CBSE Class 10 Statistics Sure Shot Questions Set G

STATISTICS

CONTENTS

• Class Mark
• Cumulative Frequency
• Histogram
• Frequency Polygon
• Mean
• Median
• Mode

IMPORTANT POINTS

• The word data means information. Statistical data are of two types: (i) Primary data (ii) Secondary data
• When an investigator collects data himself with a definite plan or design in his (her) mind, it is called Primary data.
• Data which are not originally collected rather obtained from published or unpublished sources are known as Secondary data.
• After collection of data, the investigator has to find ways to condense them in tabular form in order to study their salient features. Such an arrangement is called Presentation of data.
• Raw data (if the data is arranged in the manner as it was collected, then it is called as Raw data) when put in ascending or descending order of magnitude is called an array or arranged data.
• The number of times an observation occurs in the given data is called frequency of the observation.
• Classes/class intervals are the groups in which all the observations are divided.
• Suppose class-interval is 10-20, then 10 is called lower limit and 20 is called upper limit of the class.
• Mid-value of class-interval is called Class-mark.
  Class-mark = \( \frac{\text{lower limit} + \text{upper limit}}{2} \)
  Class-mark = lower limit + \( \frac{1}{2} \) (difference between the upper and lower limits)
• If the frequency of first class interval is added to the frequency of second class and this sum is added to third class and so on then frequencies so obtained are known as Cumulative Frequency (c.f.).
• There are two types of cumulative frequencies (a) less than, (b) greater than.

EXAMPLES

Question. Given below are the ages of 25 students of class IX in a school. Prepare a discrete frequency distribution. 15, 16, 16, 14, 17, 17, 16, 15, 15, 16, 16, 17, 15, 16, 16, 14, 16, 15, 14, 15, 16, 16, 15, 14, 15.
Answer: Frequency distribution of ages of 25 students:
Age | Tally marks | Frequency
14 | IIII | 4
15 | NII III | 8
16 | NII NII | 10
17 | III | 3
Total | | 25

Question. Form a discrete frequency distribution from the following scores:- 15, 18, 16, 20, 25, 24, 25, 20, 16, 15, 18, 18, 16, 24, 15, 20, 28, 30, 27, 16, 24, 25, 20, 18, 28, 27, 25, 24, 24, 18, 18, 25, 20, 16, 15, 20, 27, 28, 29, 16.
Answer:
Variate | Tally marks | Frequency
15 | IIII | 4
16 | NII I | 6
18 | NII I | 6
20 | NII I | 6
24 | NII | 5
25 | NII | 5
27 | III | 3
28 | III | 3
29 | I | 1
30 | I | 1
Total | | 40

Question. The water tax bills (in rupees) of 30 houses in a locality are given below. Construct a grouped frequency distribution with class size of 10. 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44.
Answer: Here the maximum and minimum values of the variate are 112 and 14 respectively. Range = 112 – 14 = 98. It is given that the class size is 10, and \( \frac{\text{Range}}{\text{Class size}} = \frac{98}{10} = 9.8 \). So, we should have 10 classes each of size 10. If we take the first class as 14-24 it includes the minimum value 14. If the last class is taken as 104-114, then it includes the maximum value 112. In the class 14-24, 14 is included but 24 is excluded. Similarly, in other classes, the lower limit is included and the upper limit is excluded. Now:
Bill (in rupees) | Tally marks | Frequency
14-24 | IIII | 4
24-34 | II | 2
34-44 | III | 3
44-54 | III | 3
54-64 | I | 1
64-74 | II | 2
74-84 | NII | 5
84-94 | III | 3
94-104 | III | 3
104-114 | IIII | 4
Total | | 30

Question. The marks obtained by 40 students of class IX in an examination are given below: 18, 8, 12, 6, 8, 16, 12, 5, 23, 2, 16, 23, 2, 10, 20, 12, 9, 7, 6, 5, 3, 5, 13, 21, 13, 15, 20, 24, 1, 7, 21, 16, 13, 18, 23, 7, 3, 18, 17, 16. Present the data in the form of a frequency distribution using the same class size, one such class being 15-20 (where 20 is not included).
Answer: Frequency Distribution of Marks:
Marks | Tally marks | Frequency
0-5 | NII I | 6
5-10 | NII NII | 10
10-15 | NII III | 8
15-20 | NII III | 8
20-25 | NII III | 8
Total | | 40

