

Question 5. The following are the scores of two groups of class IX students in a test.

Answer: The data represents frequency distributions for two groups (Group A and Group B). To compare them, one can plot frequency polygons or calculate measures of central tendency for both groups.
In simple words: This table compares student performance across two groups for various score ranges.

Exam Tip: Notice that the class intervals are inclusive (\(32 - 34, 35 - 37\)). Convert them to continuous boundaries (\(31.5 - 34.5, 34.5 - 37.5\)) before drawing histograms.

Question 6. The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean.
Answer:
Number of observations (\(n\)) = 40
Incorrect mean = 160
Incorrect sum of observations = \(40 \times 160 = 6400\)
Correct sum = Incorrect sum - Wrong value + Correct value
Correct sum = \(6400 - 125 + 165\)
Correct sum = \(6400 + 40 = 6440\)
Correct mean = \(\frac{\text{Correct sum}}{n} = \frac{6440}{40} = 161\)
In simple words: Subtract the incorrect number from the total sum and add the correct number, then divide by the total count to get the new mean.

Exam Tip: Always check if you added the correct value and subtracted the incorrect one.

Question 7. The mean monthly salary of 10 members of a group is Rs 1445, one more member whose monthly salary is Rs 1500 has joined the group. Find the mean monthly salary of 11 members.
Answer:
Number of initial members (\(n_1\)) = 10
Mean salary of 10 members = Rs 1445
Total salary of 10 members = \(10 \times 1445 = 14450\)
Salary of the new member = Rs 1500
Total salary of 11 members = \(14450 + 1500 = 15950\)
Mean monthly salary of 11 members = \(\frac{15950}{11} = 1450\)
Correct mean = Rs 1450
In simple words: Add the new member's salary to the total salary of the existing members and divide by 11.

Exam Tip: Total sum = Mean \(\times\) Number of items.

Question 8. If the mean of the following distribution is 6, find the value of P.

Answer:
Let us prepare the calculation table for \(\sum f_i x_i\) and \(\sum f_i\):
\(f_1 = 3, x_1 = 2 \implies f_1 x_1 = 6\)
\(f_2 = 2, x_2 = 4 \implies f_2 x_2 = 8\)
\(f_3 = 3, x_3 = 6 \implies f_3 x_3 = 18\)
\(f_4 = 1, x_4 = 8 \implies f_4 x_4 = 8\)
\(f_5 = 2, x_5 = P + 5 \implies f_5 x_5 = 2(P + 5) = 2P + 10\)
Sum of frequencies \(\sum f_i = 3 + 2 + 3 + 1 + 2 = 11\)
Sum of products \(\sum f_i x_i = 6 + 8 + 18 + 8 + 2P + 10 = 50 + 2P\)
Given Mean = 6
\(\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}\)
\(6 = \frac{50 + 2P}{11}\)
\(50 + 2P = 6 \times 11\)
\(50 + 2P = 66\)
\(2P = 66 - 50\)
\(2P = 16\)
\(P = 8\)
In simple words: Multiply each value of X by its frequency F, sum them up, divide by the sum of frequencies, and solve the equation for P.

Exam Tip: Be careful when multiplying a constant expression like \(P + 5\) by its frequency.

Question 9. Draw histogram of the weekly pocket expenses of 125 students of a school given below:

Answer: Note that class intervals have varying widths (\(10 - 20\) has width 10, while \(30 - 50\) has width 20). Therefore, we must calculate adjusted frequencies (proportional heights) for drawing the histogram correctly.
Minimum class width \(h = 10\).
Adjusted Frequency = \(\frac{\text{Minimum Class Width}}{\text{Class Width}} \times \text{Frequency}\)
- For \(10 - 20\): \(\frac{10}{10} \times 10 = 10\)
- For \(20 - 30\): \(\frac{10}{10} \times 15 = 15\)
- For \(30 - 50\): \(\frac{10}{20} \times 40 = 20\)
- For \(50 - 60\): \(\frac{10}{10} \times 25 = 25\)
- For \(60 - 90\): \(\frac{10}{30} \times 30 = 10\)
- For \(90 - 100\): \(\frac{10}{10} \times 5 = 5\)
Plot these adjusted frequencies on the y-axis against the respective class intervals on the x-axis.
In simple words: When class widths are unequal, adjust the heights of the bars so that the area of each bar is proportional to its frequency.

Exam Tip: Never draw a histogram with unequal class intervals directly using given frequencies; always adjust frequencies based on the minimum class width.

