CBSE Class 10 Mathematics Statistics Worksheet Set 08

Question 1. The median of the following data is 525.Find the values of x and y, if the total frequency is 100 (9, 15)

Answer: First, let us construct the cumulative frequency table from the given data:

It is given that the total frequency is 100.
\( \implies 76 + x + y = 100 \)
\( \implies x + y = 24 \) - (Equation 1)
Since the median is given as 525, it lies in the class interval 500 - 600. Therefore, the median class is 500 - 600. From this, we have:
Lower limit of median class (\( l \)) = 500
Frequency of the median class (\( f \)) = 20
Cumulative frequency of the preceding class (\( cf \)) = 36 + x
Class size (\( h \)) = 100
Total frequency (\( N \)) = 100
Now, we use the median formula:
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \implies 525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100 \)
\( \implies 525 - 500 = (14 - x) \times 5 \)
\( \implies 25 = 5(14 - x) \)
\( \implies 5 = 14 - x \)
\( \implies x = 9 \)
Substituting the value of x in Equation 1:
\( \implies 9 + y = 24 \)
\( \implies y = 15 \)
So, the missing values are \( x = 9 \) and \( y = 15 \).
In simple words: To find the missing frequencies, we first write out how the values add up step-by-step. Since we know the final total is 100 and the middle value (median) is 525, we can use the formula to find that \( x \) is 9 and \( y \) is 15.

Exam Tip: Be careful with the minus sign when subtracting cumulative frequency \( cf = 36 + x \) in the numerator; remember to distribute the negative sign to both terms so that it becomes \( 50 - 36 - x \).

Question 2. The median of the data is 28. Find the values of x and y, if the total frequency is 50 (8, 16)

Answer: Let us prepare the cumulative frequency table for the given dataset:

The total frequency is given as 50.
\( \implies 26 + x + y = 50 \)
\( \implies x + y = 24 \) - (Equation 1)
The median of this distribution is 28, which falls in the class 20 - 30. Therefore, the median class is 20 - 30. From this class, we get:
Lower limit of the class (\( l \)) = 20
Frequency of the median class (\( f \)) = 15
Cumulative frequency of the previous class (\( cf \)) = 5 + x
Class size (\( h \)) = 10
Total number of students (\( N \)) = 50
We apply the median formula:
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \implies 28 = 20 + \left( \frac{25 - (5 + x)}{15} \right) \times 10 \)
\( \implies 28 - 20 = \left( \frac{20 - x}{15} \right) \times 10 \)
\( \implies 8 = \frac{2(20 - x)}{3} \)
\( \implies 24 = 40 - 2x \)
\( \implies 2x = 16 \)
\( \implies x = 8 \)
Substituting \( x = 8 \) into Equation 1:
\( \implies 8 + y = 24 \)
\( \implies y = 16 \)
Thus, the missing frequencies are \( x = 8 \) and \( y = 16 \).
In simple words: First we add up the frequencies step-by-step to write down the cumulative frequency. Using the fact that the total frequency is 50 and the median is 28, we solve the algebraic equations to find \( x = 8 \) and \( y = 16 \).

Exam Tip: Always make sure that your identified median class contains the given median value. Since the median is 28, it lies within the 20 - 30 interval, confirming our choice of the median class.

Question 3. If the mean of the following distribution is 27, find the value of p (P = 7)

Answer: Let us find the class marks (\( x_i \)) and compute \( f_i x_i \) for each interval:

We are given that the mean is 27. Using the mean formula:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \implies 27 = \frac{1245 + 15P}{43 + P} \)
\( \implies 27(43 + P) = 1245 + 15P \)
\( \implies 1161 + 27P = 1245 + 15P \)
\( \implies 27P - 15P = 1245 - 1161 \)
\( \implies 12P = 84 \)
\( \implies P = 7 \)
So, the value of P is 7.
In simple words: First we find the middle value of each group, multiply it by the number of values in that group, and add them up. By dividing this total by the sum of frequencies, we can set up an equation with the mean of 27 to find that P equals 7.

Exam Tip: Make sure you double-check the calculations for class marks by averaging the upper and lower limits of each class interval, i.e., \( \frac{\text{Lower limit} + \text{Upper limit}}{2} \).