Question. The class marks of a distribution are: 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97, 102. Determine the class size, the class limits and the true class limits.
Answer: Here the class marks are uniformly spaced. So, the class size is the difference between any two consecutive class marks. \( \therefore \) Class size = 52 – 47 = 5. We know that, if \( a \) is the class mark of a class interval and \( h \) is its class size, then the lower and upper limits of the class interval are \( a - \frac{h}{2} \) and \( a + \frac{h}{2} \) respectively. \( \therefore \) Lower limit of first class interval = \( 47 - \frac{5}{2} = 44.5 \). And, upper limit of first class interval = \( 47 + \frac{5}{2} = 49.5 \). So, first class interval is 44.5 – 49.5. Similarly, we obtain the other class limits as given under:
Class marks | Class limits
47 | 44.5 - 49.5
52 | 49.5 - 54.5
57 | 54.5 - 59.5
62 | 59.5 - 64.5
67 | 64.5 - 69.5
72 | 69.5 - 74.5
77 | 74.5 - 79.5
82 | 79.5 - 84.5
87 | 84.5 - 89.5
92 | 89.5 - 94.5
97 | 94.5 - 99.5
102 | 99.5 - 104.5
Since the classes are exclusive, so the true class limits are same as the class limits.

Question. The class marks of a distribution are 26, 31, 36, 41, 46, 51, 56, 61, 66, 71. Find the true class limits.
Answer: Here the class marks are uniformly spaced. So, the class size is the difference between any two consecutive class marks. \( \therefore \) Class size = 31 – 26 = 5. If \( a \) is the class mark of a class interval of size \( h \), then the lower and upper limits of the class interval are \( a - \frac{h}{2} \) and \( a + \frac{h}{2} \) respectively. Here \( h = 5 \). \( \therefore \) Lower limit of first class interval = \( 26 - \frac{5}{2} = 23.5 \). And, upper limit of first class interval = \( 26 + \frac{5}{2} = 28.5 \). \( \therefore \) First class interval is 23.5 – 28.5. Thus, the class intervals are: 23.5 – 28.5, 28.5 – 33.5, 33.5 – 38.5, 38.5 – 43.5, 43.5 – 48.5, 48.5 – 53.5. Since the class are formed by exclusive method. Therefore, these limits are true class limits.

CUMULATIVE FREQUENCY TABLE

A table which displays the manner in which cumulative frequencies are distributed over various classes is called a cumulative frequency distribution or cumulative frequency table.

Question. The marks obtained by 35 students in a class are given below. Construct the cumulative frequency table:
Marks obtained: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Number of students: 1, 2, 4, 4, 3, 5, 4, 6, 3, 2, 1
Answer:
Marks | Frequency | Number of students (Cumulative Frequency)
0 | 1 | 1
1 | 2 | 3 (=1 + 2)
2 | 4 | 7 (=1 + 2 + 4)
3 | 4 | 11 (=1 + 2 + 4 + 4)
4 | 3 | 14 (=1 + 2 + 4 + 4 + 3)
5 | 5 | 19 (=1 + 2 + 4 + 4 + 3 + 5)
6 | 4 | 23 (=1 + 2 + 4 + 4 + 3 + 5 + 4)
7 | 6 | 29 (=1 + 2 + 4 + 4 + 3 + 5 + 4 + 6)
8 | 3 | 32 (=1 + 2 + 4 + 4 + 3 + 5 + 4 + 6 + 3)
9 | 2 | 34 (=1 + 2 + 4 + 4 + 3 + 5 + 4 + 6 + 3 + 2)
10 | 1 | 35 (=1 + 2 + 4 + 4 + 3 + 5 + 4 + 6 + 3 + 2 + 1)
Total | 35 |

The distribution of ages (in years) of 40 persons in a colony is given below:
Age (in years): 20-25, 25-30, 30-35, 35-40, 40-45, 45-50
Number of Persons: 7, 10, 8, 6, 4, 5

Question. Determine the class mark of each class
Answer: Class marks are \( \frac{20+25}{2}, \frac{25+30}{2}, \frac{30+35}{2}, \frac{35+40}{2}, \frac{40+45}{2}, \frac{45+50}{2} \) = 22.5, 27.5, 32.5, 37.5, 42.5, 47.5.

