CBSE Class 10 Mathematics Statistics Worksheet Set 02

Question. Find the class marks of classes 10–20 and 35–55.
(a) 10, 35
(b) 20, 55
(c) 15, 45
(d) 17.5, 45

Answer: C

Question. If d_i = x_i – 13, ∑f_id_i = 30 and ∑f_i = 120 , then mean, x is equal to
(a) 13
(b) 12.75
(c) 13.25
(d) 14.25

Answer: C

Question. The mean of first ten odd natural numbers is
(a) 5
(b) 10
(c) 20
(d) 19

Answer: B

Question. If the mean of first n natural numbers is 5n/9, then n is equal to
(a) 5
(b) 9
(c) 10
(d) 11

Answer: B

Question. If the mean of x, x + 3, x + 6, x + 9 and x + 12 is 10, then x equals
(a) 1
(b) 2
(c) 4
(d) 6

Answer: C

Question. Four observations are 2, 4, 6 and 8. The frequencies of the first three observations are 3, 2 and 1 respectively. If the mean of the observations is 4, then find the frequency of the fourth observation.
(a) 8
(b) 4
(c) 1
(d) 2

Answer: C

Question. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

(a) 18
(b) 20
(c) 22
(d) 19

Answer: B

Question. The mean of the following data is

(a) 20
(b) 24
(c) 22
(d) 25

Answer: D

Question. The mean of n observations x₁, x₂, x₃, ..., xn is x̄ . If each observation is multiplied by p, then the mean of the new observations is
(a) x̄/p
(b) p x̄
(c) x̄
(d) p + x̄

Answer: B

Question. The algebraic sum of all the deviations of all the observations from their mean is always
(a) 0
(b) +ve
(c) –ve
(d) equal to the number of observations.

Answer: A

Question. Consider the following frequency distribution.

The modal class is
(a) 10-20
(b) 20-30
(c) 30-40
(d) 40-50

Answer: C

Question. Life time of electric bulbs are given in the following frequency distribution.

Find the class mark of the modal class.
(a) 350
(b) 375
(c) 400
(d) 150

Answer: B

Question. The frequency of the class succeeding the modal class in the following frequency distribution is

(a) 3
(b) 6
(c) 9
(d) 12

Answer: D

Question. The modal class of data given below is 10–15, then

(a) f < 9
(b) f ≥ 9
(c) f > 9 only
(d) f < 3

Answer: B

Question. The mode for the following distribution is

(a) 36
(b) 35.5
(c) 36.25
(d) 35

Answer: C

Question. Consider the following table:

The mode of the above data is
(a) 23.5
(b) 24
(c) 24.4
(d) 25

Answer: C

Question. If the median of the data: 6, 7, x – 2, x, 17, 20 written in ascending order, is 16. Then x is equal to
(a) 15
(b) 16
(c) 17
(d) 18

Answer: C

Question. Find the class mark of the modal class in the following distribution.

(a) 45
(b) 55
(c) 65
(d) 63

Answer: C

Question. The median class for the following data is

(a) 20–40
(b) 40–60
(c) 60–80
(d) 80–100

Answer: C

Question. For a frequency distribution, mean, median and mode are connected by the relation
(a) Mode = 3 Mean – 2 Median
(b) Mode = 2 Median – 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean

Answer: C

Question. The mean and mode of a frequency distribution are 28 and 16 respectively. The median is
(a) 22
(b) 23.5
(c) 24
(d) 24.5

Answer: C

Question. If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a) 4
(b) 8
(c) 6
(d) 10

Answer: B

Question. The mean of 1, 2, 3, 4, ........, n is given by
(a) n(n +1)/2
(b) (n +1)/4
(c) n/2
(d) (n +1)/2

Answer: D

Question. The mean of 15 numbers is 25. If each number is multiplied by 4, mean of the new numbers is
(a) 60
(b) 100
(c) 10
(d) none of these

Answer: B

Question. Consider the following frequency distribution.

