CBSE Class 10 Mathematics Statistics Worksheet Set 10

CBSE Class 10 Maths Worksheet - Statistics (9). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Section A (1 mark each)

Question 1. Find the sum of lower limits of median class and modal class for the following distribution:

Answer: To find the sum of the lower limits, we first identify the median class and the modal class:

1. Median Class:
We calculate the cumulative frequencies (cf):
- For class 0-5: cf = 10
- For class 5-10: cf = 10 + 15 = 25
- For class 10-15: cf = 25 + 12 = 37
- For class 15-20: cf = 37 + 20 = 57
- For class 20-25: cf = 57 + 9 = 66
The total frequency \( N = 66 \).
We find \( \frac{N}{2} = \frac{66}{2} = 33 \).
The cumulative frequency just greater than 33 is 37, which corresponds to the class interval 10-15.
Therefore, the median class is 10-15, and its lower limit is 10.

2. Modal Class:
The modal class is the class with the highest frequency. Here, the maximum frequency is 20, which belongs to the class interval 15-20.
Thus, the modal class is 15-20, and its lower limit is 15.

3. Sum of Lower Limits:
Sum = Lower limit of median class + Lower limit of modal class
Sum = \( 10 + 15 = 25 \).
Therefore, the sum of the lower limits is 25.
In simple words: We find the median class (the middle of the data group) and the modal class (the group with the most items). Adding their starting values, 10 and 15, gives us 25.

Exam Tip: When finding the median class, always construct a quick cumulative frequency list first to identify where the middle value \( N/2 \) lies.

Question 2. Find the upper limit of the median class of the following frequency distribution :

Answer: Since the given class intervals are discontinuous (0-5, 6-11, 12-17, etc.), we must convert them into continuous class intervals by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits:
Continuous classes:
-0.5 - 5.5 (Frequency = 13, Cumulative Frequency = 13)
5.5 - 11.5 (Frequency = 10, Cumulative Frequency = 23)
11.5 - 17.5 (Frequency = 15, Cumulative Frequency = 38)
17.5 - 23.5 (Frequency = 8, Cumulative Frequency = 46)
23.5 - 29.5 (Frequency = 11, Cumulative Frequency = 57)

The total frequency \( N = 57 \).
We calculate \( \frac{N}{2} = \frac{57}{2} = 28.5 \).
The cumulative frequency just greater than 28.5 is 38, which lies in the continuous class interval 11.5 - 17.5.
So, the median class is 11.5 - 17.5.
Its upper limit is 17.5.
In simple words: Since the intervals have gaps, we adjust them to be continuous. The middle position is 28.5, which falls inside the 11.5 - 17.5 interval, making the upper limit 17.5.

Exam Tip: When the classes are discontinuous, always convert them to continuous form before identifying the lower or upper limits of the median class.

Question 3. Find the mean of the numbers 1,2,3,…n.
Answer: The given numbers are the first \( n \) natural numbers.
The sum of the first \( n \) natural numbers is given by the formula:
\[ S = \frac{n(n + 1)}{2} \]
The arithmetic mean is calculated by dividing the sum of the observations by the total count of observations (\( n \)):
\[ \text{Mean} = \frac{S}{n} = \frac{n(n + 1)}{2n} = \frac{n + 1}{2} \]
Therefore, the mean of the numbers is \( \frac{n + 1}{2} \).
In simple words: The average of numbers from 1 to n is found by dividing their sum by the total count, which simplifies to \( \frac{n + 1}{2} \).

Exam Tip: This is a standard formula that can be directly applied to find the mean of any consecutive sequence of natural numbers starting from 1.

Question 4. Which measure of central tendency is obtained from the abscissa of the point of intersection of the less than type and the more than type cumulative frequency curves of a grouped data?
Answer: When we plot both the 'less than' and 'more than' cumulative frequency curves (ogives) on the same coordinate axes, they intersect at a specific point. The x-coordinate (abscissa) of this intersection point gives the median of the grouped data.
Therefore, the measure of central tendency obtained is the **Median**.
In simple words: When you draw the two different cumulative graphs, the point where they cross tells you the exact middle value of the data, which is called the median.

Exam Tip: Remember that the y-coordinate (ordinate) of the intersection point of both ogives is equal to \( \frac{N}{2} \), while the x-coordinate is the median.

Question 5. For the following distribution, find the modal class.