Question. What is the upper class limit of 4th class
Answer: The fourth class interval is 35–40. Its upper limit is 40.

Question. Determine the class size
Answer: The class size is 25 – 20 = 5.

Question. Following is the distribution of marks of 40 students in a class. Construct a cumulative frequency distribution table.
Marks: 0-10, 10-20, 20-30, 30-40, 40-50
Number of students: 3, 8, 9, 15, 5
Answer:
Class interval | Frequency | Cumulative Frequency
0–10 | 3 | 3
10–20 | 8 | 11 (= 3 + 8)
20–30 | 9 | 20 (= 3 + 8 + 9)
30–40 | 15 | 35 (= 3 + 8 + 9 + 15)
40–50 | 5 | 40 (= 3 + 8 + 9 + 15 + 5)
Total | 40 |

Question. The class marks of a distribution are 25, 35, 45, 55, 65 and 75. Determine the class size and class limit.
Answer: Class size = The difference between the class marks of two adjacent classes = 35 – 25 = 10. We need classes of size 10 with class marks as 25, 35, 45, 55, 65, 75. The class limits for the first class are \( 25 - \frac{10}{2} \) and \( 25 + \frac{10}{2} \), i.e. 20 and 30. First class is, therefore, 20–30. Similarly, the other classes are 30 – 40, 40 – 50, 50 – 60, 60 – 70, 70 – 80.

Question. Given below is the cumulative frequency distribution table showing the marks secured by 40 students. Show in the class and their frequency form.
Marks: Below 20, Below 40, Below 60, Below 80, Below 100
Number of students: 5, 10, 25, 32, 40
Answer:
Marks | Cumulative frequency | Frequency
0-20 | 5 | 5
20-40 | 10 | 5 (= 10 – 5)
40-60 | 25 | 15 (= 25 – 10)
60-80 | 32 | 7 (= 32 – 25)
80-100 | 40 | 8 (= 40 – 32)

Graphical representation of data

The graphical representations such as bar graphs, histograms, frequency polygons, etc.

Bar graph (diagram) of a data

A bar graph (diagram) is a pictorial representation of the data by a series of bars or rectangles of uniform width standing on the same horizontal (or vertical) base line with equal spacing between the bars. Each rectangle or bar represents only one numerical value of the data. The height (or length in case the base is on a vertical line) of each bar is proportional to the numerical values of the data.

For example, we are given a data about the household expenditure of a family as below:
Heads of expenditure | Expenditure (in thousand rupees)
Rent | 5
Grocery | 2
Education | 3
Transport | 2
Miscellaneous | 4

The horizontal axis is generally called x-axis and the vertical axis as the y-axis.
1. Each bar of a bar diagram has same width.
2. Space between two consecutive bars is same throughout.

HISTOGRAM

A histogram is a graphical representation of frequency distribution in the form of rectangle with class interval as the bases and heights proportional to corresponding frequency. There is no gap between rectangles.

Construction of a histogram of a continuous grouped frequency distribution

(i) Take a graph paper and draw two perpendicular lines, one horizontal and one vertical, intersecting at O (say). Mark them as OX and OY.

(ii) Take horizontal line OX as X-axis and vertical line OY as Y-axis.

(iii) Choose a suitable scale and along X-axis represent class limits.

(iv) Choose a suitable scale for Y-axis and mark frequencies along Y-axis.

(v) Construct rectangles with class intervals as bases and respective frequencies as heights.

Construction of a histogram of a continuous grouped frequency distribution with unequal class intervals.

(i) Take a graph paper and draw two perpendicular lines, one horizontal and one vertical, intersecting at O (say). Mark them as OX and OY.

(ii) Take horizontal line OX as X-axis and vertical line OY as Y-axis.

(iii) Choose a suitable scale along X-axis and represent class-limits on it.

(iv) Determine a class-interval which has the minimum class size. Let the minimum class size be h.

(v) Compute the adjusted frequencies of each class by using the following formula: Adjusted Frequency of a class = \( \frac{h}{\text{Class size}} \times \text{Frequency of the class} \). These adjusted frequencies are the heights of each rectangle of histogram but widths will be according to class limits.

