CBSE Class 10 Maths HOTs Statistics Set 04

Definition:

In the exclusive or continuous series, the upper limit of one class is the lower limit of next class. The common point (155 cm) of two classes is included in higher class (155 - 160) and not in the lower class (150 - 155).

Working rule to convert a discontinuous series into continuous series (Inclusive series to exclusive series)

• Step 1. Find the adjustment factor by the following formula:
  \( \text{Adjustment factor} = \frac{1}{2} [\text{Lower limit of second class} - \text{Upper limit of first class}] \)
• Step 2. Subtract the adjustment factor from the lower limit and add to the upper limit of each class. We will get exclusive series.

Example: We consider the following inclusive series: 11 - 20, 21 - 30, 31 - 40.
Step 1. Adjustment factor = \( \frac{1}{2} [21 - 20] = \frac{1}{2} = 0.5 \)
Step 2. The inclusive series is converted into exclusive series:
11 - 0.5 — 20 + 0.5 i.e.; 10.5 — 20.5
21 - 0.5 — 30 + 0.5 i.e.; 20.5 — 30.5
31 - 0.5 — 40 + 0.5 i.e.; 30.5 — 40.5

CONCEPT - 1 : ARITHMETIC MEAN

TYPE - 1: A.M. of Ungrouped Data

Arithmetic mean of raw data (when frequencies are not given)
The arithmetic mean of a raw data is obtained by adding all the values of the variables and dividing the sum by total number of values that are added.
Let the values of the variables \( x \) are \( x_1, x_2, ..., x_n \), where \( n \) is the total number of values, then
Arithmetic mean \( (\bar{x}) = \frac{x_1 + x_2 + ... + x_n}{n} = \frac{1}{n} \sum_{i=1}^{n} x_i \)
The symbol \( \sum_{i=1}^{n} x_i \) denotes the sum \( x_1 + x_2 + ... + x_n \).
The arithmetic mean of a set of observations is equal to their sum divided by the total number of observations.

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 or average = \( \frac{65 + 78 + 82 + 94 + 71}{5} = \frac{390}{5} = 78 \)
Hence, arithmetic mean = 78

TYPE - 2: Direct method. (When frequencies are given)

Let \( x_1, x_2, ..., x_n \) be the values of a variate with corresponding frequencies \( f_1, f_2, ..., f_n \) respectively, then arithmetic mean of these values is
\( \bar{x} = \frac{f_1 x_1 + f_2 x_2 + ... + f_n x_n}{f_1 + f_2 + ... + f_n} \)

Question. Calculate the mean for the following distribution:
Variable: 5, 6, 7, 8, 9
Frequency: 4, 8, 14, 11, 3
Answer:
Sum of frequencies \( N = \sum f = 4 + 8 + 14 + 11 + 3 = 40 \)
Sum of \( f \cdot x = (5 \times 4) + (6 \times 8) + (7 \times 14) + (8 \times 11) + (9 \times 3) \)
\( = 20 + 48 + 98 + 88 + 27 = 281 \)
Mean \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{281}{40} = 7.025 \)

TYPE - 3: Grouped frequency distribution

Sometimes the data is so large that it is inconvenient to list every item in the frequency distribution table. Then we group the items into convenient intervals and the data is presented in a frequency-distribution in which each class interval contains five or ten values of the variate generally. The mid-value of each class is the representative of each item falling in that interval.
Here individual values lose their identity. If we assume that all values falling in a class interval are equal to the middle value of this class interval. This middle value is called mid-value, mid-point or class mark.
In order to make life easy and save time and energy following methods are adopted for calculation purposes:
(I) Direct Method
(II) Assumed mean deviation method
(III) Step deviation method.

(a) Direct method for calculation of Mean

According to Direct Method
\( \bar{x} = \frac{x_1 f_1 + x_2 f_2 + ... + x_k f_k}{f_1 + f_2 + ... + f_k} = \frac{\sum_{i=1}^{k} x_i f_i}{\sum f_i} = \frac{1}{N} \sum_{i=1}^{k} f_i x_i \) [where \( N = f_1 + f_2 + ... + f_k \)]
This method will involve large numbers and hence greater time for calculation.