Answer: The given data represents a cumulative frequency distribution of the 'less than' type. We need to convert it into a simple frequency distribution to find the modal class:
- Marks 0-10: Frequency = 3
- Marks 10-20: Frequency = 12 - 3 = 9
- Marks 20-30: Frequency = 27 - 12 = 15
- Marks 30-40: Frequency = 57 - 27 = 30
- Marks 40-50: Frequency = 75 - 57 = 18
- Marks 50-60: Frequency = 80 - 75 = 5

By observing the individual frequencies, we find the maximum frequency is 30, which corresponds to the class interval 30-40.
Thus, the modal class is 30-40.
In simple words: We convert the cumulative frequencies to find how many students are in each distinct 10-mark range. The 30-40 range has the most students (30), so it is our modal class.

Exam Tip: Never assume the largest value in a cumulative table represents the highest frequency; always find the individual class frequencies first.

Section B (2 marks each)

Question 6. The following is the distribution of weights (in kg) of 40 persons :

Construct a cumulative frequency distribution (less than type) table for the above data.
Answer: To construct a cumulative frequency distribution of the 'less than' type, we accumulate the frequencies of each class interval using their upper class boundaries:

In simple words: We build the 'less than' table by adding up the number of people in each weight category as we move up from the lowest weight to the highest.

Exam Tip: In 'less than' cumulative tables, always use the upper limits of the class intervals as the reference values.

Question 7. The frequency distribution table of agricultural holdings in a village is given below :

Calculate the modal agricultural holdings of the village.
Answer: To calculate the modal agricultural holdings, we determine the mode of the grouped data:
The highest frequency is 80, which lies in the class interval 5-7.
Therefore, the modal class is 5-7.
The parameters for the mode formula are:
- Lower limit of modal class (\( l \)) = 5
- Frequency of modal class (\( f_1 \)) = 80
- Frequency of preceding class (\( f_0 \)) = 45
- Frequency of succeeding class (\( f_2 \)) = 55
- Class size (\( h \)) = 2

Now, substitute these values into the mode formula:
\[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]
\[ \text{Mode} = 5 + \left(\frac{80 - 45}{2(80) - 45 - 55}\right) \times 2 \]
\[ \text{Mode} = 5 + \left(\frac{35}{160 - 100}\right) \times 2 \]
\[ \text{Mode} = 5 + \left(\frac{35}{60}\right) \times 2 \]
\[ \text{Mode} = 5 + \frac{70}{60} \approx 5 + 1.17 = 6.17\text{ ha} \]
So, the modal agricultural holdings of the village are approximately 6.17 hectares (or 6.2 hectares when rounded).
In simple words: The class 5-7 has the highest number of families. Using the mode formula with this group's values gives us an average peak of 6.17 hectares.

Exam Tip: Write down the formula for the mode clearly and state each parameter with its value before performing calculations to score full steps marks.

Question 8. Construct a cumulative frequency distribution (more than type) of the following distribution:

Answer: To construct a cumulative frequency distribution of the 'more than' type, we sum the frequencies from the highest class downwards, referencing the lower boundaries of each class interval:
The total frequency \( N = 2 + 22 + 19 + 14 + 13 = 70 \).

In simple words: We build the 'more than' table starting with the total frequency of 70, then subtract the frequency of each preceding class step-by-step as we move to higher boundaries.

Exam Tip: In 'more than or equal to' cumulative tables, always use the lower limits of the class intervals as reference values.

Section C (3 marks each)

Question 9. The arithmetic mean of the following data is 14. Find the value of k.

Answer: We calculate the sum of frequencies (\( \sum f_i \)) and the sum of products (\( \sum f_i x_i \)):
- \( \sum f_i = 7 + k + 8 + 4 + 5 = 24 + k \)
- \( \sum f_i x_i = (5 \times 7) + (10 \times k) + (15 \times 8) + (20 \times 4) + (25 \times 5) \)
- \( \sum f_i x_i = 35 + 10k + 120 + 80 + 125 = 360 + 10k \)

Given that the arithmetic mean is 14:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ 14 = \frac{360 + 10k}{24 + k} \]
\[ 14(24 + k) = 360 + 10k \]
\[ 336 + 14k = 360 + 10k \]
\[ 14k - 10k = 360 - 336 \]
\[ 4k = 24 \]
\[ k = 6 \]
Therefore, the value of \( k \) is 6.
In simple words: We multiply each value by its frequency, add them up, and set the average equal to 14. Solving this simple equation gives \( k = 6 \).