(vi) Choose a suitable scale for Y-axis and mark adjusted frequencies along Y-axis.

(vii) Construct rectangles with class intervals as bases and respective adjusted frequencies as heights.

Construction of a histogram when mid-points of class-intervals are given

(i) Compute the difference between second and first mid-point. Let it be h.

(ii) Divide the difference h by 2.

(iii) Subtract \( \frac{h}{2} \) from first mid-point to get the lower limit of first class and add \( \frac{h}{2} \) to first mid-point to get the upper limit of first class.

(iv) Repeat first three steps for all other mid-points.

HISTOGRAM

A histogram is a graphical representation of frequency distribution in the form of rectangle with class interval as the bases and heights proportional to corresponding frequency. There is no gap between rectangles.

Construction of a histogram of a continuous grouped frequency distribution

(i) Take a graph paper and draw two perpendicular lines, one horizontal and one vertical, intersecting at O (say). Mark them as OX and OY.

(ii) Take horizontal line OX as X-axis and vertical line OY as Y-axis.

(iii) Choose a suitable scale and along X-axis represent class limits.

(iv) Choose a suitable scale for Y-axis and mark frequencies along Y-axis.

(v) Construct rectangles with class intervals as bases and respective frequencies as heights.

Construction of a histogram of a continuous grouped frequency distribution with unequal class intervals.

(i) Take a graph paper and draw two perpendicular lines, one horizontal and one vertical, intersecting at O (say). Mark them as OX and OY.

(ii) Take horizontal line OX as X-axis and vertical line OY as Y-axis.

(iii) Choose a suitable scale along X-axis and represent class-limits on it.

(iv) Determine a class-interval which has the minimum class size. Let the minimum class size be h.

(v) Compute the adjusted frequencies of each class by using the following formula:
\[ \text{Adjusted Frequency of a class} = \frac{h}{\text{Class size}} \times \text{Frequency of the class} \] These adjusted frequencies are the heights of each rectangle of histogram but widths will be according to class limits.

(vi) Choose a suitable scale for Y-axis and mark adjusted frequencies along Y-axis.

(vii) Construct rectangles with class intervals as bases and respective adjusted frequencies as heights.

Construction of a histogram when mid-points of class-intervals are given

(i) Compute the difference between second and first mid-point. Let it be h.

(ii) Divide the difference h by 2.

(iii) Subtract \( \frac{h}{2} \) from first mid-point to get the lower limit of first class and add \( \frac{h}{2} \) to first mid-point to get the upper limit of first class.

(iv) Repeat first three steps for all other mid-points.

EXAMPLES

Question. What was the quantity of rice production in the year 1980-81? What is the difference between the maximum and minimum production of rice in the time span of 1978-1983?
Answer: (ii) The bar for 1980-81 has length = 55. Therefore, the production of rice in the year 1980-81 is 55 lakh tons.
(iii) In the year 1981-82, the production of rice is maximum and is equal to 65 lakh tons. In the year 1979-80, the production of rice is minimum and is equal to 25 lakh tons.
The difference between the maximum and the minimum production = 65 lakh tons – 25 lakh tons = 40 lakh tons.

Question. Represent the following data by means of histogram.
Weekly wages (in Rs): 10-15, 15-20, 20-25, 25-30, 30-40, 40-60, 60-80
No. of workers (Frequency): 7, 9, 8, 5, 12, 12, 8
Answer: Here the class intervals are of unequal width. So, we shall first compute adjusted frequencies of each class. The minimum class size is 15 – 10 = 5. The adjusted frequencies are computed by using the following formula:
Adjusted frequency of a class = \( \frac{\text{Minimum class size}}{\text{Class size}} \times \text{Frequency of the class} \)
The adjusted frequencies are computed in the following table:
10-15: \( \frac{5}{5} \times 7 = 7 \)
15-20: \( \frac{5}{5} \times 9 = 9 \)
20-25: \( \frac{5}{5} \times 8 = 8 \)
25-30: \( \frac{5}{5} \times 5 = 5 \)
30-40: \( \frac{5}{10} \times 12 = 6 \)
40-60: \( \frac{5}{20} \times 12 = 3 \)
60-80: \( \frac{5}{20} \times 8 = 2 \)
The adjusted frequencies are the heights of each rectangle of histogram but widths will be according to class limits.

