CBSE Class 10 Mathematics Statistics Worksheet Set 09

CBSE Class 10 Maths Worksheet - Statistics (8). 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.

STATICTICS

(i) Assumed Mean method or Shortcut method

Mean =

= a +

Where a = assumed mean

And di= Xi - a

(ii) Step deviation method.

Mean =

= a +

Where a = assumed mean

h = class size

And ui= (Xi – a)/h

 Median of a grouped frequency distribution can be calculated by

Median = l +

Where

l = lower limit of median class

n = number of observations

cf = cumulative frequency of class preceding the median class

f = frequency of median class

h = class size of the median class.

 Mode of grouped data can be calculated by the following formula.

Mode = l +

Where

l = lower limit of modal class

h = size of class interval

f1 = Frequency of the modal class

fo = frequency of class preceding the modal class

f2= frequency of class succeeding the modal class

 Empirical relationship between the three measures of central tendency.

3 Median = Mode + 2 Mean

Or, Mode = 3 Median – 2 Mean

 Ogive

Ogive is the graphical representation of the cumulative frequency distribution. It is of two types:

(i) Less than type ogive.

(ii) More than type ogive

 Median by graphical method

The x-coordinated of the point of intersection of ‘less than ogive’ and ‘more than ogive’ gives the median.

LEVEL – I

Slno

Question

1

What is the mean of 1st ten prime numbers?

2

What measure of central tendency is represented by the abscissa of the point where less than ogive and more than ogive intersect?

3

If the mode of a data is 45 and mean is 27, then median is ___________.

4

Find the mode of the following

Xi

35

38

40

42

44

fi

5

9

10

7

2

5

Write the median class of the following distribution.

Class

0-10

10-20

20-30

30-40

40-50

50-60

60-70

Frequency

4

4

8

10

12

8

4

6

The wickets taken by a bowler in 10 cricket matches are as follows: 2, 6 ,4 ,5, 0, 2, 1, 3, 2, 3 Find the mode of the data

7.

How one can find median of a frequency distribution graphically

8.

What important information one can get by the abscissa of the point of intersection of the less than type and the more than type commulative frequency curve of a group data

LEVEL – II

Slno

Question

Ans

1

Find the median of the following frequency distribution

Height in cm

160-162

163-165

166-168

169-171

172-174

Frequency

15

117

136

118

14

167

2

Given below is the distribution of IQ of the 100 students. Find the median IQ

IQ

75-84

85-94

95-104

105-114

115-124

125-134

135-144

Frequency

8

11

26

31

18

4

2

106.1

3

Find the median of the following distribution

Class interval

0-10

10-20

20-30

30-40

40-50

50-60

Frequency

5

8

20

15

7

5

28.5

4

A class teacher has the following absentee record of 40 students of a class for the whole

term.

No. of days

0-6

6-10

10-14

14-20

20-28

28-38

38-40

No. of students

11

10

7

4

4

3

1

Write the above distribution as less than type cumulative frequency distribution.

5

Using the assumed mean method find the mean of the following data.

Class interval

0-10

10-20

20-30

30-40

40-50

frequency

7

8

12

13

10

Ans

27.2

6

Name the keyword used for central tendency

Mean , median , mode

LEVEL – III

SN

Question

Ans

1

If the mean distribution is 25

Class

0-10

10-20

20-30

30-40

40-50

Frequency

5

18

15

P

6

Then find p.

P=16

2

Find the mean of the following frequency distribution using step deviation method

Class

0-10

10-20

20-30

30-40

40-50

Frequency

7

12

13

10

8

25

3

Find the value of p if the median of the following frequency distribution is 50

Class

20-30

30-40

40-50

50-60

60-70

70-80

80-90

Frequency

25

15

P

6

24

12

8

P=10

4

Find the median of the following data

Marks

Less Than 10

Less Than 30

Less Than 50

Less Than 70

Less Than 90

Less Than 110

Less Than 130

Less than 150

Frequency

0

10

25

43

65

87

96

100

.

76.36

5

Compare the modal ages of two groups of students appearing for entrance examination.

Age in yrs

16-18

18-20

20-22

22-24

24-26

Group A

50

78

46

28

23

Key Points

Measures of Central Tendency

1. Mean of Grouped Data

• Assumed Mean Method (Shortcut Method):
  \[ \text{Mean } (\bar{X}) = a + \frac{\sum f_i d_i}{\sum f_i} \] where \( a \) is the assumed mean, and \( d_i = x_i - a \) is the deviation of each class mark from \( a \).
• Step Deviation Method:
  \[ \text{Mean } (\bar{X}) = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h \] where \( a \) is the assumed mean, \( h \) is the class size, and \( u_i = \frac{x_i - a}{h} \).

2. Median of Grouped Data

The median is calculated using the following formula:

\[ \text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h \]

where:

• \( l \) = lower limit of the median class
• \( n \) = total number of observations
• \( cf \) = cumulative frequency of the class preceding the median class
• \( f \) = frequency of the median class
• \( h \) = class size of the median class

3. Mode of Grouped Data

The mode is calculated using the following formula:

\[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]

where:

• \( l \) = lower limit of the modal class
• \( h \) = size of the class interval
• \( f_1 \) = frequency of the modal class
• \( f_0 \) = frequency of the class preceding the modal class
• \( f_2 \) = frequency of the class succeeding the modal class

4. Empirical Relationship

The relationship between the three measures of central tendency is given by:

\[ 3 \text{ Median} = \text{Mode} + 2 \text{ Mean} \]

Or, equivalently:

\[ \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \]

5. Graphical Representation (Ogive)

An ogive is a graphical representation of a cumulative frequency distribution. It is categorized into two types:

• (i) Less than type ogive
• (ii) More than type ogive

The x-coordinate of the point of intersection of the "less than ogive" and the "more than ogive" gives the median value.