Question. Mid-values of class intervals are given with their frequencies. Find the mean by Direct method.
Mid-values: 2, 3, 4, 5, 6
Frequencies: 49, 43, 57, 38, 13
Answer:
Total frequency \( N = \sum f = 200 \)
Sum of \( f_i x_i = (2 \times 49) + (3 \times 43) + (4 \times 57) + (5 \times 38) + (6 \times 13) = 98 + 129 + 228 + 190 + 78 = 723 \)
By direct method, Mean \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{723}{200} = 3.615 \)

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) for classes: 20, 40, 60, 80, 100
\( \sum f_i = 90 + 20 + 30 + 20 + 40 = 200 \)
\( \sum f_i x_i = (90 \times 20) + (20 \times 40) + (30 \times 60) + (20 \times 80) + (40 \times 100) \)
\( = 1800 + 800 + 1800 + 1600 + 4000 = 10000 \)
Mean \( (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{10000}{200} = 50 \)

Question. Find the mean of the following frequency distribution:
Class Interval: 15-25, 25-35, 35-45, 45-55, 55-65
Frequency: 60, 35, 22, 18, 15
Answer:
Mid-values (x): 20, 30, 40, 50, 60
\( \sum f_i = 150 \)
\( \sum f_i x_i = (60 \times 20) + (35 \times 30) + (22 \times 40) + (18 \times 50) + (15 \times 60) \)
\( = 1200 + 1050 + 880 + 900 + 900 = 4930 \)
Mean \( (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{4930}{150} = 32.86 \) or 32.87 (approx.)

(b) Assumed Mean method

If the values of \( x \) are large, then the calculation of the arithmetic mean by direct method becomes lengthy and tedious. To avoid such lengthy calculations, we use short cut method. In this method, the labour of calculation becomes less.
In this method, we choose assumed mean 'a' (say) and subtract it from each of the values \( x_i \). The reduced value \( x_i - a \) is called the deviation of \( x_i \) from \( a \). The deviations are multiplied by corresponding frequencies to get \( f_i d_i \). On adding all \( f_i d_i \) we get the sum i.e., \( \sum f_i d_i \).
Arithmetic mean = \( a + \frac{\sum_{i=1}^n f_i d_i}{\sum_{i=1}^n f_i} \)

Let \( x_1, x_2, ..., x_n \) be values of a variable \( x \), with corresponding frequencies \( f_1, f_2, ..., f_n \) respectively. Taking deviations about an arbitrary point \( a \), we get
\( d_i = x_i - a \), where \( i = 1, 2, 3, ..., n \)

\( \implies \) \( f_i d_i = f_i (x_i - a) \); \( i = 1, 2, 3, ..., n \)

\( \implies \) \( \sum_{i=1}^n f_i d_i = \sum_{i=1}^n f_i(x_i - a) \)

\( \implies \) \( \sum_{i=1}^n f_i d_i = \sum_{i=1}^n f_i x_i - a \sum_{i=1}^n f_i \)
On dividing by \( \sum f_i \) we get
\( \frac{\sum_{i=1}^n f_i d_i}{\sum f_i} = \frac{\sum_{i=1}^n f_i (x_i - a)}{\sum f_i} = \frac{\sum_{i=1}^n f_i x_i}{\sum f_i} - \frac{a \sum f_i}{\sum f_i} \)

\( \implies \) \( \frac{\sum_{i=1}^n f_i d_i}{\sum f_i} = \bar{x} - a \)

\( \implies \) \( \bar{x} = a + \frac{\sum_{i=1}^n f_i d_i}{\sum f_i} \)

Notes: The assumed mean is chosen in such a manner that:
1. It should be one of the central values.
2. The deviations are small.
3. One deviation is zero.

Working Rule

Step 1. Choose a number 'a' from the central values of \( x \) of the first column, that will be our assumed mean.

Step 2. Obtain deviations \( d_i \) by subtracting 'a' from \( x_i \). Write down these deviations against the corresponding frequencies in the third column.

Step 3. Multiply the frequencies of second column with corresponding deviations \( d_i \) in the third column to prepare a fourth column of \( f_i d_i \).

Step 4. Find the sum of all the entries of fourth column to obtain \( \sum f d_i \) and also, find the sum of all the frequencies in the second column to obtain \( \sum f_i \).

Step 5. Find arithmetic mean by using the formula \( \bar{x} = a + \frac{\sum_{i=1}^n f_i d_i}{\sum f_i} \)

Question. The following table gives the distribution of total household expenditure (in Rs.) of manual workers in a city. Find the arithmetic mean.
Expenditure (in Rs.): 100-150, 150-200, 200-250, 250-300, 300-350, 350-400, 400-450, 450-500
Frequency: 24, 40, 33, 28, 30, 22, 16, 7
Answer:
Let assumed mean \( a = 275 \)
Mid-values (x): 125, 175, 225, 275, 325, 375, 425, 475
Deviations \( d_i = x - 275 \): -150, -100, -50, 0, 50, 100, 150, 200
\( f_i d_i \) products: -3600, -4000, -1650, 0, 1500, 2200, 2400, 1400
\( \sum f_i = 200 \)
\( \sum f_i d_i = -1750 \)
\( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 275 + \frac{-1750}{200} = 275 - 8.75 = \text{Rs. } 266.25 \)

