CBSE Class 10 Mathematics Statistics Worksheet Set 07

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

Answer:
First, let us construct the cumulative frequency (cf) table for the given distribution:

The total frequency is given as \( N = 100 \). Therefore:
\( 76 + x + y = 100 \)
\( x + y = 24 \) - (i)

The median is given as 525, which lies in the class interval 500 - 600. Thus, 500 - 600 is the median class.
From this median class, we have:
Lower limit, \( l = 500 \)
Frequency of the median class, \( f = 20 \)
Cumulative frequency of the preceding class, \( cf = 36 + x \)
Class size, \( h = 100 \)

Now, using the median formula:
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( 525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100 \)
\( 25 = (14 - x) \times 5 \)
\( 5 = 14 - x \)
\( x = 9 \)

Substituting \( x = 9 \) into equation (i):
\( 9 + y = 24 \)
\( y = 15 \)

Therefore, the missing values are \( x = 9 \) and \( y = 15 \).

In simple words: First, we write the sum of all frequencies as 100 to get our first equation. Since the median is 525, we locate its group and use the median formula to solve for both unknown values step-by-step.

Exam Tip: Be extra careful with the brackets when subtracting the cumulative frequency term, i.e., write it as \( -(36 + x) \) to prevent simple sign errors that can completely ruin your calculations.

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

Answer:
First, let us construct the cumulative frequency (cf) table for the given distribution:

The total frequency is given as \( N = 50 \). Therefore:
\( 26 + x + y = 50 \)
\( x + y = 24 \) - (i)

The median is 28, which falls in the class interval 20 - 30. Therefore, 20 - 30 is our median class.
From this class, we identify the following variables:
Lower limit, \( l = 20 \)
Frequency of the median class, \( f = 15 \)
Cumulative frequency of the preceding class, \( cf = 5 + x \)
Class size, \( h = 10 \)

Using the median formula:
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( 28 = 20 + \left( \frac{25 - (5 + x)}{15} \right) \times 10 \)
\( 8 = \left( \frac{20 - x}{15} \right) \times 10 \)
\( 8 = \frac{2(20 - x)}{3} \)
\( 24 = 40 - 2x \)
\( 2x = 16 \)
\( x = 8 \)

Substituting \( x = 8 \) into equation (i):
\( 8 + y = 24 \)
\( y = 16 \)

Therefore, the values of the missing frequencies are \( x = 8 \) and \( y = 16 \).

In simple words: By summing up all the given frequencies and equating them to 50, we get our first relation. Since the median is 28, we locate the 20-30 group and solve for both x and y using the standard formula.

Exam Tip: Be precise when simplifying fractions containing the class size and frequency to avoid making mistakes in your algebraic steps.

Question 3. If the mean of the following distribution is 27, find the value of p

Answer:
First, let us construct the table to calculate the mean by finding the class marks (\( x_i \)) and the products (\( f_i x_i \)):

We are given that the Mean is 27. Using the mean formula:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( 27 = \frac{1245 + 15p}{43 + p} \)
\( 27(43 + p) = 1245 + 15p \)
\( 1161 + 27p = 1245 + 15p \)
\( 12p = 84 \)
\( p = 7 \)

Therefore, the value of \( p \) is 7.

In simple words: We calculate the midpoint for each interval and multiply it by its frequency. By dividing the sum of these products by the total frequency and setting it equal to 27, we solve a simple linear equation to get p = 7.

Exam Tip: Double-check your arithmetic when summing up large products like 455 and 450 to avoid carrying forward an incorrect total.

Question 4. Find the missing frequency: mean = 50, Total frequency = 120

Answer:
We are given that the total frequency is 120. Therefore:
\( 17 + F_1 + 32 + F_2 + 19 = 120 \)
\( 68 + F_1 + F_2 = 120 \)
\( F_1 + F_2 = 52 \) - (i)

Next, we calculate the sum of the products of \( f_i \) and \( x_i \):
\( \sum f_i x_i = (10 \times 17) + (30 \times F_1) + (50 \times 32) + (70 \times F_2) + (90 \times 19) \)
\( \sum f_i x_i = 170 + 30F_1 + 1600 + 70F_2 + 1710 \)
\( \sum f_i x_i = 3480 + 30F_1 + 70F_2 \)

The mean is given as 50. Using the formula for the mean:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( 50 = \frac{3480 + 30F_1 + 70F_2}{120} \)
\( 6000 = 3480 + 30F_1 + 70F_2 \)
\( 30F_1 + 70F_2 = 2520 \)
Divide the entire equation by 10:
\( 3F_1 + 7F_2 = 252 \) - (ii)

Now, multiply equation (i) by 3:
\( 3F_1 + 3F_2 = 156 \) - (iii)

Subtract equation (iii) from equation (ii):
\( 4F_2 = 96 \)
\( F_2 = 24 \)

Substitute \( F_2 = 24 \) into equation (i):
\( F_1 + 24 = 52 \)
\( F_1 = 28 \)

Therefore, the missing frequencies are \( F_1 = 28 \) and \( F_2 = 24 \).