LEVEL - I

Question 1. What is the mean of 1st ten prime numbers?
Answer: The first ten prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
The sum of these prime numbers is:
\[ 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129 \]
Dividing this sum by the total count of observations (10):
\[ \text{Mean} = \frac{129}{10} = 12.9 \]
Therefore, the mean of the first ten prime numbers is 12.9.
In simple words: We list the first ten prime numbers, add them up to get 129, and then divide by 10 to find the average of 12.9.

Exam Tip: Do not include 1 in your list of prime numbers, and remember that 2 is the only even prime number.

Question 2. What measure of central tendency is represented by the abscissa of the point where less than ogive and more than ogive intersect?
Answer: The x-coordinate (abscissa) of the point where the "less than" cumulative frequency curve and the "more than" cumulative frequency curve intersect represents the **Median** of the grouped data.
In simple words: The point where the two cumulative frequency curves cross has an x-value that is exactly the median.

Exam Tip: This property is frequently asked as a one-mark theoretical question; remember that the abscissa refers to the X-axis value.

Question 3. If the mode of a data is 45 and mean is 27, then median is ___________.
Answer: We use the empirical formula relating the three measures of central tendency:
\[ \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \]
Given that \( \text{Mode} = 45 \) and \( \text{Mean} = 27 \):
\[ 45 = 3 \text{ Median} - 2(27) \]
\[ \implies 45 = 3 \text{ Median} - 54 \]
\[ \implies 3 \text{ Median} = 45 + 54 \]
\[ \implies 3 \text{ Median} = 99 \]
\[ \implies \text{Median} = 33 \]
Therefore, the median is 33.
In simple words: Using the formula \( \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \), we plug in our values and solve to find the median is 33.

Exam Tip: Memorize the empirical formula \( 3 \text{ Median} = \text{Mode} + 2 \text{ Mean} \) because it is extremely useful for quick calculations.

Question 4. Find the mode of the following

Answer: The mode is the value of the observation that has the highest frequency:
- In this table, the maximum frequency is 10.
- The corresponding value of \( X_i \) for this frequency is 40.
Therefore, the mode of the data is 40.
In simple words: We look at the table to find which number has the highest frequency. Since 40 has a frequency of 10 (the highest), the mode is 40.

Exam Tip: The mode is always the value of the observation (\( X_i \)), not the frequency (\( f_i \)) itself.

Question 5. Write the median class of the following distribution.

Answer: Let's find the cumulative frequencies (\( cf \)) for each class interval:
- 0-10: \( cf = 4 \)
- 10-20: \( cf = 4 + 4 = 8 \)
- 20-30: \( cf = 8 + 8 = 16 \)
- 30-40: \( cf = 16 + 10 = 26 \)
- 40-50: \( cf = 26 + 12 = 38 \)
- 50-60: \( cf = 38 + 8 = 46 \)
- 60-70: \( cf = 46 + 4 = 50 \)

The total frequency \( N = 50 \), so \( \frac{N}{2} = 25 \).
The cumulative frequency just greater than 25 is 26, which belongs to the class interval 30-40.
Therefore, the median class is 30-40.
In simple words: We add up the frequencies progressively. The middle of our 50 total observations is 25. The class interval containing this 25th observation is 30-40.

Exam Tip: Always construct a list of cumulative frequencies to identify the median class accurately.

Question 6. The wickets taken by a bowler in 10 cricket matches are as follows: 2, 6 ,4 ,5, 0, 2, 1, 3, 2, 3 Find the mode of the data
Answer: Let's tabulate the frequency of each number of wickets taken:
- 0 wickets: 1 match
- 1 wicket: 1 match
- 2 wickets: 3 matches
- 3 wickets: 2 matches
- 4 wickets: 1 match
- 5 wickets: 1 match
- 6 wickets: 1 match

Since the observation 2 has the highest frequency (occurring 3 times), the mode of the data is 2.
In simple words: We count how many times each number of wickets appears. Since 2 wickets appears most often (3 times), the mode is 2.

Exam Tip: For ungrouped data, count the occurrences of each unique value; the value with the highest count is the mode.

Question 7. How one can find median of a frequency distribution graphically
Answer: We can find the median graphically by drawing both the 'less than' type and 'more than' type cumulative frequency curves (ogives) on the same coordinate axes. The x-coordinate (abscissa) of the point where these two curves intersect gives the median of the distribution.
In simple words: Plot both the less-than and more-than ogives. The point where they cross tells you the median on the horizontal axis.

Exam Tip: Alternatively, we can draw a single ogive, locate \( \frac{N}{2} \) on the y-axis, draw a horizontal line to the curve, and project it down to the x-axis to read the median.

Question 8. What important information one can get by the abscissa of the point of intersection of the less than type and the more than type commulative frequency curve of a group data
Answer: The x-coordinate (abscissa) of the point of intersection of the "less than" type and "more than" type cumulative frequency curves of grouped data provides the **Median** value of the dataset.
In simple words: The horizontal position of the point where both cumulative curves intersect gives us the median value.

Exam Tip: Be familiar with graphical terms: "abscissa" refers to the x-coordinate, and "ordinate" refers to the y-coordinate.