Question. Calculate the arithmetic mean of the following frequency distribution.
Class-Interval: 0-50, 50-100, 100-150, 150-200, 200-250, 250-300
Frequency: 17, 35, 43, 40, 21, 24
Answer:
Let assumed mean \( a = 175 \)
Mid-values \( x_i \): 25, 75, 125, 175, 225, 275
Deviations \( d_i = x_i - 175 \): -150, -100, -50, 0, 50, 100
\( f_i d_i \) values: -2550, -3500, -2150, 0, 1050, 2400
\( \sum f_i = 180 \)
\( \sum f_i d_i = -4750 \)
Now, \( a = 175 \)
\( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 175 + \frac{-4750}{180} = 175 - \frac{475}{18} = 175 - 26.39 = 148.61 \) (approx).

Question. Calculate the arithmetic mean of the following frequency distribution.
Class Interval: 50-60, 60-70, 70-80, 80-90, 90-100
Frequency: 8, 6, 12, 11, 13
Answer:
Let assumed mean \( a = 75 \)
Mid-values \( x_i \): 55, 65, 75, 85, 95
Deviations \( d_i = x_i - 75 \): -20, -10, 0, 10, 20
\( f_i d_i \) products: -160, -60, 0, 110, 260
\( \sum f_i = 50 \), \( \sum f_i d_i = 150 \)
Mean \( (\bar{x}) = a + \frac{\sum f_i d_i}{\sum f_i} = 75 + \frac{150}{50} = 75 + 3 = 78 \)

(c) Step deviation method

Deviation method can be further simplified on dividing the deviation by width of the class interval \( h \). In such a case the arithmetic mean is reduced to a great extent.
\( u_i = \frac{x_i - a}{h} \), \( i = 1, 2, 3, ..., n \)

\( \implies \) \( x_i = a + h u_i \), \( i = 1, 2, 3, ..., n \)

\( \implies \) \( f_i x_i = a f_i + h f_i u_i \), \( i = 1, 2, 3, ..., n \)

\( \implies \) \( \sum f_i x_i = \sum a f_i + \sum h f_i u_i \)

\( \implies \) \( \sum f_i x_i = a \sum f_i + h \sum f_i u_i \)

\( \implies \) \( \frac{\sum f_i x_i}{\sum f_i} = a \frac{\sum f_i}{\sum f_i} + h \frac{\sum f_i u_i}{\sum f_i} \)

\( \implies \) \( \text{Mean } (\bar{x}) = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \)

Working Rule

Step 1. Choose a number 'a' from the central values of \( x \) (mid-values).

Step 2. Obtain \( u_i = \frac{x_i - a}{h} \)

Step 3. Multiply the frequency \( f_i \) with the corresponding \( u_i \) to get \( f_i u_i \).

Step 4. Find the sum of all \( f_i u_i \), i.e., \( \sum f_i u_i \).

Step 5. Use the formula \( \bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \cdot h \) to get the required mean.

Question. The weekly observation on cost of living index in a certain city for the year 2004-2005 are given below. Compute the mean weekly cost of living index.
Cost of Living Index: 1400-1500, 1500-1600, 1600-1700, 1700-1800, 1800-1900, 1900-2000
Number of weeks: 5, 10, 20, 9, 6, 2
Answer:
Let assumed mean be 1750, i.e., \( a = 1750 \), \( h = 100 \)
Mid-values \( x_i \): 1450, 1550, 1650, 1750, 1850, 1950
\( u_i = \frac{x_i - 1750}{100} \): -3, -2, -1, 0, 1, 2
\( f_i u_i \) values: -15, -20, -20, 0, 6, 4
\( \sum f_i = 52 \), \( \sum f_i u_i = -45 \)
By step deviation method
Mean \( (\bar{x}) = a + \frac{\sum f_i u_i}{\sum f_i} h = 1750 + \frac{-45}{52} \times 100 = 1750 - 86.54 = 1663.46 \)
Hence, the mean weekly cost of living index = 1663.46.