Question 4. Find the missing frequency: mean = 50, Total frequency = 120 (28, 24)

Answer: Let us prepare the frequency table to compute the mean:

We are given that the total frequency is 120.
\( \implies 68 + F_1 + F_2 = 120 \)
\( \implies F_1 + F_2 = 52 \) - (Equation 1)
Also, the mean is 50. Using the mean formula:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \implies 50 = \frac{3480 + 30F_1 + 70F_2}{120} \)
\( \implies 50 \times 120 = 3480 + 30F_1 + 70F_2 \)
\( \implies 6000 = 3480 + 30F_1 + 70F_2 \)
\( \implies 30F_1 + 70F_2 = 2520 \)
Dividing both sides by 10, we get:
\( \implies 3F_1 + 7F_2 = 252 \) - (Equation 2)
From Equation 1, substitute \( F_1 = 52 - F_2 \) into Equation 2:
\( \implies 3(52 - F_2) + 7F_2 = 252 \)
\( \implies 156 - 3F_2 + 7F_2 = 252 \)
\( \implies 4F_2 = 252 - 156 \)
\( \implies 4F_2 = 96 \)
\( \implies F_2 = 24 \)
Substitute \( F_2 = 24 \) back into Equation 1:
\( \implies F_1 + 24 = 52 \)
\( \implies F_1 = 28 \)
Thus, the missing frequencies are \( F_1 = 28 \) and \( F_2 = 24 \).
In simple words: We find the sum of all values multiplied by their frequencies, and use both the sum of frequencies (120) and the mean (50) to create a system of two equations. Solving these gives \( F_1 = 28 \) and \( F_2 = 24 \).

Exam Tip: Simplify the linear equations by dividing by their common factors (like dividing by 10 in this step) to reduce the calculation size and prevent arithmetic errors.

Question 5. The mean of the following frequency distribution is 132 and the sum of the observations is 50. Find the Missing frequencies f1 and f2 (10, 8)

Answer: Let us prepare the calculation table by finding the class marks (\( x_i \)) and finding the product \( f_i x_i \):

The sum of the frequencies is given as 50.
\( \implies 32 + f_1 + f_2 = 50 \)
\( \implies f_1 + f_2 = 18 \) - (Equation 1)
Also, the mean is 132. Using the mean formula:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \implies 132 = \frac{4160 + 100f_1 + 180f_2}{50} \)
\( \implies 132 \times 50 = 4160 + 100f_1 + 180f_2 \)
\( \implies 6600 = 4160 + 100f_1 + 180f_2 \)
\( \implies 100f_1 + 180f_2 = 6600 - 4160 \)
\( \implies 100f_1 + 180f_2 = 2440 \)
Dividing by 20, we get:
\( \implies 5f_1 + 9f_2 = 122 \) - (Equation 2)
Using Equation 1, substitute \( f_1 = 18 - f_2 \) into Equation 2:
\( \implies 5(18 - f_2) + 9f_2 = 122 \)
\( \implies 90 - 5f_2 + 9f_2 = 122 \)
\( \implies 4f_2 = 122 - 90 \)
\( \implies 4f_2 = 32 \)
\( \implies f_2 = 8 \)
Substitute \( f_2 = 8 \) in Equation 1:
\( \implies f_1 + 8 = 18 \)
\( \implies f_1 = 10 \)
So, the missing frequencies are \( f_1 = 10 \) and \( f_2 = 8 \).
In simple words: First we sum up the given frequencies to set up our first equation. Next we find the class marks, calculate \( f_i x_i \), and use the mean formula to establish the second equation. Solving these two together yields \( f_1 = 10 \) and \( f_2 = 8 \).

Exam Tip: Be mindful of placing the correct variables in your calculations. Ensure that \( f_1 \) and \( f_2 \) correspond exactly to the correct class intervals (80 - 120 and 160 - 200 respectively) in the formula.