The upper limit of the median class is
(a) 14.5
(b) 14.5
(c) 28
(d) 28.5

Answer: D

Question. Extreme value of a given data
(a) affect the median
(b) do not affect the median
(c) nothing can be said
(d) none of the options

Answer: B

Question. One of the properties of mode is
(a) Not easy to calculate
(b) It is not affected by greatest and least values
(c) Algebraic
(d) Difference of greatest and least values

Answer: B

Question. The mean of n observations is x̄. If the first item is increased by 1, second by 2 and so on, then the new mean is
(a) x̄ + n
(b) x̄ n + 2
(c) x̄ + (n + 1/2)
(d) None of the options

Answer: C

Question. Look at the frequency distribution table given below.

The median of the above distribution is
(a) 56.5
(b) 57.5
(c) 58.5
(d) 59

Answer: B

Question. The mean, mode and median of the observations, 7, 7, 5, 7 and x are the same. Then the observation x is
(a) 10
(b) 9
(c) 8
(d) 7

Answer: B

Question. Mean of 20 observations is 15. If each observation is multiplied by 2/3, then the mean of new observations is
(a) 10
(b) 30
(c) 45
(d) 15

Answer: A

Question. The mean of six numbers : x – 5, x – 1, x, x + 2, x + 4 and x + 12 is 15. Find the mean of first four numbers.
(a) 11
(b) 12
(c) 13
(d) 14

Answer: B

Question. The numbers are arranged in the descending order : 108, 94, 88, 82, x + 7, x – 7, 60, 58, 42, 39. If the median is 73, the value of x is
(a) 72
(b) 73
(c) 76
(d) 75

Answer: B

Question. If the mean of the following distribution is 2.6, then the value of y is

(a) 3
(b) 8
(c) 13
(d) 24

Answer: B

Question. The mean of x₁, x₂,.......,x_n is M. If x_i, i = 1,2,......, n is replaced by 5xi, the mean becomes M₁, then M₁ is equal to
(a) 5M
(b) M + 5
(c) M + 100
(d) 10 M

Answer: A

Question. If mean of ten consecutive odd numbers is 120, then the mean of first five odd numbers among them is
(a) 113
(b) 115
(c) 114
(d) 116

Answer: B

Question. The numbers 3, 5, 7 and 9 have their respectively frequencies x – 2, x + 2, and x – 3, x + 3. If the mean is 6.5, then the value of x is
(a) 3
(b) 4
(c) 5
(d) 6

Answer: C

Short Answer Type Questions

Question. Write down less than type cumulative frequency and greater than type cumulative frequency.

Answer: We have

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: Let us prepare the cumulative frequency table from the given grouped data:

Since the total frequency is given as 100:
\( \implies 76 + x + y = 100 \)
\( \implies x + y = 24 \) - (Equation 1)
The median is 525, which lies in the class interval 500 - 600. Therefore, the median class is 500 - 600. We identify the following parameters:
Lower limit (\( l \)) = 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
Using 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 \( x = 9 \) into Equation 1:
\( \implies 9 + y = 24 \)
\( \implies y = 15 \)
So, the missing values are \( x = 9 \) and \( y = 15 \).
In simple words: First we find the cumulative totals. Since we know the final total is 100 and the median value is 525, we use the formula to find that \( x \) is 9 and \( y \) is 15.

Exam Tip: Pay close attention to the negative sign when subtracting the cumulative frequency term \( (36 + x) \) to ensure it is distributed correctly as \( 50 - 36 - x \).

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 verify that the calculated median class matches the interval that actually contains the given median value (28 lies between 20 and 30).

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: We find the midpoint of each group, multiply it by the group's count, and sum them up. Dividing this total by the sum of frequencies allows us to solve for P, which is 7.

Exam Tip: Be careful to perform the multiplication \( 27 \times 43 \) carefully; small arithmetic errors in these basic products will carry over to the final variable calculation.

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

Answer: Let us set up 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: Using the total count of 120 and the mean of 50, we write out two equations linking the missing values. Solving these tells us that \( F_1 = 28 \) and \( F_2 = 24 \).

Exam Tip: Whenever possible, divide equations by their common factor (like 10 in this case) to simplify the coefficients before executing substitution or elimination steps.