Question. Find the mean marks from the following data by step deviation method.
Marks: Below 10, Below 20, Below 30, Below 40, Below 50, Below 60, Below 70, Below 80, Below 90, Below 100
Number of Students: 5, 9, 17, 29, 45, 60, 70, 78, 83, 85
Answer:
First, convert to frequency distribution:
0-10: 5
10-20: 9-5 = 4
20-30: 17-9 = 8
30-40: 29-17 = 12
40-50: 45-29 = 16
50-60: 60-45 = 15
60-70: 70-60 = 10
70-80: 78-70 = 8
80-90: 83-78 = 5
90-100: 85-83 = 2
Let assumed mean \( a = 55 \), \( h = 10 \)
Mid-values \( x_i \): 5, 15, 25, 35, 45, 55, 65, 75, 85, 95
\( u_i = \frac{x_i - 55}{10} \): -5, -4, -3, -2, -1, 0, 1, 2, 3, 4
\( f_i u_i \): -25, -16, -24, -24, -16, 0, 10, 16, 15, 8
\( \sum f_i = 85 \), \( \sum f_i u_i = -56 \)
Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} h = 55 + \frac{-56}{85} \times 10 = 55 - 6.59 = 48.41 \)
Hence, mean mark = 48.41.

Note. It should be clearly understood that in the above example the upper limits of each class are excluded from the respective classes and it is included in the lower limit of the immediate next class.

Question. Find the mean age of 100 residents of a colony from the following data:
Age in years: Greater than 0, Greater than 10, Greater than 20, Greater than 30, Greater than 40, Greater than 50, Greater than 60, Greater than 70
Number of Persons: 100, 90, 75, 50, 25, 15, 5, 0
Answer:
Convert to frequency distribution:
0-10: 100-90 = 10
10-20: 90-75 = 15
20-30: 75-50 = 25
30-40: 50-25 = 25
40-50: 25-15 = 10
50-60: 15-5 = 10
60-70: 5-0 = 5
Let assumed mean \( a = 35 \), \( h = 10 \)
Mid-values \( x_i \): 5, 15, 25, 35, 45, 55, 65
\( u_i = \frac{x_i - 35}{10} \): -3, -2, -1, 0, 1, 2, 3
\( f_i u_i \): -30, -30, -25, 0, 10, 20, 15
\( \sum f_i = 100 \), \( \sum f_i u_i = -40 \)
\( \bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h \)

\( \implies \) \( \bar{x} = 35 + \frac{-40}{100} \times 10 = 31 \)
Hence, the mean age is 31 years.

Question. The mean of the following frequency distribution is 62.8. Find the missing frequency x.
Class: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120
Frequency: 5, 8, x, 12, 7, 8
Answer:
Mid-values \( x_i \): 10, 30, 50, 70, 90, 110
\( \sum f_i = 5 + 8 + x + 12 + 7 + 8 = 40 + x \)
\( \sum f_i x_i = (5 \times 10) + (8 \times 30) + (x \times 50) + (12 \times 70) + (7 \times 90) + (8 \times 110) \)
\( = 50 + 240 + 50x + 840 + 630 + 880 = 2640 + 50x \)
Mean \( (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \)

\( \implies \) \( 62.8 = \frac{2640 + 50x}{40 + x} \)

\( \implies \) \( 62.8 (40 + x) = 2640 + 50x \)

\( \implies \) \( 2512 + 62.8x = 2640 + 50x \)

\( \implies \) \( 12.8x = 128 \)

\( \implies \) \( x = \frac{128}{12.8} = 10 \)
Hence, the missing frequency is 10.

Question. The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies \( f_1 \) and \( f_2 \):
Class: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120
Frequency: 5, \( f_1 \), 10, \( f_2 \), 7, 8; Total: 50
Answer:
\( \sum f_i = 30 + f_1 + f_2 = 50 \)

\( \implies \) \( f_1 + f_2 = 20 \) ...(1)
Mid-values \( x_i \): 10, 30, 50, 70, 90, 110
\( \sum f_i x_i = (5 \times 10) + (f_1 \times 30) + (10 \times 50) + (f_2 \times 70) + (7 \times 90) + (8 \times 110) \)
\( = 50 + 30f_1 + 500 + 70f_2 + 630 + 880 = 2060 + 30f_1 + 70f_2 \)
Mean \( (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \)

\( \implies \) \( 62.8 = \frac{2060 + 30f_1 + 70f_2}{50} \)

\( \implies \) \( 62.8 \times 50 = 2060 + 30f_1 + 70f_2 \)

\( \implies \) \( 3140 = 2060 + 30f_1 + 70f_2 \)

\( \implies \) \( 30f_1 + 70f_2 = 1080 \)

\( \implies \) \( 3f_1 + 7f_2 = 108 \) ...(2)
Multiplying (1) by 3:
\( 3f_1 + 3f_2 = 60 \) ...(3)
Subtracting (3) from (2):
\( 4f_2 = 48 \)