LEVEL - II

Question 1. Find the median of the following frequency distribution

Answer: First, convert the discontinuous class intervals to continuous ones by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits:
- 159.5 - 162.5: frequency = 15, cf = 15
- 162.5 - 165.5: frequency = 117, cf = 132
- 165.5 - 168.5: frequency = 136, cf = 268
- 168.5 - 171.5: frequency = 118, cf = 386
- 171.5 - 174.5: frequency = 14, cf = 400

The total frequency \( N = 400 \), so \( \frac{N}{2} = 200 \).
The cumulative frequency just greater than 200 is 268, which lies in the continuous class 165.5 - 168.5.
Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
Substitute the values: \( l = 165.5 \), \( cf = 132 \), \( f = 136 \), \( h = 3 \):
\[ \text{Median} = 165.5 + \left(\frac{200 - 132}{136}\right) \times 3 \]
\[ \text{Median} = 165.5 + \left(\frac{68}{136}\right) \times 3 \]
\[ \text{Median} = 165.5 + 0.5 \times 3 = 165.5 + 1.5 = 167 \]
Therefore, the median height is 167 cm.
In simple words: We adjust the intervals to make them continuous. The midpoint of our 400 total observations is 200, which falls in the 165.5 - 168.5 group. Plugging these values into the median formula gives us 167 cm.

Exam Tip: When class intervals have gaps (like 162 and 163), always convert them to continuous classes before identifying the lower limit \( l \) and class size \( h \).

Question 2. Given below is the distribution of IQ of the 100 students. Find the median IQ

Answer: Let's convert the class intervals to continuous ones:
- 74.5 - 84.5: frequency = 8, cf = 8
- 84.5 - 94.5: frequency = 11, cf = 19
- 94.5 - 104.5: frequency = 26, cf = 45
- 104.5 - 114.5: frequency = 31, cf = 76
- 114.5 - 124.5: frequency = 18, cf = 94
- 124.5 - 134.5: frequency = 4, cf = 98
- 134.5 - 144.5: frequency = 2, cf = 100

Here, \( N = 100 \implies \frac{N}{2} = 50 \).
The cumulative frequency just greater than 50 is 76, which corresponds to the continuous class 104.5 - 114.5.
Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
Substitute \( l = 104.5 \), \( cf = 45 \), \( f = 31 \), \( h = 10 \):
\[ \text{Median} = 104.5 + \left(\frac{50 - 45}{31}\right) \times 10 \]
\[ \text{Median} = 104.5 + \frac{50}{31} \approx 104.5 + 1.61 = 106.11 \]
Therefore, the median IQ of the students is approximately 106.1.
In simple words: We adjust the IQ gaps to make them continuous. The middle value is 50, which falls into the 104.5 - 114.5 group. Using our formula gives a median IQ of 106.1.

Exam Tip: Make sure the continuous conversion step is clearly written out; skipping it changes the lower limit \( l \) and class size \( h \) and leads to errors.

Question 3. Find the median of the following distribution

Answer: Let's find the cumulative frequencies for each class interval:
- 0-10: frequency = 5, cf = 5
- 10-20: frequency = 8, cf = 13
- 20-30: frequency = 20, cf = 33
- 30-40: frequency = 15, cf = 48
- 40-50: frequency = 7, cf = 55
- 50-60: frequency = 5, cf = 60

Total frequency \( N = 60 \), so \( \frac{N}{2} = 30 \).
The cumulative frequency just greater than 30 is 33, which corresponds to the class interval 20-30.
Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
Substitute \( l = 20 \), \( cf = 13 \), \( f = 20 \), \( h = 10 \):
\[ \text{Median} = 20 + \left(\frac{30 - 13}{20}\right) \times 10 \]
\[ \text{Median} = 20 + \frac{17}{2} = 20 + 8.5 = 28.5 \]
Therefore, the median of the distribution is 28.5.
In simple words: We list cumulative frequencies. The 30th student falls into the 20-30 interval. Plugging these values into the formula yields a median of 28.5.

Exam Tip: For continuous classes like these, we do not need any boundary adjustments; we can apply the formula directly.

Question 4. A class teacher has the following absentee record of 40 students of a class for the whole term.

Write the above distribution as less than type cumulative frequency distribution.
Answer: To convert the given frequency distribution into a "less than" type cumulative frequency distribution, we accumulate the frequencies at the upper limit of each class interval:

In simple words: We sum up the frequencies progressively at each class interval's upper limit. This tells us the total number of students who were absent less than a certain number of days.

Exam Tip: When writing "less than" distributions, always use the upper limits of each interval as the boundary values.

Question 5. Using the assumed mean method find the mean of the following data.

Answer: First, find the class marks (\( x_i \)) for each interval, which is the midpoint:
- 0-10: \( x_i = 5 \)
- 10-20: \( x_i = 15 \)
- 20-30: \( x_i = 25 \)
- 30-40: \( x_i = 35 \)
- 40-50: \( x_i = 45 \)

Let the assumed mean \( a = 25 \).
Next, compute the deviations \( d_i = x_i - a = x_i - 25 \) and the product \( f_i d_i \):
- For 0-10: \( d_i = 5 - 25 = -20 \implies f_i d_i = 7 \times (-20) = -140 \)
- For 10-20: \( d_i = 15 - 25 = -10 \implies f_i d_i = 8 \times (-10) = -80 \)
- For 20-30: \( d_i = 25 - 25 = 0 \implies f_i d_i = 12 \times 0 = 0 \)
- For 30-40: \( d_i = 35 - 25 = 10 \implies f_i d_i = 13 \times 10 = 130 \)
- For 40-50: \( d_i = 45 - 25 = 20 \implies f_i d_i = 10 \times 20 = 200 \)

Calculate the sums:
- \( \sum f_i = 7 + 8 + 12 + 13 + 10 = 50 \)
- \( \sum f_i d_i = -140 - 80 + 0 + 130 + 200 = 110 \)

Using the assumed mean method formula:
\[ \bar{X} = a + \frac{\sum f_i d_i}{\sum f_i} \]
\[ \bar{X} = 25 + \frac{110}{50} \]
\[ \bar{X} = 25 + 2.2 = 27.2 \]
Therefore, the mean of the given data is 27.2.
In simple words: We find the middle number of each interval. By picking 25 as our "guess average" (assumed mean) and using the formula to correct it, we find the true average is 27.2.