Question. Construct a histogram from the following distribution of total marks obtained by 65 students of IX class in the final examination.
Marks (mid-points): 150, 160, 170, 180, 190, 200
No. of students: 8, 10, 25, 12, 7, 3
Answer: Ascertainment of lower and upper class limits: Since the difference between the second and first mid-point is 160 – 150 = 10.
\( \therefore h = 10 \Rightarrow \frac{h}{2} = 5 \).
So, lower and upper limits of the first class are 150 – 5 and 150 + 5 i.e. 145 and 155 respectively.
\( \therefore \) First class interval is 145 – 155.
Using the same procedure, we get the classes of other mid-points as under:
145-155 (8), 155-165 (10), 165-175 (25), 175-185 (12), 185-195 (7), 195-205 (3).
[The histogram is then constructed based on these class intervals.]

FREQUENCY POLYGON

A frequency polygon is the polygon obtained by joining the mid-points of upper horizontal sides of all the rectangles in a histogram.

Construction of a frequency polygon with Histogram

(i) Obtain the frequency distribution and draw a histogram representing it.

(ii) Obtain the mid-points of the upper horizontal side of each rectangle.

(iii) Join these mid-points of the adjacent rectangles of the histogram by dotted line segments.

(iv) Obtain the mid-points of two class-intervals of zero frequency i.e. on X-axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class intervals.

(v) Complete the polygon by joining the mid-points of first and last class intervals to the mid-points of imagined class-intervals adjacent to them.

Construction of a frequency polygon without using a histogram

(i) Obtain the frequency distribution.

(ii) Compute the mid-points of class intervals i.e. class marks.

(iii) Represent class marks on X-axis on a suitable scale.

(iv) Represent frequencies on Y-axis on a suitable scale.

(v) Plot the points \( (x_i, f_i) \) where \( x_i \) denotes class mark and \( f_i \) corresponding frequency.

(vi) Join the points plotted in step V by line segments.

(vii) Take two class intervals of zero frequency, one at the beginning and the other at the end. Obtain their mid-points. These classes are known as imagined classes.

(viii) Complete the frequency polygon by joining the mid-points of first and last class intervals to the mid-points of the imagined classes adjacent to them.

Question. Construct a frequency polygon for the following data:
Age (in years): 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14, 14-16, 16-18
Frequency: 2, 4, 6, 8, 9, 6, 5, 3, 1
Answer: First we obtain the class marks:
(0-2): 1, (2-4): 3, (4-6): 5, (6-8): 7, (8-10): 9, (10-12): 11, (12-14): 13, (14-16): 15, (16-18): 17.
We plot the points (1, 2), (3, 4), (5, 6), (7, 8), (9, 9), (11, 6), (13, 5), (15, 3), (17, 1) and join them to form the frequency polygon.

MEAN

If \( x_1, x_2, x_3, \dots, x_n \) are \( n \) values of a variable X, then the arithmetic mean or simply the mean of these values is denoted by \( \bar{X} \) and is defined as:
\[ \bar{X} = \frac{x_1 + x_2 + x_3 + \dots + x_n}{n} = \frac{1}{n} \left( \sum_{i=1}^{n} x_i \right) \]

• If \( \bar{X} \) is the mean of \( n \) observations \( x_1, x_2, \dots, x_n \), then the mean of \( x_1 + a, x_2 + a, \dots, x_n + a \) is \( \bar{X} + a \).
• If \( \bar{X} \) is the mean of \( x_1, x_2, \dots, x_n \), then the mean of \( ax_1, ax_2, \dots, ax_n \) is \( a\bar{X} \).
• If \( \bar{X} \) is the mean of \( n \) observations, the mean of \( \frac{x_1}{a}, \frac{x_2}{a}, \dots, \frac{x_n}{a} \) is \( \frac{\bar{X}}{a} \).
• If \( \bar{X} \) is the mean of \( n \) observations, the mean of \( x_1 - a, x_2 - a, \dots, x_n - a \) is \( \bar{X} - a \).