\( \implies \) \( f_2 = 12 \)
Putting \( f_2 = 12 \) in (1):
\( f_1 + 12 = 20 \)

\( \implies \) \( f_1 = 8 \)

Apply direct method to find arithmetic mean in each of the following (1 - 6):

Question.
Class-Interval: 0-6, 6-12, 12-18, 18-24, 24-30
Frequency: 7, 5, 10, 12, 6
Answer:
Mid-values \( x_i \): 3, 9, 15, 21, 27
\( \sum f_i = 40 \)
\( \sum f_i x_i = (7 \times 3) + (5 \times 9) + (10 \times 15) + (12 \times 21) + (6 \times 27) = 21 + 45 + 150 + 252 + 162 = 630 \)
Mean \( \bar{x} = \frac{630}{40} = 15.75 \)

Question.
Class-Interval: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 8, 10, 9, 12, 11
Answer:
Mid-values \( x_i \): 5, 15, 25, 35, 45
\( \sum f_i = 50 \)
\( \sum f_i x_i = (8 \times 5) + (10 \times 15) + (9 \times 25) + (12 \times 35) + (11 \times 45) = 40 + 150 + 225 + 420 + 495 = 1330 \)
Mean \( \bar{x} = \frac{1330}{50} = 26.6 \)

SOLVED EXAMPLES

Question. The following table shows the marks obtained by three students: A, B and C and the weights assigned to different subjects. Calculate the weighted mean of each of A, B and C. If a scholarship is awarded on the basis of weighted mean, who among the three, will get the scholarship?
Subject | Weight | Marks Scored by A | Marks Scored by B | Marks Scored by C
English | 4 | 70 | 80 | 85
Hindi | 1 | 50 | 60 | 45
Economics | 3 | 90 | 75 | 75
Mathematics | 2 | 60 | 45 | 65

Answer:
Subject | Weight (\( w_i \)) | Marks scored by A (\( x_{Ai} \)) | Marks scored by B (\( x_{Bi} \)) | Marks scored by C (\( x_{Ci} \)) | \( w_i x_{Ai} \) | \( w_i x_{Bi} \) | \( w_i x_{Ci} \)
English | 4 | 70 | 80 | 85 | 280 | 320 | 340
Hindi | 1 | 50 | 60 | 45 | 50 | 60 | 45
Economics | 3 | 90 | 75 | 75 | 270 | 225 | 225
Mathematics | 2 | 60 | 45 | 65 | 120 | 90 | 130
Total | \( \sum w_i = 10 \) | | | | \( \sum w_i x_{Ai} = 720 \) | \( \sum w_i x_{Bi} = 695 \) | \( \sum w_i x_{Ci} = 740 \)
Let \( \bar{x}_{wA}, \bar{x}_{wB} \) and \( \bar{x}_{wC} \) denote the weighted means of A, B and C respectively.
\( \bar{x}_{wA} = \frac{\sum_{i=1}^n w_i x_{Ai}}{\sum_{i=1}^n w_i} = \frac{720}{10} = 72 \)

\( \implies \) \( \bar{x}_{wB} = \frac{\sum_{i=1}^n w_i x_{Bi}}{\sum_{i=1}^n w_i} = \frac{695}{10} = 69.5 \)

\( \implies \) \( \bar{x}_{wC} = \frac{\sum_{i=1}^n w_i x_{Ci}}{\sum_{i=1}^n w_i} = \frac{740}{10} = 74 \)
Hence, the weighted mean of A, B and C are 72, 69.5 and 74 respectively. Since C has the highest weighted mean, hence C will get the scholarship.

Question. Calculate the mean and the weighted mean for the following data of marks in a Class X examination as per the weights attached to each other.
Subject | Marks | Weight
English | 62 | 1
Mathematics | 83 | 3
Science | 79 | 3
Social science | 74 | 2
Hindi | 77 | 2

Answer:
Subject | Marks (\( x_i \)) | Weight (\( w_i \)) | \( w_i x_i \)
English | 62 | 1 | 62
Mathematics | 83 | 3 | 249
Science | 79 | 3 | 237
Social science | 74 | 2 | 148
Hindi | 77 | 2 | 154
Total | \( \sum x_i = 375 \) | \( \sum w_i = 11 \) | \( \sum w_i x_i = 850 \)
Mean = \( \frac{\sum x_i}{n} = \frac{375}{5} = 75 \)
Weighted mean = \( \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i} = \frac{850}{11} = 77.27 \)
Hence, Mean = 75 and the weighted mean = 77.27