Exam Tip: Make sure to create a neat table containing columns for \( x_i \), \( d_i \), and \( f_i d_i \) to make your calculations clear and easy for the examiner to follow.

Question 6. Name the keyword used for central tendency
Answer: The three principal measures used to describe the central tendency of a frequency distribution are the **Mean**, the **Median**, and the **Mode**.
In simple words: The terms used to measure the center of data are mean, median, and mode.

Exam Tip: These three terms summarize different ways of finding the middle point of a dataset.

LEVEL - III

Question 1. If the mean distribution is 25

Then find p.
Answer: Let's find the class marks (\( x_i \)) and compute the products \( f_i x_i \):
- For 0-10: \( x_i = 5 \implies f_i x_i = 5 \times 5 = 25 \)
- For 10-20: \( x_i = 15 \implies f_i x_i = 18 \times 15 = 270 \)
- For 20-30: \( x_i = 25 \implies f_i x_i = 15 \times 25 = 375 \)
- For 30-40: \( x_i = 35 \implies f_i x_i = p \times 35 = 35p \)
- For 40-50: \( x_i = 45 \implies f_i x_i = 6 \times 45 = 270 \)

Calculate the sums:
- \( \sum f_i = 5 + 18 + 15 + p + 6 = 44 + p \)
- \( \sum f_i x_i = 25 + 270 + 375 + 35p + 270 = 940 + 35p \)

Using the mean formula:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \]
Substitute the given mean of 25:
\[ 25 = \frac{940 + 35p}{44 + p} \]
\[ \implies 25(44 + p) = 940 + 35p \]
\[ \implies 1100 + 25p = 940 + 35p \]
\[ \implies 1100 - 940 = 35p - 25p \]
\[ \implies 160 = 10p \]
\[ \implies p = 16 \]
Therefore, the value of \( p \) is 16.
In simple words: We multiply the midpoints of the classes by their frequencies and find the total sum. Setting up the average formula with the given mean of 25, we solve the algebraic equation to find \( p = 16 \).

Exam Tip: Write out the algebraic steps carefully when cross-multiplying to avoid any mistake with the variable \( p \).

Question 2. Find the mean of the following frequency distribution using step deviation method

Answer: Let the assumed mean \( a = 25 \) and the class size \( h = 10 \).
Let's find the class marks (\( x_i \)), step deviations (\( u_i = \frac{x_i - a}{h} \)), and the products \( f_i u_i \):
- For 0-10: \( x_i = 5 \implies u_i = \frac{5 - 25}{10} = -2 \implies f_i u_i = 7 \times (-2) = -14 \)
- For 10-20: \( x_i = 15 \implies u_i = \frac{15 - 25}{10} = -1 \implies f_i u_i = 12 \times (-1) = -12 \)
- For 20-30: \( x_i = 25 \implies u_i = \frac{25 - 25}{10} = 0 \implies f_i u_i = 13 \times 0 = 0 \)
- For 30-40: \( x_i = 35 \implies u_i = \frac{35 - 25}{10} = 1 \implies f_i u_i = 10 \times 1 = 10 \)
- For 40-50: \( x_i = 45 \implies u_i = \frac{45 - 25}{10} = 2 \implies f_i u_i = 8 \times 2 = 16 \)

Calculate the sums:
- \( \sum f_i = 7 + 12 + 13 + 10 + 8 = 50 \)
- \( \sum f_i u_i = -14 - 12 + 0 + 10 + 16 = 0 \)

Using the step deviation formula:
\[ \bar{X} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h \]
\[ \bar{X} = 25 + \left(\frac{0}{50}\right) \times 10 \]
\[ \bar{X} = 25 + 0 = 25 \]
Therefore, the mean of the given distribution is 25.
In simple words: We use the step deviation method to find the average. Since the deviations above and below our assumed middle point of 25 perfectly balance each other out (the sum is 0), the true mean is exactly 25.

Exam Tip: The step deviation method is highly recommended for larger numbers as it simplifies calculations to very small, manageable integers.

Question 3. Find the value of p if the median of the following frequency distribution is 50

Answer: First, construct the cumulative frequency (cf) table:
- 20-30: frequency = 25, cf = 25
- 30-40: frequency = 15, cf = 40
- 40-50: frequency = \( p \), cf = \( 40 + p \)
- 50-60: frequency = 6, cf = \( 46 + p \)
- 60-70: frequency = 24, cf = \( 70 + p \)
- 70-80: frequency = 12, cf = \( 82 + p \)
- 80-90: frequency = 8, cf = \( 90 + p \)

The total frequency \( N = 90 + p \).
Given that the median is 50, which lies on the boundary, we use the median class 50-60:
- \( l = 50 \), \( f = 6 \), \( cf = 40 + p \), and \( h = 10 \).

Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
Substitute the values:
\[ 50 = 50 + \left(\frac{\frac{90+p}{2} - (40+p)}{6}\right) \times 10 \]
\[ \implies 0 = \frac{90+p}{2} - (40+p) \]
\[ \implies \frac{90+p}{2} = 40+p \]
\[ \implies 90 + p = 2(40 + p) \]
\[ \implies 90 + p = 80 + 2p \]
\[ \implies p = 10 \]
Therefore, the value of \( p \) is 10.
In simple words: We set up the cumulative frequency formula with the variable p. Plugging our known median of 50 into the formula simplifies the equation, leading to \( p = 10 \).