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

Answer: Let us prepare the calculation table:

We are given that the sum of the observations (total frequency) is 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 both sides by 20, we get:
\( \implies 5f_1 + 9f_2 = 122 \) - (Equation 2)
From 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 \( f_2 = 8 \) into Equation 1:
\( \implies f_1 + 8 = 18 \implies f_1 = 10 \)
Thus, the missing frequencies are \( f_1 = 10 \) and \( f_2 = 8 \).
In simple words: We find the cumulative total of the frequencies to make our first equation, and then use the mean of 132 to get our second equation. Solving them together yields \( f_1 = 10 \) and \( f_2 = 8 \).

Exam Tip: Be careful when calculating class marks (\( x_i \)) for large class intervals like 0-40, 40-80, etc. The midpoint is always \( \frac{\text{Lower limit} + \text{Upper limit}}{2} \).

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 compute all three statistical measures:

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 30 - 40.
Therefore, the median class is 30 - 40.
Here, \( l = 30, \ cf = 24, \ f = 15, \ h = 10 \).
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \text{Median} = 30 + \left( \frac{25 - 24}{15} \right) \times 10 \)
\( \text{Median} = 30 + \frac{10}{15} \approx 30.67 \)

3. Calculation of Mode:
The modal class is the interval with the highest frequency, which is 30 - 40 (frequency = 15).
Here, \( l = 30, \ f_1 = 15, \ f_0 = 10, \ f_2 = 5, \ h = 10 \).
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( \text{Mode} = 30 + \left( \frac{15 - 10}{2(15) - 10 - 5} \right) \times 10 \)
\( \text{Mode} = 30 + \left( \frac{5}{15} \right) \times 10 = 30 + 3.33 \approx 33.33 \)

So, Mean = 30, Median = 30.67, and Mode = 33.3.
In simple words: We list all the values in a single clean table. The mean is computed as 30, the median is calculated as 30.67, and the mode is found to be 33.3.

Exam Tip: Constructing a single joint table with cumulative frequencies, midpoints, and products saves valuable time and keeps your work well-organized.

Question 7. The mode of the following frequency distribution is 55, find the values of x and y, If the total frequency is 50 (X = 7, Y = 5)

Answer: We are 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 class, we get:
Lower limit (\( l \)) = 45
Frequency of modal class (\( f_1 \)) = 15
Frequency of preceding class (\( f_0 \)) = Y
Frequency of succeeding class (\( f_2 \)) = 10
Class size (\( h \)) = 15
Applying the mode formula:
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( \implies 55 = 45 + \left( \frac{15 - Y}{2(15) - Y - 10} \right) \times 15 \)
\( \implies 55 - 45 = \left( \frac{15 - Y}{20 - Y} \right) \times 15 \)
\( \implies 10 = \left( \frac{15 - Y}{20 - Y} \right) \times 15 \)
\( \implies \frac{10}{15} = \frac{15 - Y}{20 - Y} \implies \frac{2}{3} = \frac{15 - Y}{20 - Y} \)
By cross-multiplication:
\( \implies 2(20 - Y) = 3(15 - Y) \)
\( \implies 40 - 2Y = 45 - 3Y \)
\( \implies Y = 5 \)
Also, the total frequency is given as 50:
\( \implies 6 + 7 + Y + 15 + 10 + X = 50 \)
\( \implies 38 + 5 + X = 50 \)
\( \implies 43 + X = 50 \implies X = 7 \)
Thus, the missing frequencies are \( X = 7 \) and \( Y = 5 \).
In simple words: Since the mode is 55, it belongs to the 45-60 interval. Working through the mode formula gives \( Y = 5 \). Knowing that the total sum of frequencies is 50 allows us to find \( X = 7 \).

Exam Tip: The modal class is defined by where the mode value lies, not by the largest visible frequency in the table since 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 (x-coordinate) of the point of intersection \( P(x, y) \) of the less-than and more-than ogives represents the Median of the given frequency distribution.
In simple words: On a cumulative frequency graph, the x-coordinate of the point where the two lines cross gives us the median value of the data.