In simple words: Since the sum of all frequencies is 120, we get our first equation. Using the mean formula with the given average of 50 gives us a second equation, and we solve both to find our missing frequencies.

Exam Tip: Simplifying linear equations by dividing common multiples (like dividing by 10 here) helps to keep calculations clean and prevents errors during substitution.

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

Answer:
First, let us construct the table to find the class marks (\( x_i \)) and products (\( f_i x_i \)):

We are given that the sum of the frequencies is 50. Therefore:
\( 32 + f_1 + f_2 = 50 \)
\( f_1 + f_2 = 18 \) - (i)

The mean is given as 132. Using the mean formula:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( 132 = \frac{4160 + 100f_1 + 180f_2}{50} \)
\( 6600 = 4160 + 100f_1 + 180f_2 \)
\( 100f_1 + 180f_2 = 2440 \)
Divide the entire equation by 20:
\( 5f_1 + 9f_2 = 122 \) - (ii)

Now, multiply equation (i) by 5:
\( 5f_1 + 5f_2 = 90 \) - (iii)

Subtract equation (iii) from equation (ii):
\( 4f_2 = 32 \)
\( f_2 = 8 \)

Substitute \( f_2 = 8 \) into equation (i):
\( f_1 + 8 = 18 \)
\( f_1 = 10 \)

Therefore, the missing frequencies are \( f_1 = 10 \) and \( f_2 = 8 \).

In simple words: The sum of all students' counts is 50, giving us our first linear relationship. Combining the mean formula with the average of 132 gives us the second equation, letting us solve for both unknown groups.

Exam Tip: Be sure to compute midpoints carefully for large class widths like 0-40, 40-80, etc. before carrying out multiplications.

Question 6. Find the mean, median and mode of the following data

Answer:
First, let us organize the calculation tables for Mean, Median, and Mode:

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

2. Calculation of Median:
Total frequency, \( N = 50 \implies \frac{N}{2} = 25 \)
The cumulative frequency just greater than 25 is 39, which corresponds to the class interval 30 - 40. Thus, 30 - 40 is our median class.
From this class, we have:
\( l = 30 \), \( f = 15 \), \( cf = 24 \), \( h = 10 \)
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \text{Median} = 30 + \left( \frac{25 - 24}{15} \right) \times 10 \)
\( \text{Median} = 30 + \frac{10}{15} \approx 30 + 0.67 = 30.67 \)

3. Calculation of Mode:
The highest frequency is 15, which falls in the class 30 - 40. Hence, 30 - 40 is our modal class.
From this class, we have:
\( l = 30 \), \( f_1 = 15 \) (frequency of modal class), \( f_0 = 10 \) (preceding frequency), \( f_2 = 5 \) (succeeding frequency), \( h = 10 \)
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( \text{Mode} = 30 + \left( \frac{15 - 10}{30 - 10 - 5} \right) \times 10 \)
\( \text{Mode} = 30 + \left( \frac{5}{15} \right) \times 10 \)
\( \text{Mode} = 30 + 3.33 = 33.33 \)

Therefore, the Mean is 30, the Median is 30.67, and the Mode is 33.33.

In simple words: We calculate all three statistics using their standard definitions: the average of weighted midpoints (Mean), the point dividing the frequency list in half (Median), and the most frequent group (Mode).

Exam Tip: Be precise with the formulas of all three measures of central tendency, as keeping track of pre-existing and post-existing frequencies for the mode formula is highly critical.