Exam Tip: Be extra careful with bracket signs when subtracting the cumulative frequency term \( (40+p) \) in the numerator of the formula.

Question 4. Find the median of the following data

Answer: The given data is of "less than" cumulative frequency type. Let's convert it to a standard class interval frequency distribution:
- 10-30: frequency = 10, cf = 10
- 30-50: frequency = 25 - 10 = 15, cf = 25
- 50-70: frequency = 43 - 25 = 18, cf = 43
- 70-90: frequency = 65 - 43 = 22, cf = 65
- 90-110: frequency = 87 - 65 = 22, cf = 87
- 110-130: frequency = 96 - 87 = 9, cf = 96
- 130-150: frequency = 100 - 96 = 4, cf = 100

Here, \( N = 100 \implies \frac{N}{2} = 50 \).
The cumulative frequency just greater than 50 is 65, which corresponds to the class interval 70-90.
So, Median Class = 70-90.
- \( l = 70 \), \( cf = 43 \), \( f = 22 \), and \( h = 20 \).

Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
\[ \text{Median} = 70 + \left(\frac{50 - 43}{22}\right) \times 20 \]
\[ \text{Median} = 70 + \frac{140}{22} \approx 70 + 6.36 = 76.36 \]
Therefore, the median of the given data is 76.36.
In simple words: We convert the "less than" data to normal classes. Finding the 50th student leads us to the 70-90 group, and using the formula gives a median score of 76.36.

Exam Tip: Note that the class size here is \( h = 20 \) because the intervals have a width of 20.

Question 5. Compare the modal ages of two groups of students appearing for entrance examination.

Answer: Let's calculate the modal age for each group separately:

1. For Group A:
The highest frequency is 78, which corresponds to the class 18-20.
- \( l = 18 \), \( f_1 = 78 \), \( f_0 = 50 \), \( f_2 = 46 \), and \( h = 2 \).
Using the mode formula:
\[ \text{Mode}_A = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]
\[ \text{Mode}_A = 18 + \left(\frac{78 - 50}{2(78) - 50 - 46}\right) \times 2 \]
\[ \text{Mode}_A = 18 + \left(\frac{28}{156 - 96}\right) \times 2 \]
\[ \text{Mode}_A = 18 + \frac{56}{60} \approx 18 + 0.93 = 18.93\text{ years} \]

2. For Group B:
The highest frequency is 89, which also corresponds to the class 18-20.
- \( l = 18 \), \( f_1 = 89 \), \( f_0 = 54 \), \( f_2 = 40 \), and \( h = 2 \).
Using the mode formula:
\[ \text{Mode}_B = 18 + \left(\frac{89 - 54}{2(89) - 54 - 40}\right) \times 2 \]
\[ \text{Mode}_B = 18 + \left(\frac{35}{178 - 94}\right) \times 2 \]
\[ \text{Mode}_B = 18 + \frac{70}{84} \approx 18 + 0.83 = 18.83\text{ years} \]

Comparison:
The modal age of Group A is 18.93 years, while the modal age of Group B is 18.83 years. Therefore, Group A has a slightly higher modal age than Group B.
In simple words: We calculate the peak (modal) age for both groups. Group A has a modal age of 18.93 years and Group B has 18.83 years, making Group A's modal age slightly higher.

Exam Tip: Always state the comparison conclusion clearly at the end of your answer to earn full marks.

Question 6. The mean of the following frequency distribution is 57.6 and the sum of the observations is 50. Find the missing frequencies f1 and f2.

Answer: First, write down the sum of frequencies equation:
\[ 7 + f_1 + 12 + f_2 + 8 + 5 = 50 \]
\[ \implies f_1 + f_2 + 32 = 50 \implies f_1 + f_2 = 18 \tag{Equation 1} \]

Next, determine the class marks (\( x_i \)) for each interval:
- 0-20: \( x_i = 10 \implies f_i x_i = 70 \)
- 20-40: \( x_i = 30 \implies f_i x_i = 30f_1 \)
- 40-60: \( x_i = 50 \implies f_i x_i = 600 \)
- 60-80: \( x_i = 70 \implies f_i x_i = 70f_2 \)
- 80-100: \( x_i = 90 \implies f_i x_i = 720 \)
- 100-120: \( x_i = 110 \implies f_i x_i = 550 \)

Using the mean formula:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \]
\[ 57.6 = \frac{70 + 30f_1 + 600 + 70f_2 + 720 + 550}{50} \]
\[ 57.6 \times 50 = 1940 + 30f_1 + 70f_2 \]
\[ 2880 = 1940 + 30f_1 + 70f_2 \]
\[ 30f_1 + 70f_2 = 940 \implies 3f_1 + 7f_2 = 94 \tag{Equation 2} \]

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

Subtract Equation 3 from Equation 2:
\[ 4f_2 = 40 \implies f_2 = 10 \]
Substitute \( f_2 = 10 \) back 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: By setting up equations using the total frequency (50) and the average value (57.6), we solve the linear system to find the missing counts are 8 and 10.

Exam Tip: Missing frequency calculations are very standard; double check each step to ensure no algebraic mistakes are made.

Question 7. The following distribution give the daily income of 65 workers of a factory

Convert the above to a more than type cumulative frequency distribution and draw its ogive.
Answer: First, convert the given distribution into a "more than" type cumulative frequency distribution:

To draw the ogive, we plot the coordinates \( (100, 65), (120, 51), (140, 35), (160, 25), \) and \( (180, 9) \) on the grid and join them with a smooth, freehand curve:
In simple words: We construct a "more than" table by starting with all 65 workers and subtracting the frequency of each class. Plotting these points produces a downward-sloping cumulative frequency curve.