Exam Tip: Remember that while the x-coordinate (abscissa) represents the median, the y-coordinate (ordinate) represents \( \frac{N}{2} \) (half of the total frequency).

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

Answer: The empirical relationship connecting the three measures of central tendency is:
\( \text{Mode} = 3 \ \text{Median} - 2 \ \text{Mean} \)
This can also be expressed as:
\( 3 \ \text{Median} = \text{Mode} + 2 \ \text{Mean} \)
In simple words: The mode of a dataset is equal to three times its median minus two times its mean.

Exam Tip: This formula is highly scoring in short-answer questions. Memorize it carefully to solve statistical problems quickly.

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

Answer: Given values:
\( \text{Median} = 15 \)
\( \text{Mean} = 16 \)
Using the empirical relationship formula:
\( \text{Mode} = 3 \ \text{Median} - 2 \ \text{Mean} \)
\( \implies \text{Mode} = 3(15) - 2(16) \)
\( \implies \text{Mode} = 45 - 32 = 13 \)
Thus, the mode of the distribution is 13.
In simple words: By putting the median of 15 and mean of 16 into our formula, we find that the mode is 13.

Exam Tip: Write down the algebraic formula before substituting any values; this helps secure partial marks in case of an arithmetic oversight.

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

Answer: First, let us convert the cumulative "more than" table to a regular frequency distribution table:
- Frequency of class 0-10: \( 60 - 56 = 4 \)
- Frequency of class 10-20: \( 56 - 40 = 16 \)
- Frequency of class 20-30: \( 40 - 20 = 20 \)
- Frequency of class 30-40: \( 20 - 10 = 10 \)
- Frequency of class 40-50: \( 10 - 3 = 7 \)
- Frequency of class 50-60: 3

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

Using the arithmetic mean formula:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1590}{60} = 26.5 \)
Thus, the arithmetic mean of the distribution is 26.5.
In simple words: First we turn the cumulative "more than" table into regular intervals. Then we find the midpoints, multiply them by the group frequencies, and divide the sum by 60 to get 26.5.

Exam Tip: When converting from cumulative to normal tables, double-check that the sum of the derived frequencies equals the total cumulative value (60) to confirm your conversion is accurate.

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

Answer: Let us construct the cumulative frequency tables for both curve types:

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:
Total frequency \( N = 30 \), so \( \frac{N}{2} = 15 \). The intersection of the two curves yields the median.
By formula verification:
The median class is 240-260 since its cumulative frequency is 16.
With \( l = 240, \ cf = 10, \ f = 6, \ h = 20 \):
\( \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \)
\( \text{Median} = 240 + \left( \frac{15 - 10}{6} \right) \times 20 = 240 + \frac{100}{6} \approx 256.67 \)

Below is the graphical representation of both curves:
In simple words: We plot both cumulative curves together. The point where they cross each other projects straight down to 256.7 on the horizontal axis, which gives the median.

Exam Tip: Drawing dotted lines from the intersection point to the axes helps highlight your visual solution on the graph clearly to the examiner.

Question 13. Compute the arithmetic mean for the following data:

Answer: Let us convert this "less than" cumulative table to a regular frequency distribution table and calculate the arithmetic mean:
- Frequency of class 0-10: 14
- Frequency of class 10-20: \( 22 - 14 = 8 \)
- Frequency of class 20-30: \( 37 - 22 = 15 \)
- Frequency of class 30-40: \( 58 - 37 = 21 \)
- Frequency of class 40-50: \( 67 - 58 = 9 \)
- Frequency of class 50-60: \( 75 - 67 = 8 \)

Now, let us calculate the mean using class marks (\( x_i \)):

Using the arithmetic mean formula:
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2145}{75} = 28.6 \)
Thus, the arithmetic mean is 28.6.
In simple words: First we turn the "less than" table into regular intervals by finding the difference between each step. Next we calculate midpoints, multiply them by the frequencies, and divide the total by 75 to get 28.6.

Exam Tip: When converting cumulative frequencies, remember that the individual class frequencies must sum up to the total number of students in the final class (75).