Exam Tip: Keep your algebraic steps organized; make sure not to drop the variable \( k \) from either the numerator or denominator when setting up the mean formula.

Question 10. Candidates of four schools appeared in mathematics test. The data were as follows :

If the average score of the candidates of all the four schools was 66, find the no. of candidates appeared from school B.
Answer: Let the number of candidates from school B be \( x \).
Using the average score formula, the sum of scores for each school is equal to (Number of candidates \(\times\) Average score):
- Sum of scores for School A = \( 60 \times 75 = 4500 \)
- Sum of scores for School B = \( x \times 55 = 55x \)
- Sum of scores for School C = \( 48 \times 80 = 3840 \)
- Sum of scores for School D = \( 40 \times 50 = 2000 \)

Total number of candidates across all schools = \( 60 + x + 48 + 40 = 148 + x \).
Total sum of scores across all schools = \( 4500 + 55x + 3840 + 2000 = 10340 + 55x \).

The combined average score is given as 66:
\[ \frac{10340 + 55x}{148 + x} = 66 \]
\[ 10340 + 55x = 66(148 + x) \]
\[ 10340 + 55x = 9768 + 66x \]
\[ 10340 - 9768 = 66x - 55x \]
\[ 572 = 11x \]
\[ x = 52 \]
Therefore, the number of candidates who appeared from school B is 52.
In simple words: We find the total sum of scores by multiplying candidate numbers by their averages. Setting up the combined average equation reveals that school B had 52 candidates.

Exam Tip: The combined mean formula is \( \bar{X} = \frac{N_1\bar{X}_1 + N_2\bar{X}_2 + \dots}{N_1 + N_2 + \dots} \). This is extremely useful for combining multiple group means.

Question 11. The following table gives the no. of pages written by sarika for completing her own work for 30 days :

Find the average no. of pages written by her.
Answer: Since the class intervals are discontinuous, we calculate the class mark (\( x_i \)) for each class directly as the average of the lower and upper limits of that class:
- 16-18: \( x_i = 17, f_i = 1 \implies f_i x_i = 17 \)
- 19-21: \( x_i = 20, f_i = 3 \implies f_i x_i = 60 \)
- 22-24: \( x_i = 23, f_i = 4 \implies f_i x_i = 92 \)
- 25-27: \( x_i = 26, f_i = 9 \implies f_i x_i = 234 \)
- 28-30: \( x_i = 29, f_i = 13 \implies f_i x_i = 377 \)

Calculating sums:
- Total days (\( \sum f_i \)) = \( 1 + 3 + 4 + 9 + 13 = 30 \)
- Total pages product (\( \sum f_i x_i \)) = \( 17 + 60 + 92 + 234 + 377 = 780 \)

Now, compute the mean:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{780}{30} = 26 \]
Therefore, the average number of pages written by her per day is 26.
In simple words: We find the midpoint of each page range and multiply it by the number of days. Dividing the total of 780 by 30 days gives an average of 26 pages daily.

Exam Tip: For calculating the mean, we do not need to convert discontinuous classes into continuous classes, as the midpoint (class mark) remains exactly the same.

Question 12. Find the mean, median and mode of the following frequency distribution.

Answer: We construct a consolidated table containing class marks (\( x_i \)), products (\( f_i x_i \)), and cumulative frequencies (\( cf \)):

1. Mean:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2850}{80} = 35.625 \]

2. Median:
Here, \( N = 80 \implies \frac{N}{2} = 40 \).
The cumulative frequency just greater than 40 is 50, which lies in the class interval 30-40.
So, Median Class = 30-40.
- \( l = 30, cf = 30, f = 20, h = 10 \)
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
\[ \text{Median} = 30 + \left(\frac{40 - 30}{20}\right) \times 10 = 30 + 5 = 35 \]

3. Mode:
The highest frequency is 20, corresponding to the modal class 30-40.
- \( l = 30, f_1 = 20, f_0 = 15, f_2 = 12, h = 10 \)
\[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]
\[ \text{Mode} = 30 + \left(\frac{20 - 15}{2(20) - 15 - 12}\right) \times 10 \]
\[ \text{Mode} = 30 + \left(\frac{5}{13}\right) \times 10 \approx 30 + 3.85 = 33.85 \]
In simple words: We compile all our values to find the three averages. The mean is 35.625, the median is exactly 35, and the mode is 33.85.