Advantages of Arithmetic Mean

(i) Arithmetic mean is simple to understand and easy to calculate.

(ii) It is rigidly defined.

(iii) It is suitable for further algebraic treatment.

(iv) It is least affected by fluctuation of sampling.

(v) It takes into account all the values in the series.

Disadvantages of Arithmetic Mean

(i) It is highly affected by the presence of a few abnormally high or abnormally low scores.

(ii) In absence of a single item, its value becomes inaccurate.

(iii) It cannot be determined by inspection.

Question. If the mean of \( n \) observations \( ax_1, ax_2, ax_3, \dots, ax_n \) is \( a\bar{X} \), show that \( (ax_1 - a\bar{X}) + (ax_2 - a\bar{X}) + \dots + (ax_n - a\bar{X}) = 0 \).
Answer: We have \( a\bar{X} = \frac{ax_1 + ax_2 + \dots + ax_n}{n} \Rightarrow ax_1 + ax_2 + \dots + ax_n = n(a\bar{X}) \dots(i) \).
Now, \( (ax_1 - a\bar{X}) + (ax_2 - a\bar{X}) + \dots + (ax_n - a\bar{X}) = (ax_1 + ax_2 + \dots + ax_n) - (a\bar{X} + a\bar{X} + \dots + a\bar{X}) \)
\( = n(a\bar{X}) - n(a\bar{X}) = 0 \).

Question. The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean.
Answer: Here, \( n = 40 \), \( \bar{X} = 160 \).
\( \sum x_i = 160 \times 40 = 6400 \) (Incorrect sum).
Correct sum = \( \text{Incorrect sum} - \text{Incorrect item} + \text{Correct item} \)
\( = 6400 - 125 + 165 = 6440 \).
\( \therefore \text{Correct Mean} = \frac{6440}{40} = 161 \).

Question. The mean of 10 numbers is 20. If 5 is subtracted from every number, what will be the new mean?
Answer: Let \( x_1, x_2, \dots, x_{10} \) be 10 numbers. \( \bar{X} = \frac{\sum x_i}{10} = 20 \Rightarrow \sum x_i = 200 \).
New numbers are \( x_1 - 5, x_2 - 5, \dots, x_{10} - 5 \).
New Mean \( \bar{X}' = \frac{(x_1 - 5) + (x_2 - 5) + \dots + (x_{10} - 5)}{10} = \frac{\sum x_i - 50}{10} \)
\( = \frac{200 - 50}{10} = \frac{150}{10} = 15 \).

Question. Neeta and her four friends secured 65, 78, 82, 94 and 71 marks in a test of mathematics. Find the average (arithmetic mean) of their marks.
Answer: Arithmetic mean = \( \frac{65 + 78 + 82 + 94 + 71}{5} = \frac{390}{5} = 78 \).

Question. The mean of 5, 7, p, 11, 15, 17, and 20 is 12, find p.
Answer: Mean = \( \frac{5 + 7 + p + 11 + 15 + 17 + 20}{7} = 12 \Rightarrow \frac{75 + p}{7} = 12 \).
\( p + 75 = 84 \Rightarrow p = 9 \).

Question. If the mean of 5 observations is 15 and that of another 10 observations is 20, find the mean of all 15 observations.
Answer: Sum of 5 observations = \( 15 \times 5 = 75 \).
Sum of 10 observations = \( 20 \times 10 = 200 \).
Mean of 15 observations = \( \frac{75 + 200}{15} = \frac{275}{15} = 18.33 \).

Question. If a variate X takes values \( x_1, x_2, \dots, x_n \) with corresponding frequencies \( f_1, f_2, \dots, f_n \), then arithmetic mean is:
\( \bar{X} = \frac{f_1x_1 + f_2x_2 + \dots + f_nx_n}{f_1 + f_2 + \dots + f_n} = \frac{\sum f_ix_i}{N} \), where \( N = \sum f_i \). Find the mean of:
x: 4, 6, 9, 10, 15
f: 5, 10, 10, 7, 8
Answer: Calculation Table:
4x5=20, 6x10=60, 9x10=90, 10x7=70, 15x8=120.
\( \sum f_i = 40 \), \( \sum f_ix_i = 360 \).
Mean = \( \frac{360}{40} = 9 \).