Exam Tip: In more than type cumulative frequency distributions, the cumulative frequencies are plotted corresponding to the lower limits of each class interval.

Question 8. Draw a less than type and more than type ogives for the following distribution on the same graph. Also find the median from the graph.

Answer: First, convert the discontinuous class intervals into continuous intervals:
- 29.5 - 39.5: Frequency = 14
- 39.5 - 49.5: Frequency = 6
- 49.5 - 59.5: Frequency = 10
- 59.5 - 69.5: Frequency = 20
- 69.5 - 79.5: Frequency = 30
- 79.5 - 89.5: Frequency = 8
- 89.5 - 99.5: Frequency = 12

Now, construct both cumulative frequency tables:

Plotting both the "less than" and "more than" ogives on the same coordinate axes, they intersect at a point.
Since \( N = 100 \implies \frac{N}{2} = 50 \), the curves intersect exactly at \( (69.5, 50) \).
The x-coordinate of the point of intersection is 69.5.
Therefore, the median is 69.5.
In simple words: We plot both cumulative frequency graphs together. They intersect exactly at the 50th student height, giving an x-value of 69.5, which is our median.

Exam Tip: Remember that if the intervals are discontinuous, you must find the continuous class boundaries first to plot the points at the correct coordinates on the graph.

SELF - EVALUATION

Question 1. What is the value of the median of the data using the graph in figure of less than ogive and more than ogive?
Answer: The "less than" and "more than" ogives shown in the figure intersect at a specific point on the graph.
The coordinates of this point of intersection are \( (4, 15) \).
The x-coordinate (abscissa) represents the median of the distribution.
Therefore, the median of the data is 4.
In simple words: From the given graph, the point where the two ogive curves cross has an x-value of 4, meaning the median is 4.

Exam Tip: Graphical reading is quick; simply identify the intersection point's projection on the horizontal axis (X-axis).

Question 2. If mean =60 and median =50, then find mode using empirical relationship.
Answer: We use the standard empirical relation between the measures of central tendency:
\[ \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \]
Substitute the given values of \( \text{Mean} = 60 \) and \( \text{Median} = 50 \):
\[ \text{Mode} = 3(50) - 2(60) \]
\[ \text{Mode} = 150 - 120 = 30 \]
Therefore, the mode is 30.
In simple words: Using the formula \( \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \), we calculate \( 3 \times 50 - 2 \times 60 \), which gives 30.

Exam Tip: The empirical formula is \( 3 \text{ Median} = \text{Mode} + 2 \text{ Mean} \); be sure to rearrange it correctly before solving.

Question 3. Find the value of p, if the mean of the following distribution is 18.

Answer: Let's write down the product of each variate \( x_i \) and its frequency \( f_i \):
- \( 13 \times 8 = 104 \)
- \( 15 \times 2 = 30 \)
- \( 17 \times 3 = 51 \)
- \( 19 \times 4 = 76 \)
- \( (20 + p) \times 5p = 100p + 5p^2 \)
- \( 23 \times 6 = 138 \)

Calculate the sums:
- \( \sum f_i = 8 + 2 + 3 + 4 + 5p + 6 = 23 + 5p \)
- \( \sum f_i x_i = 104 + 30 + 51 + 76 + 100p + 5p^2 + 138 = 399 + 100p + 5p^2 \)

Using the mean formula:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \]
Substitute the given mean of 18:
\[ 18 = \frac{399 + 100p + 5p^2}{23 + 5p} \]
\[ \implies 18(23 + 5p) = 399 + 100p + 5p^2 \]
\[ \implies 414 + 90p = 399 + 100p + 5p^2 \]
Rearranging terms into a quadratic equation:
\[ 5p^2 + 10p - 15 = 0 \]
Divide by 5:
\[ p^2 + 2p - 3 = 0 \]
Factorize the quadratic equation:
\[ (p + 3)(p - 1) = 0 \implies p = 1 \text{ or } p = -3 \]
Since frequency must be a positive quantity, \( p = -3 \) is not possible.
Therefore, \( p = 1 \).
In simple words: We multiply the values by their frequencies to set up the mean equation. This leads to a quadratic equation that simplifies to \( (p+3)(p-1) = 0 \). Since frequencies can't be negative, \( p = 1 \) is the only correct answer.

Exam Tip: When factorizing quadratic equations in statistics, always write a line explaining why you discarded the negative value.

Question 4. Find the mean, mode and median for the following data.

Answer: Let's construct the complete frequency distribution table with class marks (\( x_i \)), products (\( f_i x_i \)), and cumulative frequencies (\( cf \)):

1. Mean Calculation: \[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2615}{75} \approx 34.87 \] 2. Median Calculation: Here, \( N = 75 \implies \frac{N}{2} = 37.5 \). The cumulative frequency just greater than 37.5 is 48, which lies in the class interval 30-40. So, Median Class = 30-40. - \( l = 30 \), \( cf = 28 \), \( f = 20 \), and \( h = 10 \). \[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \] \[ \text{Median} = 30 + \left(\frac{37.5 - 28}{20}\right) \times 10 = 30 + \frac{9.5}{2} = 30 + 4.75 = 34.75 \] 3. Mode Calculation: The maximum frequency is 20, corresponding to the modal class 30-40. - \( l = 30 \), \( f_1 = 20 \), \( f_0 = 15 \), \( f_2 = 14 \), and \( h = 10 \). \[ \text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \] \[ \text{Mode} = 30 + \left(\frac{20 - 15}{2(20) - 15 - 14}\right) \times 10 \] \[ \text{Mode} = 30 + \left(\frac{5}{11}\right) \times 10 \approx 30 + 4.55 = 34.55 \] In simple words: We compile all our values to find the three averages. The mean is 34.87, the median is 34.75, and the mode is 34.55.