Exam Tip: Practicing step-by-step table construction is the best way to tackle comprehensive questions that ask for mean, median, and mode together.

Section D (4 marks each)

Question 13. The mean of the following frequency distribution is 57.6 and the sum of observation is 50. Find the missing frequency 𝑓1 and 𝑓2.

Answer: The sum of all frequencies is given as 50:
\[ 7 + f_1 + 12 + f_2 + 8 + 5 = 50 \]
\[ \implies f_1 + f_2 + 32 = 50 \]
\[ \implies f_1 + f_2 = 18 \tag{Equation 1} \]

Now, let's find the class marks (\( x_i \)) and calculate \( f_i x_i \):
- 0-20: \( x_i = 10, f_i = 7 \implies f_i x_i = 70 \)
- 20-40: \( x_i = 30, f_i = f_1 \implies f_i x_i = 30f_1 \)
- 40-60: \( x_i = 50, f_i = 12 \implies f_i x_i = 600 \)
- 60-80: \( x_i = 70, f_i = f_2 \implies f_i x_i = 70f_2 \)
- 80-100: \( x_i = 90, f_i = 8 \implies f_i x_i = 720 \)
- 100-120: \( x_i = 110, f_i = 5 \implies f_i x_i = 550 \)

Adding the terms:
\[ \sum f_i x_i = 70 + 30f_1 + 600 + 70f_2 + 720 + 550 = 1940 + 30f_1 + 70f_2 \]

Using the mean formula:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ 57.6 = \frac{1940 + 30f_1 + 70f_2}{50} \]
\[ 2880 = 1940 + 30f_1 + 70f_2 \]
\[ 30f_1 + 70f_2 = 940 \]
Divide the equation by 10:
\[ 3f_1 + 7f_2 = 94 \tag{Equation 2} \]

From Equation 1, multiply by 3:
\[ 3f_1 + 3f_2 = 54 \tag{Equation 3} \]

Subtract Equation 3 from Equation 2:
\[ 4f_2 = 40 \implies f_2 = 10 \]
Substitute \( f_2 = 10 \) into Equation 1:
\[ f_1 + 10 = 18 \implies f_1 = 8 \]
Therefore, the missing frequencies are \( f_1 = 8 \) and \( f_2 = 10 \).
In simple words: We set up two equations: one using the total sum of frequencies (50) and the other using the arithmetic mean (57.6). Solving this system of equations gives \( f_1 = 8 \) and \( f_2 = 10 \).

Exam Tip: Missing frequency problems are very common; always begin by setting up the sum of frequencies equation first to simplify the system of equations.

Question 14. Compute the arithmetic mean for the following distribution.

Answer: The given cumulative frequencies can be transformed into a standard frequency table with class intervals and class marks (\( x_i \)):

Now, compute the arithmetic mean:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4115}{85} \approx 48.41 \]
Therefore, the arithmetic mean is 48.41.
In simple words: We subtract consecutive values in the cumulative frequency list to find the actual frequencies. Then, we use the standard mean formula to get 48.41.

Exam Tip: Always make sure the sum of your converted frequencies equals the final cumulative frequency value (which is 85 here) to ensure no arithmetic mistakes were made.

Question 15. Form a frequency distribution table and compute arithmetic mean for the following frequency distribution :

Answer: The given cumulative frequencies of the 'above' type can be converted into a standard frequency table with class intervals and class marks (\( x_i \)):

Now, compute the mean:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3215}{50} = 64.3\text{ kg} \]
Therefore, the arithmetic mean weight is 64.3 kg.
In simple words: We rewrite the cumulative list to find the actual number of people in each weight group, then find the average weight, which is 64.3 kg.

Exam Tip: Subtract succeeding cumulative frequencies to convert 'above' or 'more than' cumulative tables into simple frequency tables.

Question 16. The median of the following data is 50. Find the value of p and q, if the sum of all the frequencies is 90.