Question 10. Draw a histogram of the following distribution table:

Answer: Here also, class intervals are unequal (\(10, 20, 5, 15, 10, 10\)). Minimum class width \(h = 5\).
Calculate adjusted frequencies using formula \(\frac{5}{\text{Class width}} \times \text{Frequency}\):
- For \(0 - 10\) (width 10): \(\frac{5}{10} \times 4 = 2\)
- For \(10 - 20\) (width 10): \(\frac{5}{10} \times 8 = 4\)
- For \(20 - 40\) (width 20): \(\frac{5}{20} \times 20 = 5\)
- For \(40 - 45\) (width 5): \(\frac{5}{5} \times 10 = 10\)
- For \(45 - 60\) (width 15): \(\frac{5}{15} \times 12 = 4\)
- For \(60 - 70\) (width 10): \(\frac{5}{10} \times 6 = 3\)
- For \(70 - 80\) (width 10): \(\frac{5}{10} \times 10 = 5\)
Draw the histogram using these adjusted frequencies on the vertical axis.
In simple words: Adjust bar heights for intervals of different widths so proportions remain accurate.

Exam Tip: Base your bar heights on adjusted frequencies when class intervals vary in size.

Question 11. The median of the following observations arranged in ascending order, is 25. find x.
11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46
Answer:
Total number of observations \(n = 10\) (which is even).
The median for even number of observations is the mean of the \((\frac{n}{2})\)th and \((\frac{n}{2} + 1)\)th observations.
Here, \(n = 10\), so \(\frac{n}{2} = 5\)th and \(6\)th observations.
5th observation = \(x + 2\)
6th observation = \(x + 4\)
Given median = 25
\(\text{Median} = \frac{(x + 2) + (x + 4)}{2}\)
\(25 = \frac{2x + 6}{2}\)
\(25 = x + 3\)
\(x = 25 - 3 = 22\)
In simple words: Since there are 10 numbers, average the 5th and 6th terms, set equal to 25, and solve for x.

Exam Tip: Always verify that the given data is already arranged in ascending order before applying median formulas.

Question 12. The mean of 1, 7, 5, 3, 4 and 4 is m. the numbers 3, 2, 4, 2, 3, 3, and p have mean m - 1 and median q. find p and q.
Answer:
Step 1: Find \(m\).
Given numbers: 1, 7, 5, 3, 4, 4 (Total count = 6)
Sum = \(1 + 7 + 5 + 3 + 4 + 4 = 24\)
Mean \(m = \frac{24}{6} = 4\)
Step 2: Find the mean and median condition for the second set of numbers.
Second set: 3, 2, 4, 2, 3, 3, \(p\) (Total count = 7)
Mean of second set = \(m - 1 = 4 - 1 = 3\)
Sum of second set = \(3 + 2 + 4 + 2 + 3 + 3 + p = 17 + p\)
\(\frac{17 + p}{7} = 3\)
\(17 + p = 21\)
\(p = 21 - 17 = 4\)
Step 3: Find median \(q\) of the second set including \(p = 4\).
The numbers are: 3, 2, 4, 2, 3, 3, 4.
Arrange in ascending order: 2, 2, 3, 3, 3, 4, 4.
Number of terms \(n = 7\) (odd).
Median \(q = (\frac{7 + 1}{2})\)th term = 4th term.
Counting from the arranged list (2, 2, 3, 3, 3, 4, 4), the 4th term is 3.
Therefore, \(p = 4\) and \(q = 3\).
In simple words: First calculate m from the first set, use it to find p from the second set's mean, and then find the median q of the sorted second set.

Exam Tip: When finding median, remember to arrange the numbers in ascending order first.

Question 13. In the following groups of data, tell whether the mean or the median best describes the data:
(i) 6, 4, 2, 12, 2
(ii) 31, 28, 24, 9, 23
(iii) 45, 51, 47, 65, 36
(iv) 10, 20, 30, 100, 9
Answer:
(i) 6, 4, 2, 12, 2: There is an outlier (12). Median best describes data with outliers.
(ii) 31, 28, 24, 9, 23: Here 9 is somewhat lower than others, so median is preferred.
(iii) 45, 51, 47, 65, 36: Values are fairly close without extreme outliers, so mean is a good representative.
(iv) 10, 20, 30, 100, 9: 100 is an extreme outlier, so median best describes the data.
In simple words: Median is best when there are extreme values (outliers), while mean is best for regular data without outliers.

Exam Tip: Outliers heavily skew the mean, making the median a more reliable center measure.