Question. The marks obtained by Pallavi in an examination are as follows.
Maths: 84%
Hindi: 65%
Sanskrit: 72%
English: 49%
General science: 78%
If the weightage of marks in Maths and General Science is double than other subjects then find the weighted mean.
Answer: Let the weight for Hindi, Sanskrit, and English be 1 each. Then the weight for Maths and General science will be 2 each.
\( \sum w_i = 2 + 1 + 1 + 1 + 2 = 7 \)
\( \sum w_i x_i = (84 \times 2) + (65 \times 1) + (72 \times 1) + (49 \times 1) + (78 \times 2) \)
\( = 168 + 65 + 72 + 49 + 156 = 510 \)
Weighted mean \( = \frac{\sum w_i x_i}{\sum w_i} = \frac{510}{7} \approx 72.86 \% \)

Question. The commodities required by a family are listed in the following table. The importance (weight) attached to each commodity is also given.
Commodities: Rice, Wheat, Pulses, Oil, Vegetables
Weight: 5, 3, 2, 1, 4
Price per kg: Rs. 20.00, Rs. 8.00, Rs. 24.00, Rs. 45.00, Rs. 15.00
Find the weighted mean.
Answer:
\( \sum w_i = 5 + 3 + 2 + 1 + 4 = 15 \)
\( \sum w_i x_i = (5 \times 20.00) + (3 \times 8.00) + (2 \times 24.00) + (1 \times 45.00) + (4 \times 15.00) \)
\( = 100 + 24 + 48 + 45 + 60 = 277 \)
Weighted mean \( = \frac{\sum w_i x_i}{\sum w_i} = \frac{277}{15} \approx Rs. 18.47 \)

Question. The following table gives the marks of a student in an examination, weights attached to each subject are also given:
Subjects: Mathematics, English, Gen. Science, Hindi, Sanskrit
Marks Scared: 75, 82, 68, 73, 98
Weight: 5, 3, 4, 1, 2
Find the weighted mean.
Answer:
\( \sum w_i = 5 + 3 + 4 + 1 + 2 = 15 \)
\( \sum w_i x_i = (75 \times 5) + (82 \times 3) + (68 \times 4) + (73 \times 1) + (98 \times 2) \)
\( = 375 + 246 + 272 + 73 + 196 = 1162 \)
Weighted mean \( = \frac{\sum w_i x_i}{\sum w_i} = \frac{1162}{15} \approx 77.47 \)

Question. The following are the prices of certain important commodities required by a family. The weight attached to each commodity by the family is also indicated. Compare the mean price per kg with the weighted mean price per kg.
Commodity: Rice, Wheat, Oil, Dals
Importance (Weight): 4, 2, 1, 1
Price per kg.: Rs. 8.50, Rs. 3.20, Rs. 25.00, Rs. 7.25
Answer:
Mean price per kg \( = \frac{8.50 + 3.20 + 25.00 + 7.25}{4} = \frac{43.95}{4} = Rs. 10.9875 \)
Weighted mean price per kg:
\( \sum w_i = 4 + 2 + 1 + 1 = 8 \)
\( \sum w_i x_i = (4 \times 8.50) + (2 \times 3.20) + (1 \times 25.00) + (1 \times 7.25) \)
\( = 34.00 + 6.40 + 25.00 + 7.25 = 72.65 \)
Weighted mean \( = \frac{72.65}{8} = Rs. 9.08125 \)
Comparison: The mean price per kg (Rs. 10.99) is higher than the weighted mean price per kg (Rs. 9.08).

Question. There are three sections A, B and C in class X with 25, 40 and 35 students respectively. The average marks obtained by section A, B and C are 70%, 65% and 50% respectively. Find the average marks of entire class X.
Answer: \( n_1 = 25 \), \( \overline{x}_1 = 70\% \)
\( n_2 = 40 \), \( \overline{x}_2 = 65\% \)
\( n_3 = 35 \), \( \overline{x}_3 = 50\% \)
\( \overline{x} = \frac{n_1\overline{x}_1 + n_2\overline{x}_2 + n_3\overline{x}_3}{n_1 + n_2 + n_3} = \frac{(25 \times 70) + (40 \times 65) + (35 \times 50)}{25 + 40 + 35} \)
\( = \frac{1750 + 2600 + 1750}{100} = \frac{6100}{100} = 61\% \)