Question 7. The mode of the following frequency distribution is 55. Find the values of x and y

Answer:
The mode is given as 55, which falls in the class interval 45 - 60. Therefore, 45 - 60 is our modal class.
From this, we identify the following variables:
Lower limit of modal class, \( l = 45 \)
Frequency of the modal class, \( f_1 = 15 \)
Frequency of the preceding class, \( f_0 = y \)
Frequency of the succeeding class, \( f_2 = 10 \)
Class size, \( h = 15 \)

Using the formula for mode:
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( 55 = 45 + \left( \frac{15 - y}{30 - y - 10} \right) \times 15 \)
\( 10 = \left( \frac{15 - y}{20 - y} \right) \times 15 \)
Divide both sides by 5:
\( 2 = 3 \times \frac{15 - y}{20 - y} \)
\( 2(20 - y) = 3(15 - y) \)
\( 40 - 2y = 45 - 3y \)
\( y = 5 \)

Assuming the standard total frequency of 50 for this distribution:
\( 6 + 7 + y + 15 + 10 + x = 50 \)
\( 38 + y + x = 50 \)
\( 38 + 5 + x = 50 \)
\( x = 7 \)

Therefore, the values of the missing frequencies are \( x = 7 \) and \( y = 5 \).

In simple words: Since the mode is 55, we use the 45-60 class to set up our equation. Solving it gives y = 5, and assuming the total frequency is 50 allows us to easily find x = 7.

Exam Tip: If the total frequency is not explicitly printed in the question text, check standard CBSE question banks as these problems almost always assume a total sum like 50.

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:
When we draw both a "less than" ogive and a "more than" ogive on the same coordinate axes, they intersect at a unique point.
The x-coordinate (abscissa) of this point of intersection represents the **Median** of the given grouped data, while the y-coordinate represents \( \frac{N}{2} \).

Therefore, the abscissa represents the **Median** of the given distribution.

In simple words: The intersection point of both kinds of cumulative frequency curves aligns perfectly with the median value on the horizontal x-axis.

Exam Tip: Memorize this intersection property of ogives, as it is a very common one-mark theory question in board examinations.

Question 9. Write the empirical relationship between the three measures of central tendency
Answer:
The empirical formula relating Mode, Median, and Mean is given as:
\( \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \)

In simple words: This equation acts as a rule of thumb relating our three main statistical measures together, letting us find any one if we know the other two.

Exam Tip: Write this formula down on your scratch sheet immediately at the start of the exam, as it is highly useful for verifying calculations across different sections.

Question 10. If median = 15 and mean = 16, find mode of the distribution
Answer:
We are given:
\( \text{Median} = 15 \)
\( \text{Mean} = 16 \)

Using the empirical relationship:
\( \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \)
\( \text{Mode} = 3(15) - 2(16) \)
\( \text{Mode} = 45 - 32 \)
\( \text{Mode} = 13 \)

Therefore, the mode of the distribution is 13.

In simple words: Substituting our known values into the statistical formula lets us find that the Mode is equal to 13.

Exam Tip: Ensure you do not swap the coefficients of Mean and Median, as a very common error is calculating \( 3 \text{ Mean} - 2 \text{ Median} \) instead.

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

Answer:
First, we convert the cumulative "more than" table into a standard grouped frequency distribution:

Now, we calculate the arithmetic mean:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \text{Mean} = \frac{1590}{60} = 26.5 \)

Therefore, the arithmetic mean of the distribution is 26.5.

In simple words: We convert the cumulative "more than" groups into standard class intervals by subtracting adjacent frequency totals, then compute the average of our class midpoints.

Exam Tip: Be meticulous during the conversion of a cumulative distribution to a continuous frequency distribution, as a subtraction mistake in any one interval will throw off the entire mean calculation.

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

Answer:
First, let us construct both "less than" and "more than" cumulative frequency tables to determine the points for our ogives:

1. Less than cumulative frequency table:

The plotted points for the less than ogive are: \( (220, 7), (240, 10), (260, 16), (280, 24), (300, 26), (320, 30) \).

2. More than cumulative frequency table:

The plotted points for the more than ogive are: \( (200, 30), (220, 23), (240, 20), (260, 14), (280, 6), (300, 4) \).

To find the median mathematically:
Here, \( N = 30 \implies \frac{N}{2} = 15 \). The cumulative frequency just greater than 15 is 16, which corresponds to the class interval 240 - 260. Hence, 240 - 260 is the median class.
From this, we have:
\( l = 240 \), \( f = 6 \), \( cf = 10 \), \( h = 20 \)
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \text{Median} = 240 + \left( \frac{15 - 10}{6} \right) \times 20 \)
\( \text{Median} = 240 + \frac{100}{6} \approx 240 + 16.67 = 256.67 \).

The curves intersect at a point whose x-coordinate represents the median value of approximately 256.67.

In simple words: We generate coordinates for both standard cumulative curves. Plotted on a graph, the intersection point of these curves corresponds to our calculated median value of 256.67 on the horizontal axis.

Exam Tip: When drawing ogive curves, ensure you use a free-hand smooth curve rather than joining the plotted points with straight line segments.