Question 6. Find the mean, median and mode of the following data (30, 30.67, 33.3)

Answer: Let us construct a single comprehensive table to calculate the mean, median, and mode:

1. Calculation of Mean:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1500}{50} = 30 \)

2. Calculation of Median:
Total frequency \( N = 50 \), so \( \frac{N}{2} = 25 \).
The cumulative frequency just greater than 25 is 39, which falls in the class interval 30 - 40. Therefore, the median class is 30 - 40.
Using the parameters:
Lower limit (\( l \)) = 30
Cumulative frequency of preceding class (\( cf \)) = 24
Frequency of median class (\( f \)) = 15
Class size (\( h \)) = 10
Applying the median formula:
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \text{Median} = 30 + \left( \frac{25 - 24}{15} \right) \times 10 \)
\( \text{Median} = 30 + \frac{10}{15} = 30 + 0.67 = 30.67 \)

3. Calculation of Mode:
The class with the highest frequency is 30 - 40 (frequency = 15). Therefore, the modal class is 30 - 40.
Using the parameters:
Lower limit (\( l \)) = 30
Frequency of modal class (\( f_1 \)) = 15
Frequency of preceding class (\( f_0 \)) = 10
Frequency of succeeding class (\( f_2 \)) = 5
Class size (\( h \)) = 10
Applying the mode formula:
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( \text{Mode} = 30 + \left( \frac{15 - 10}{2(15) - 10 - 5} \right) \times 10 \)
\( \text{Mode} = 30 + \left( \frac{5}{30 - 15} \right) \times 10 = 30 + \frac{50}{15} = 30 + 3.33 = 33.33 \approx 33.3 \)

So, Mean = 30, Median = 30.67, and Mode = 33.3.
In simple words: We create a comprehensive table containing the group values, cumulative frequency, and products. The mean is calculated to be 30, the median is calculated using the midpoint of frequencies to be 30.67, and the mode is found using the peak class to be 33.3.

Exam Tip: Constructing a combined table for mean, median, and mode saves time during exams and keeps your working space tidy and organized.

Question 7. The mode of the following frequency distribution is 55. Find the values of x and y (X = 7, Y = 5)

Answer: Given that the mode of the distribution is 55.
Since 55 lies in the class interval 45 - 60, the modal class is 45 - 60. From this, we extract:
Lower limit of modal class (\( l \)) = 45
Frequency of modal class (\( f_1 \)) = 15
Frequency of preceding class (\( f_0 \)) = Y
Frequency of succeeding class (\( f_2 \)) = 10
Class width (\( h \)) = 15
Applying the mode formula:
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( \implies 55 = 45 + \left( \frac{15 - Y}{2(15) - Y - 10} \right) \times 15 \)
\( \implies 55 - 45 = \left( \frac{15 - Y}{30 - Y - 10} \right) \times 15 \)
\( \implies 10 = \left( \frac{15 - Y}{20 - Y} \right) \times 15 \)
\( \implies \frac{10}{15} = \frac{15 - Y}{20 - Y} \)
\( \implies \frac{2}{3} = \frac{15 - Y}{20 - Y} \)
By cross-multiplying:
\( \implies 2(20 - Y) = 3(15 - Y) \)
\( \implies 40 - 2Y = 45 - 3Y \)
\( \implies 3Y - 2Y = 45 - 40 \)
\( \implies Y = 5 \)
To find the value of X, we use the total frequency of 50:
\( \implies \sum f_i = 6 + 7 + Y + 15 + 10 + X = 50 \)
\( \implies 38 + 5 + X = 50 \)
\( \implies 43 + X = 50 \)
\( \implies X = 7 \)
Thus, the missing frequencies are \( X = 7 \) and \( Y = 5 \).
In simple words: Since the mode is 55, it belongs to the class interval 45-60. Using the mode formula, we calculate \( Y = 5 \). Assuming the total frequency is 50, we add up the rest to solve for \( X = 7 \).

Exam Tip: Remember that the modal class is always determined by the class interval in which the given mode value lies, rather than just the largest visible frequency when some frequencies are unknown.

Question 8. For a given data less than ogive and more than ogive intersect at a point P(x, y). Then what does abscissa of the Point represents

Answer: The abscissa (which is the x-coordinate) of the point of intersection \( P(x, y) \) of a less than ogive and a more than ogive represents the Median of the given frequency distribution.
In simple words: When we draw both types of cumulative frequency curves on a graph, the point where they cross each other points directly down to the median on the horizontal axis.

Exam Tip: Keep in mind that while the x-coordinate (abscissa) represents the median, the y-coordinate (ordinate) represents \( N/2 \) (half of the total frequency).