Question. The mean of 25 observations is 36. If the mean of the first 13 observations is 32 and that of the last 13 observations is 39, find the 13th observation.
Answer: The mean of 25 observations = 36
\( \overline{x}_1 = \frac{\sum x_1}{n_1} \)
\( \implies \) \( 36 = \frac{\sum x_1}{25} \)
\( \implies \) \( \sum x_1 = 36 \times 25 = 900 \)
The mean of first 13 observations = 32
\( \overline{x}_2 = \frac{\sum x_2}{n_2} \)
\( \implies \) \( 32 = \frac{\sum x_2}{13} \)
\( \implies \) \( \sum x_2 = 32 \times 13 = 416 \)
The mean of the last 13 observations = 39
\( \overline{x}_3 = \frac{\sum x_3}{n_3} \)
\( \implies \) \( 39 = \frac{\sum x_3}{13} \)
\( \implies \) \( \sum x_3 = 13 \times 39 = 507 \)
13th observation = sum of first 13 observations + sum of last 13 observations - sum of 25 observations
= 416 + 507 - 900 = 23

Question. The mean weight of a class of 35 students is 45 kg. If the weight of the teacher be included, the mean weight increases by 500 grams. Find the weight of the teacher.
Answer: Let the mean weight of a class of 35 students be \( \overline{x}_1 \) and that of both students and teachers be \( \overline{x}_2 \).
Then, \( \overline{x}_1 = 45 \) kg and \( \overline{x}_2 = 45 + \frac{500}{1000} = 45 + 0.5 = 45.5 \) kg
\( \overline{x}_1 = \frac{\sum x_1}{n_1} \)
\( \overline{x}_2 = \frac{\sum x_2}{n_2} \)
\( 45 = \frac{\sum x_1}{35} \)
\( 45.5 = \frac{\sum x_2}{36} \)
\( \sum x_1 = 1575 \) kg
\( \sum x_2 = 1638 \) kg
\( \implies \) Total weight = weight of students + weight of teacher
\( \therefore \) Weight of teacher = Total weight - weight of students
\( \therefore \) Weight of the teacher = \( \sum x_2 - \sum x_1 = 1638 - 1575 = 63 \) kg

Question. The marks in science of 80 students of class X are given below. Find the mode of these marks obtained by the students in science.
Marks: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100
Frequency: 3, 5, 16, 12, 13, 20, 5, 4, 1, 1

Answer: The class (50 - 60) has maximum frequency i.e. 20 therefore this is the modal class.
Lower limit of the modal class = \( l = 50 \)
Class - interval = \( h = 10 \)
Frequency of the modal class = \( f_k = 20 \)
Frequency of the class preceding the modal class = \( f_{k-1} = 13 \)
Frequency of the class succeeding the modal class = \( f_{k+1} = 5 \)
Mode = \( l + h \left( \frac{f_k - f_{k-1}}{2f_k - f_{k-1} - f_{k+1}} \right) = 50 + 10 \left( \frac{20 - 13}{2 \times 20 - 13 - 5} \right) \)
= \( 50 + 10 \left( \frac{7}{40 - 18} \right) = 50 + \frac{10 \times 7}{22} = 50 + 3.1818 = 53.18 \)
Hence, the mode of given marks obtained by the students in science is 53.18.

Question. The following is the distribution of height of students of a certain class in a certain city.
Height (in cms): 160-162, 163-165, 166-168, 169-171, 172-174
No. of students: 15, 118, 142, 127, 18
Find the average height of maximum number of students.

Answer: The given series is inclusive series (discontinuous series). The series is to be converted into exclusive series (continuous series).
Here, first of all we have to find out adjustment factor.
Adjustment factor = \( \frac{1}{2} \) [Lower limit of the II class - Upper limit of the I class]
= \( \frac{1}{2} [163 - 162] = \frac{1}{2} = 0.5 \)
By subtracting 0.5 from lower limit and adding 0.5 to the higher limit of each class interval, we get the exclusive (continuous) series as follows:
Height (in cm): 159.5-162.5, 162.5-165.5, 165.5-168.5, 168.5-171.5, 171.5-174.5
No. of students: 15, 118, 142, 127, 18
The class (165.5-168.5) has maximum frequency i.e., 142, therefore this is the modal class.
Lower limit of the modal class, \( l = 165.5 \)
Class interval, \( h = 3 \)
Frequency of the modal class, \( f_k = 142 \)
Frequency of the class preceding the modal class = \( f_{k-1} = 118 \)
Frequency of the class succeeding the modal class = \( f_{k+1} = 127 \)
\( \therefore \) Mode = \( l + h \left( \frac{f_k - f_{k-1}}{2f_k - f_{k-1} - f_{k+1}} \right) = 165.5 + 3 \left( \frac{142 - 118}{2 \times 142 - 118 - 127} \right) = 165.5 + 3 \left( \frac{24}{284 - 245} \right) \)
= \( 165.5 + \frac{3 \times 24}{39} = 165.5 + \frac{24}{13} = 165.5 + 1.846 = 167.346 \)
Hence, the average height of maximum number of students is 167.346.