Exam Tip: Make sure to write out the individual steps and calculations for each of the three statistics as they carry separate marks in long-answer questions.

Self-Evaluation Questions

Question 5. The median of the following data is 52.5. find the value of x and y, if the total frequency is 100.

Answer: First, write down the sum of all frequencies:
\[ 2 + 5 + X + 12 + 17 + 20 + Y + 9 + 7 + 4 = 100 \]
\[ \implies 76 + X + Y = 100 \]
\[ \implies X + Y = 24 \tag{Equation 1} \]

Next, construct the cumulative frequency table:
- 0-10: cf = 2
- 10-20: cf = 7
- 20-30: cf = 7 + X
- 30-40: cf = 19 + X
- 40-50: cf = 36 + X
- 50-60: cf = 56 + X
- 60-70: cf = 56 + X + Y
- 70-80: cf = 65 + X + Y
- 80-90: cf = 72 + X + Y
- 90-100: cf = 76 + X + Y = 100

Since the median is given as 52.5, the median class is 50-60.
Using the median parameters:
- \( l = 50 \), \( f = 20 \), \( cf = 36 + X \), \( h = 10 \), and \( \frac{N}{2} = 50 \).

Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
\[ 52.5 = 50 + \left(\frac{50 - (36 + X)}{20}\right) \times 10 \]
\[ 2.5 = \frac{14 - X}{2} \]
\[ 5 = 14 - X \]
\[ X = 9 \]

Substitute \( X = 9 \) into Equation 1:
\[ 9 + Y = 24 \implies Y = 15 \]
Therefore, the values of \( X \) and \( Y \) are 9 and 15 respectively.
In simple words: By setting up the total frequency equation, we find that the sum of X and Y is 24. Using the median of 52.5 in the median formula allows us to solve directly for X = 9, which then gives Y = 15.

Exam Tip: Pay close attention to parentheses when subtracting the cumulative frequency term \( (36 + X) \) in the numerator of the median formula to avoid sign errors.

Question 6. Draw ‘less than ogive’ and ‘more than ogive’ for the following distribution and hence find its median.

Answer: Let's find the cumulative frequencies for both "less than" and "more than" type distributions:
Total frequency \( N = 100 \).

1. **'Less than' type cumulative frequencies:**
- Less than 30: 10
- Less than 40: 18
- Less than 50: 30
- Less than 60: 54
- Less than 70: 60
- Less than 80: 85
- Less than 90: 100

2. **'More than' type cumulative frequencies:**
- More than or equal to 20: 100
- More than or equal to 30: 90
- More than or equal to 40: 82
- More than or equal to 50: 70
- More than or equal to 60: 46
- More than or equal to 70: 40
- More than or equal to 80: 15
- More than or equal to 90: 0

Plotting both curves on a coordinate grid, they intersect at a point where the y-value is \( \frac{N}{2} = 50 \).
The x-coordinate (abscissa) at this point of intersection represents the median of the distribution. From the curves, the coordinates of the intersection point are \( (58.33, 50) \).
Therefore, the median is 58.33.

In simple words: Plotting both curves on a graph shows they intersect at a point where cumulative frequency is exactly 50. Projecting this point down to the horizontal axis gives a median value of 58.33.

Exam Tip: Plot the 'less than' points using the upper class limits and the 'more than' points using the lower class limits to ensure a mathematically correct curve.

Question 7. Find the mean marks for the following data.

Answer: Let's convert the given 'less than' cumulative distribution to a standard class interval frequency table:

Now, calculate the arithmetic mean:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4115}{85} \approx 48.41 \]
Therefore, the mean marks are 48.41.
In simple words: We find the actual frequency for each class interval by subtracting consecutive values, calculate the sum of the products, and divide by the total number of students (85) to get 48.41.

Exam Tip: Be sure to compute individual frequencies by subtracting the previous cumulative frequency from the current one before calculating the class marks.

Question 8. The following table shows age distribution of persons in a particular region. Calculate the median age.

Answer: Let's convert the given 'less than' cumulative distribution to a standard class interval frequency table:

Total number of persons \( N = 1560 \), so \( \frac{N}{2} = \frac{1560}{2} = 780 \).
The cumulative frequency just greater than 780 is 900, which belongs to the class interval 20-30.
Therefore, the median class is 20-30.
- \( l = 20 \), \( cf = 500 \), \( f = 400 \), and \( h = 10 \).

Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
\[ \text{Median} = 20 + \left(\frac{780 - 500}{400}\right) \times 10 \]
\[ \text{Median} = 20 + \frac{280}{40} = 20 + 7 = 27\text{ years} \]
Therefore, the median age is 27 years.
In simple words: We find the actual frequencies and use the cumulative table to locate the middle observer (780th person), who falls inside the 20-30 age group. Applying the median formula gives us a median age of 27.

Exam Tip: Be sure to keep track of the standard units (years) in your final solution, as dimensional context is highly valued by examiners.

Question 9. If the median of the following data is 32.5. Find the value of x and y.

Answer: First, write down the sum of all frequencies:
\[ x + 5 + 9 + 12 + y + 3 + 2 = 40 \]
\[ \implies x + y + 31 = 40 \]
\[ \implies x + y = 9 \tag{Equation 1} \]

Next, construct the cumulative frequency table:
- 0-10: cf = \( x \)
- 10-20: cf = \( x + 5 \)
- 20-30: cf = \( x + 14 \)
- 30-40: cf = \( x + 26 \)
- 40-50: cf = \( x + y + 26 \)
- 50-60: cf = \( x + y + 29 \)
- 60-70: cf = \( x + y + 31 = 40 \)

Since the median is given as 32.5, the median class is 30-40.
Using the median parameters:
- \( l = 30 \), \( f = 12 \), \( cf = x + 14 \), \( h = 10 \), and \( \frac{N}{2} = 20 \).

Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
\[ 32.5 = 30 + \left(\frac{20 - (x + 14)}{12}\right) \times 10 \]
\[ 2.5 = \left(\frac{6 - x}{12}\right) \times 10 \]
\[ \frac{2.5 \times 12}{10} = 6 - x \]
\[ 3 = 6 - x \]
\[ x = 3 \]

Substitute \( x = 3 \) into Equation 1:
\[ 3 + y = 9 \implies y = 6 \]
Therefore, the values of \( x \) and \( y \) are 3 and 6 respectively.
In simple words: The total count of observations is 40, meaning x + y must equal 9. Using the median of 32.5 in the median formula, we solve for x = 3, which then tells us y = 6.

Exam Tip: Be extra careful with bracket signs when subtracting cumulative frequency \( (x + 14) \) in the median formula.

Question 10. The following are ages of 300 patients getting medical treatment in a hospital on a particular day.

Draw:
1. Less than type cumulative frequency distribution
2. More than type cumulative frequency distribution
Answer: Let's construct both the cumulative frequency distributions:

1. **'Less than' type cumulative frequency distribution table:**

2. **'More than' type cumulative frequency distribution table:**

In simple words: We construct two cumulative lists. The 'less than' list sums up the frequencies starting from the youngest group, while the 'more than' list starts with the total 300 patients and subtracts each group's frequency progressively.

Exam Tip: In 'less than' cumulative frequency distributions, the cumulative frequencies are plotted corresponding to upper class limits. In 'more than' cumulative frequency distributions, they are plotted corresponding to lower class limits.

Value Based Question

Question Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality.
Monthly consumption (in units) 65 - 85, 85 - 105, 105 - 125, 125- 145, 145- 165, 165 - 185, 185 - 205
Number of consumers 4, 5, 13, 20, 14, 8, 4
Mr. Sharma always saves electricity by switching of all the electrical equipment just immediately after their uses. So , his family belongs to the group 65- 85 .
(i) Find the median of the above data
(ii) How many families consumed 125 or more units of electricity during a month?
(iii) What moral values of Mr. Sharma have been depicted in this situation?
Answer: Let's solve the sub-questions step-by-step:

(i) First, construct the cumulative frequency (cf) table:
- 65-85: frequency = 4, cf = 4
- 85-105: frequency = 5, cf = 9
- 105-125: frequency = 13, cf = 22
- 125-145: frequency = 20, cf = 42
- 145-165: frequency = 14, cf = 56
- 165-185: frequency = 8, cf = 64
- 185-205: frequency = 4, cf = 68

Total number of consumers \( N = 68 \), so \( \frac{N}{2} = 34 \).
The cumulative frequency just greater than 34 is 42, which corresponds to the class interval 125-145.
Therefore, the median class is 125-145.
- \( l = 125 \), \( cf = 22 \), \( f = 20 \), and \( h = 20 \).

Using the median formula:
\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]
\[ \text{Median} = 125 + \left(\frac{34 - 22}{20}\right) \times 20 \]
\[ \text{Median} = 125 + 12 = 137\text{ units} \]

(ii) To find the number of families that consumed 125 or more units:
We sum the frequencies of all classes starting from 125-145:
\( \text{Number of families} = 20 + 14 + 8 + 4 = 46 \) families.

(iii) The moral values depicted by Mr. Sharma's habits include civic responsibility, environmental awareness, and a conscious effort to conserve energy.
In simple words: The median electricity consumption is 137 units. There are 46 families who consumed 125 units or more. Mr. Sharma's habits show environmental care and responsibility.

Exam Tip: Value-based questions require both accurate mathematical calculation and a brief explanation of the moral/environmental values reflected in the scenario.

Question Q2. The mileage (km per litre) of 50 cars of the same models is tested by manufacturers and details are tabulated as given below:-
Mileage (km per litre) 10 - 12, 12 - 14, 14 - 16, 16- 18
No. of cars 7, 12, 18, 13
i. Find the mean mileage.
ii. The manufacturer claims that the mileage of the model is 16km/litre. Do you agree with this claim?
iii. Which values do you think the manufacturer should imbibe in his life?
Answer: Let's find the class marks (\( x_i \)) and compute the products \( f_i x_i \):
- For 10-12: \( x_i = 11 \implies f_i x_i = 7 \times 11 = 77 \)
- For 12-14: \( x_i = 13 \implies f_i x_i = 12 \times 13 = 156 \)
- For 14-16: \( x_i = 15 \implies f_i x_i = 18 \times 15 = 270 \)
- For 16-18: \( x_i = 17 \implies f_i x_i = 13 \times 17 = 221 \)

Calculate the sums:
- \( \sum f_i = 50 \)
- \( \sum f_i x_i = 77 + 156 + 270 + 221 = 724 \)

Now, addressing each sub-part:

i. Find the mean mileage:
\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{724}{50} = 14.48\text{ km/litre} \]

ii. Since the actual mean mileage is 14.48 km/litre, which is less than 16 km/litre, we do not agree with the manufacturer's claim.

iii. The manufacturer should imbibe values of honesty, truthfulness, integrity, transparency, and ethical business practices.
In simple words: The average mileage of the tested cars is 14.48 km/litre. Since this is less than 16, the manufacturer's claim is inaccurate. The manufacturer should practice honesty and transparency in business.

Exam Tip: In part (ii), always support your agreement/disagreement conclusion with the numerical average calculated in part (i) to earn full credit.