Question. Find the mean of the following distribution:
x: 10, 30, 50, 70, 89
f: 7, 8, 10, 15, 10
Answer: Calculation Table:
10x7=70, 30x8=240, 50x10=500, 70x15=1050, 89x10=890.
\( \sum f_i = 50 \), \( \sum f_ix_i = 2750 \).
Mean = \( \frac{2750}{50} = 55 \).

Question. Find the value of p, if the mean of following distribution is 7.5.
x: 3, 5, 7, 9, 11, 13
y(f): 6, 8, 15, p, 8, 4
Answer: \( \sum f_i = 41 + p \), \( \sum f_ix_i = 303 + 9p \).
Mean = \( \frac{303 + 9p}{41 + p} = 7.5 \Rightarrow 303 + 9p = 7.5(41 + p) \).
\( 303 + 9p = 307.5 + 7.5p \Rightarrow 1.5p = 4.5 \Rightarrow p = 3 \).

Question. Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 1.46.
Number of accidents (x): 0, 1, 2, 3, 4, 5, Total=200
Frequency (f): 46, f1, f2, 25, 10, 5
Answer: \( N = 46 + f_1 + f_2 + 25 + 10 + 5 = 200 \Rightarrow f_1 + f_2 = 114 \dots(i) \).
\( \sum f_ix_i = 0(46) + 1(f_1) + 2(f_2) + 3(25) + 4(10) + 5(5) = f_1 + 2f_2 + 140 \).
Mean = \( \frac{f_1 + 2f_2 + 140}{200} = 1.46 \Rightarrow f_1 + 2f_2 + 140 = 292 \Rightarrow f_1 + 2f_2 = 152 \dots(ii) \).
Solving (i) and (ii), we get \( f_1 = 76 \) and \( f_2 = 38 \).

Question. If the mean of the following data be 9.2, find the value of p.
x: 4, 6, 7, p + 4, 12, 14
f: 5, 6, 4, 10, 8, 7
Answer: \( \sum f_i = 40 \), \( \sum f_ix_i = 20 + 36 + 28 + 10(p + 4) + 96 + 98 = 318 + 10p \).
Mean = \( \frac{318 + 10p}{40} = 9.2 \Rightarrow 318 + 10p = 368 \Rightarrow 10p = 50 \Rightarrow p = 5 \).

Question. The marks of 30 students are given below, find the mean marks.
Marks (x): 10, 11, 12, 13, 14, 15
Students (f): 4, 3, 8, 6, 7, 2
Answer: \( \sum f_i = 30 \), \( \sum f_ix_i = 40 + 33 + 96 + 78 + 98 + 30 = 375 \).
Mean = \( \frac{375}{30} = 12.5 \).

GROUPED FREQUENCY DISTRIBUTION

There are 3 methods for calculation of mean:
1. Direct Method
2. Assumed mean deviation method
3. Step deviation method.

Direct Method for Calculation of Mean

\[ \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{1}{N} \sum f_ix_i \]

Question. Mid-values: 2, 3, 4, 5, 6; Frequencies: 49, 43, 57, 38, 13. Find the mean by direct method.
Answer: Table: 2x49=98, 3x43=129, 4x57=228, 5x38=190, 6x13=78.
\( \sum f_i = 200 \), \( \sum f_ix_i = 723 \).
Mean = \( \frac{723}{200} = 3.615 \).

Question. Find the mean of the following frequency distribution:
Class Interval: 10–30, 30–50, 50–70, 70–90, 90–110
Frequency: 90, 20, 30, 20, 40
Answer: Mid-values (x): 20, 40, 60, 80, 100.
\( fx \): 1800, 800, 1800, 1600, 4000.
\( \sum f = 200 \), \( \sum fx = 10000 \).
Mean = \( \frac{10000}{200} = 50 \).

Question. A survey was conducted regarding the number of plants in 20 houses. Find the mean number of plants per house.
Number of plants: 0–2, 2–4, 4–6, 6–8, 8–10, 10–12, 12–14
No. of houses: 1, 2, 1, 5, 6, 2, 3
Answer: Mid-values (x): 1, 3, 5, 7, 9, 11, 13.
\( fx \): 1, 6, 5, 35, 54, 22, 39.
\( \sum f = 20 \), \( \sum fx = 162 \).
Mean = \( \frac{162}{20} = 8.1 \).