Answer: The sum of all frequencies is given as 90:
\[ p + 15 + 25 + 20 + q + 8 + 10 = 90 \]
\[ \implies p + q + 78 = 90 \]
\[ \implies p + q = 12 \tag{Equation 1} \]

Let's construct the cumulative frequency (cf) table:
- 20-30: \( f = p, cf = p \)
- 30-40: \( f = 15, cf = p + 15 \)
- 40-50: \( f = 25, cf = p + 40 \)
- 50-60: \( f = 20, cf = p + 60 \)
- 60-70: \( f = q, cf = p + 60 + q \)
- 70-80: \( f = 8, cf = p + 68 + q \)
- 80-90: \( f = 10, cf = p + 78 + q \)

The median is given as 50.
Since the median is 50, it lies on the boundary of the class 50-60.
Thus, we take the median class as 50-60.
- \( l = 50, f = 20, cf = p + 40, h = 10, N = 90 \)

Now, substitute these parameters into the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
\[ 50 = 50 + \left(\frac{45 - (p + 40)}{20}\right) \times 10 \]
\[ 0 = \frac{45 - p - 40}{2} \]
\[ 0 = 5 - p \]
\[ p = 5 \]

Substitute \( p = 5 \) into Equation 1:
\[ 5 + q = 12 \implies q = 7 \]
Therefore, the values are \( p = 5 \) and \( q = 7 \).
In simple words: Knowing the total frequency is 90 gives us a simple relation between p and q. Using the median of 50 in our median formula lets us solve directly for \( p = 5 \) and \( q = 7 \).

Exam Tip: When the median value lies exactly on a class boundary (like 50 here), the median class is identified as the class whose lower limit is that value.

Question 17. Draw ‘less than ogive’ and ‘more than ogive’ for the following distribution and hence find its median.

Answer: To find the median using the graphical method, we first construct both the 'less than' and 'more than' type cumulative frequency tables:

1. 'Less than' type table:
- Less than 30: 25
- Less than 40: 40
- Less than 50: 50
- Less than 60: 56
- Less than 70: 80
- Less than 80: 92
- Less than 90: 100

2. 'More than' type table:
- More than or equal to 20: 100
- More than or equal to 30: 75
- More than or equal to 40: 60
- More than or equal to 50: 50
- More than or equal to 60: 44
- More than or equal to 70: 20
- More than or equal to 80: 8

When both of these cumulative frequency curves (ogives) are plotted on a graph, they intersect at a point where the y-value is \( \frac{N}{2} = 50 \).
The x-coordinate (abscissa) corresponding to this point of intersection is 50.
Therefore, the median of the given distribution is 50.
In simple words: We plot both cumulative curves on a graph. The point where they cross corresponds to the 50th mark on the horizontal axis, showing the median is 50.

Exam Tip: Plotting both ogives on the same graph and finding their intersection point is the most common graphical method to determine the median.

Question 18. 50 students enter for a school javelin throw competition. The distance (in metres) thrown are recorded below :

a) Construct a cumulative frequency table .
b) Draw a cumulative frequency curve (less than type) and calculate the median distance thrown by using the curve.
c) Calculate the median distance by using the formula for median.
d) Are the median distance calculated in (b) and (c) same?
Answer:
a) Let's construct the 'less than' cumulative frequency table:

b) To construct the curve, we plot the points \( (20, 6), (40, 17), (60, 34), (80, 46), (100, 50) \) and join them with a smooth curve.
Since \( N = 50 \implies \frac{N}{2} = 25 \), we locate 25 on the y-axis, draw a horizontal line to meet the curve, and drop a perpendicular to the x-axis.
The value on the x-axis corresponds to approximately 49.4 meters.

c) Using the median formula:
Since \( \frac{N}{2} = 25 \), the median class is 40-60.
- \( l = 40, cf = 17, f = 17, h = 20 \)
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
\[ \text{Median} = 40 + \left(\frac{25 - 17}{17}\right) \times 20 \]
\[ \text{Median} = 40 + \left(\frac{8}{17}\right) \times 20 \]
\[ \text{Median} = 40 + 9.41 = 49.41\text{ m} \]

d) Yes, the median distance calculated graphically in part (b) and analytically using the formula in part (c) are approximately the same.
In simple words: We create a cumulative table and a graph. Both the graph and the formula show that the median distance thrown is approximately 49.41 meters, confirming both methods match.

Exam Tip: Graphical estimation yields a very close approximation, but the formula gives the exact mathematical value. Always write both results to show your understanding of the slight difference.