Question 9. Write the empirical relationship between the three measures of central tendency

Answer: The empirical relationship linking the three main measures of central tendency is:
\( \text{Mode} = 3 \ \text{Median} - 2 \ \text{Mean} \)
This can also be written in other equivalent mathematical forms:
\( 3 \ \text{Median} = \text{Mode} + 2 \ \text{Mean} \)
In simple words: The mode is equal to three times the median minus two times the mean. This is a handy rule that connects all three averages together.

Exam Tip: This formula is highly scoring in multiple-choice questions. Remember it carefully to solve related numeric problems quickly.

Question 10. If median = 15 and mean = 16, find mode of the distribution (13)

Answer: We are given:
\( \text{Median} = 15 \)
\( \text{Mean} = 16 \)
Using the empirical relationship formula:
\( \text{Mode} = 3 \ \text{Median} - 2 \ \text{Mean} \)
Substitute the given values into the formula:
\( \implies \text{Mode} = 3(15) - 2(16) \)
\( \implies \text{Mode} = 45 - 32 \)
\( \implies \text{Mode} = 13 \)
Thus, the mode of the distribution is 13.
In simple words: Using the empirical connection formula, we multiply the median by 3 to get 45, and the mean by 2 to get 32. Subtracting 32 from 45 gives us a mode of 13.

Exam Tip: Write down the empirical formula before substituting the numbers. Doing this ensures you get partial marking even if a calculation mistake happens later.

Question 11. Following is the distribution of marks obtained by 60 students: Calculate the arithmetic mean (26.5)

Answer: We first convert this "more than" cumulative frequency distribution into a normal grouped frequency distribution:
- For Class Interval 0 - 10: Frequency = \( 60 - 56 = 4 \)
- For Class Interval 10 - 20: Frequency = \( 56 - 40 = 16 \)
- For Class Interval 20 - 30: Frequency = \( 40 - 20 = 20 \)
- For Class Interval 30 - 40: Frequency = \( 20 - 10 = 10 \)
- For Class Interval 40 - 50: Frequency = \( 10 - 3 = 7 \)
- For Class Interval 50 - 60: Frequency = 3

Now, let us calculate the arithmetic mean using the direct method:

Using the arithmetic mean formula:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \text{Mean} = \frac{1590}{60} = \frac{159}{6} = 26.5 \)
Thus, the arithmetic mean is 26.5.
In simple words: First we turn the cumulative "more than" table into regular intervals by finding the difference between each step. Then we calculate the average marks for each interval, multiply them by the frequencies, and divide the total by the 60 students to get 26.5.

Exam Tip: When converting cumulative distributions, always make sure your final frequencies sum up to the total number of observations (60 in this case) as a quick validation step.

Question 12. From the following data draw the two types of curves and find the median

Answer: Let us prepare both the "less than" and "more than" cumulative frequency tables:

1. Less Than Type Cumulative Frequency Table:

Points to plot: \( (220, 7), (240, 10), (260, 16), (280, 24), (300, 26), (320, 30) \).

2. More Than Type Cumulative Frequency Table:

Points to plot: \( (200, 30), (220, 23), (240, 20), (260, 14), (280, 6), (300, 4) \).

Finding the Median:
The total frequency is \( N = 30 \), so \( \frac{N}{2} = 15 \). The intersection point of both curves corresponds to the y-value of 15. The x-coordinate of this point of intersection gives the median.
By plotting these curves on a graph, they intersect at point \( P \). The x-value of this intersection is:
\( \text{Median} \approx 256.67 \)

We can also calculate the median mathematically to verify:
Median class is 240 - 260.
\( l = 240, \ f = 6, \ cf = 10, \ h = 20 \)
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \text{Median} = 240 + \left( \frac{15 - 10}{6} \right) \times 20 = 240 + 16.67 = 256.67 \)

Here is the graphic plot of both curves showing their point of intersection:
In simple words: We plot both cumulative frequency graphs, less than and more than. The point of intersection directly maps down to 256.7 on the x-axis, which is our median value.

Exam Tip: Label the axes properly with appropriate scales. A dashed line drawn from the intersection point to the x-axis helps the examiner easily identify your visual solution for the median.