MEDIAN

Median of a distribution is the value of the variable which divides the distribution into two equal parts i.e. it is the value such that the number of observations above it is equal to the number of observations below it.

Algorithm:

• Step I: Arrange the observations in ascending or descending order.
• Step II: Determine the total number of observations, n.
• Step III: If n is odd, then Median = value of \( \left(\frac{n+1}{2}\right)^{\text{th}} \) observation.
  If n is even, then Median = \( \frac{\text{value of } (\frac{n}{2})^{\text{th}} \text{ observation} + \text{value of } (\frac{n}{2} + 1)^{\text{th}} \text{ observation}}{2} \).

Note: Median is not affected by extreme values.

Question. Find the median of the following data: 25, 34, 31, 23, 22, 26, 35, 28, 20, 32.
Answer: Ascending order: 20, 22, 23, 25, 26, 28, 31, 32, 34, 35.
n = 10 (even).
Median = \( \frac{\text{5th obs} + \text{6th obs}}{2} = \frac{26 + 28}{2} = 27 \).

Question. Find the median of the following values: 37, 31, 42, 43, 46, 25, 39, 45, 32.
Answer: Ascending order: 25, 31, 32, 37, 39, 42, 43, 45, 46.
n = 9 (odd).
Median = value of \( \frac{9+1}{2} = \text{5th obs} = 39 \).

Question. The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.
Answer: n = 10 (even). Median = \( \frac{\text{5th obs} + \text{6th obs}}{2} = \frac{(x + 2) + (x + 4)}{2} = 24 \).
\( \frac{2x + 6}{2} = 24 \Rightarrow x + 3 = 24 \Rightarrow x = 21 \).

Question. Find the median of 19, 25, 59, 48, 35, 31, 30, 32, 51. If 25 is replaced by 52, what will be the new median.
Answer: Case 1: n = 9 (odd). Ascending: 19, 25, 30, 31, 32, 35, 48, 51, 59. Median = 5th obs = 32.
Case 2: 25 replaced by 52. Ascending: 19, 30, 31, 32, 35, 48, 51, 52, 59. New Median = 5th obs = 35.

Question. Calculate the median for: Weight(kg): 46, 47, 48, 49, 50, 51, 52; Freq: 3, 2, 4, 6, 5, 2, 1.
Answer: Cumulative frequencies: 3, 5, 9, 15, 20, 22, 23.
n = 23 (odd). Median = \( \text{12th observation} = 49 \).

Question. The data 75, 70, 68, x + 2, x – 2, 50, 45, 40 arranged in descending order has median 60. Find x.
Answer: n = 8. Median = \( \frac{(x + 2) + (x - 2)}{2} = 60 \Rightarrow \frac{2x}{2} = 60 \Rightarrow x = 60 \).

MODE

Mode is also known as norm. Mode is the value which occurs most frequently in a set of observations.

Algorithm:

• Step I: Obtain the set of observations.
• Step II: Prepare the frequency distribution.
• Step III: Obtain the value which has the maximum frequency.
• Step IV: The value obtained in step III is the mode.

Question. Find the mode from: 110, 120, 130, 120, 110, 140, 130, 120, 140, 120.
Answer: Value 120 occurs maximum number of times (4). Mode = 120.

Question. Compute mode for: 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 11, 12, 13, 13.
Answer: Scores 8 and 10 occur thrice. Using empirical formula: Mode = 3 Median – 2 Mean.
Mean = \( \frac{146}{15} = 9.73 \). Median = \( t_8 = 10 \).
Mode = \( 3(10) - 2(9.73) = 30 - 19.46 = 10.54 \).

Question. Calculate mode for weights: 54, 72, 80, 64, 62, 60, 58, 56, 63 with frequencies 6, 6, 1, 2, 6, 5, 5, 4, 5.
Answer: Scores 54, 72 and 62 occur 6 times. Empirical formula: Mean = \( \frac{2465}{40} = 61.625 \). Median = \( \frac{60 + 62}{2} = 61 \).
Mode = \( 3(61) - 2(61.625) = 183 - 123.25 = 59.75 \text{ kg} \).