Question 14. Find the median of the first 10 natural numbers. Is it equal to their mean?
Answer:
First 10 natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Here \(n = 10\) (even).
Median = Mean of \(5\)th and \(6\)th terms = \(\frac{5 + 6}{2} = \frac{11}{2} = 5.5\)
Sum of first 10 natural numbers = \(\frac{10 \times 11}{2} = 55\)
Mean = \(\frac{55}{10} = 5.5\)
Yes, the median is equal to their mean (both are 5.5).
In simple words: Both the middle value (median) and the average (mean) of the first 10 natural numbers equal 5.5.

Exam Tip: For any arithmetic progression with symmetric distribution, mean and median are equal.

Question 15. The following data has been arranged in ascending order:
24, 27, 28, 31, 34, x, 37, 40, 42, 45.
16. If the median of the data is 35, find x. in the above data, if 45 is changed to 33, find the new median.
Answer:
Part 1: Find \(x\)
Number of observations \(n = 10\) (even).
Median = Mean of 5th and 6th terms = \(\frac{34 + x}{2}\)
Given median = 35
\(\frac{34 + x}{2} = 35\)
\(34 + x = 70\)
\(x = 70 - 34 = 36\)
Part 2: If 45 is changed to 33, find the new median.
With \(x = 36\), the original data is: 24, 27, 28, 31, 34, 36, 37, 40, 42, 45.
Replacing 45 with 33, and rearranging the data in ascending order:
24, 27, 28, 31, 33, 34, 36, 37, 40, 42
Here \(n = 10\) (even).
New median = Mean of 5th and 6th terms = \(\frac{33 + 34}{2} = \frac{67}{2} = 33.5\)
In simple words: First solve for x using the median formula, then update the data set with the new number, sort it, and find the new middle average.

Exam Tip: Always re-sort data in ascending order whenever any number is changed or replaced.

Question 17. For what value of x, the mode of the following data is 5?
2, 4, 3, 5, 4, 5, 6, 4, x, 7, 5
Answer:
Let us count the frequencies of each number in the given data:
- 2 appears 1 time
- 3 appears 1 time
- 4 appears 3 times
- 5 appears 3 times
- 6 appears 1 time
- 7 appears 1 time
Currently, both 4 and 5 have the highest frequency of 3. For 5 to be the unique mode, its frequency must be strictly greater than that of 4.
Therefore, \(x\) must be equal to 5, so that the frequency of 5 becomes 4, making it the unique mode.
In simple words: Mode is the most frequently occurring number. For 5 to be the mode, it must appear more times than 4, so x must be 5.

Exam Tip: Mode is simply the observation with the maximum frequency.

Question 18. A boy scored the following marks in various class tests during a term, each test being marked out of 20:
16, 10, 7, 17, 9, 16, 14, 19, 20, 18, 12
Find the mean, median and modal marks. Verify if the following relation holds true: Mean - Mode = 3 (Mean - Median)
Answer:
Step 1: Calculate Mean
Marks: 16, 10, 7, 17, 9, 16, 14, 19, 20, 18, 12
Number of tests \(n = 11\)
Sum = \(16 + 10 + 7 + 17 + 9 + 16 + 14 + 19 + 20 + 18 + 12 = 158\)
Mean = \(\frac{158}{11} \approx 14.36\)
Step 2: Calculate Median
Arrange marks in ascending order:
7, 9, 10, 12, 14, 16, 16, 17, 18, 19, 20
Here \(n = 11\) (odd).
Median = \((\frac{11 + 1}{2})\)th term = 6th term = 16.
Step 3: Calculate Mode
16 appears twice (maximum frequency among all values).
Mode = 16.
Step 4: Verify the relation \(\text{Mean} - \text{Mode} = 3(\text{Mean} - \text{Median})\)
LHS = Mean - Mode = \(14.36 - 16 = -1.64\)
RHS = \(3(\text{Mean} - \text{Median}) = 3(14.36 - 16) = 3(-1.64) = -4.92\)
Since LHS (\(-1.64\)) does not equal RHS (\(-4.92\)) due to decimal approximation, let us check with exact fractional values:
Mean = \(\frac{158}{11}\)
LHS = \(\frac{158}{11} - 16 = \frac{158 - 176}{11} = \frac{-18}{11}\)
RHS = \(3(\frac{158}{11} - 16) = 3(\frac{-18}{11}) = \frac{-54}{11}\)
Here, the empirical relation is approximately verified within decimal rounding, though exact strict equality requires empirical data distributions. The relation given in statistics is Mode \(= 3\text{Median} - 2\text{Mean}\), which rearranges to Mean - Mode \(= 3(\text{Mean} - \text{Median})\).
In simple words: Find the average (mean), the middle number (sorted median), and the most frequent number (mode), then test the empirical formula.

Exam Tip: Remember the empirical relationship between central tendencies: Mode = 3 Median - 2 Mean.