Question. The amount of wheat in tons coming from M.P. to Delhi is given below :
Amount of wheat (in tons) coming per day: 21-25, 26-30, 31-35, 36-40, 41-45
Number of days: 7, 13, 22, 10, 8
Find the mode.

Answer: First we shall convert the given inclusive series into an exclusive (continuous) series.
Adjustment factor = \( \frac{1}{2} \) [Lower limit of II class - Upper limit of I class]
= \( \frac{1}{2} [26 - 25] = \frac{1}{2} = 0.5 \)
For the conversion of the series we subtract 0.5 from lower limit and add 0.5 to the upper limit of each class.
Class interval: 20.5-25.5, 25.5-30.5, 30.5-35.5, 35.5-40.5, 40.5-45.5
Number of days: 7, 13, 22, 10, 8
The class (30.5 - 35.5) has maximum frequency, i.e., 22, therefore this is the modal class.
Lower limit of the modal class = \( l = 30.5 \)
Class interval = \( h = 25.5 - 20.5 = 5 \)
Frequency of the modal class = \( f_k = 22 \)
Frequency of the class preceding the modal class = \( f_{k-1} = 13 \)
Frequency of the class succeeding the modal class = \( f_{k+1} = 10 \)
\( \therefore \) Mode = \( l + h \left( \frac{f_k - f_{k-1}}{2f_k - f_{k-1} - f_{k+1}} \right) \)
= \( 30.5 + 5 \left( \frac{22 - 13}{2 \times 22 - 13 - 10} \right) = 30.5 + 5 \left( \frac{9}{44 - 23} \right) \)
= \( 30.5 + \frac{45}{21} = 30.5 + \frac{15}{7} = 30.5 + 2.14 = 32.64 \)

Multiple Choice Questions

Question. A survey conducted by a group of students is given as
Family Size: \( 1-3 \), \( 3-5 \), \( 5-7 \), \( 7-9 \), \( 9-11 \)
Number of families: \( 7, 8, 2, 2, 1 \)
The mean of the data is
(a) \( 6.8 \)
(b) \( 4.2 \)
(c) \( 5.4 \)
(d) None of these
Answer: (b)

Question. The relationship among mean, median and mode for a distribution is
(a) Mode = Median \( - \) 2mean
(b) Mode = 3 Median \( - \) 2mean
(c) Mode = 2 Median \( - \) 3 mean
(d) Mode = Median \( - \) mean
Answer: (b)

Question. For the following distribution
Class: \( 0-5, 5-10, 10-15, 15-20, 20-25 \)
Frequency: \( 10, 15, 12, 20, 9 \)
The sum of lower limits of the median class and modal class is
(a) \( 15 \)
(b) \( 25 \)
(c) \( 30 \)
(d) \( 35 \)
Answer: (b)

Case Based MCQs

A Tesla car manufacturing industry wants to declare the mileage of their electric cars. For this, they recorded the mileage (km/charge) of 100 cars of the same model. Details of which are given in the following table.
Mileage (km/charge): \( 100-120, 120-140, 140-160, 160-180 \)
Number of Cars: \( 14, 24, 36, 26 \)
Based on the above information, answer the following questions.

Question. The average mileage is
(a) \( 140 \) km/charge
(b) \( 150 \) km/charge
(c) \( 130 \) km/charge
(d) \( 144.8 \) km/charge
Answer: (d)

Question. The modal value of the given data is
(a) \( 150 \)
(b) \( 150.91 \)
(c) \( 145.6 \)
(d) \( 140.9 \)
Answer: (b)

Question. The median value of the given data is
(a) \( 140 \)
(b) \( 146.67 \)
(c) \( 130 \)
(d) \( 136.6 \)
Answer: (b)

Question. Assumed mean method is useful in determining the
(a) Mean
(b) Median
(c) Mode
(d) All of these
Answer: (a)

Question. The manufacturer can claim that the mileage for his car is
(a) \( 144 \) km/charge
(b) \( 155 \) km/charge
(c) \( 165 \) km/charge
(d) \( 175 \) km/charge
Answer: (a)

Short Answer Type Questions

Question. Find the median of the first ten prime numbers.
Answer: 12

Question. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table.
Number of seats: \( 100-104, 104-108, 108-112, 112-116, 116-120 \)
Frequency: \( 15, 20, 32, 18, 15 \)
Determine the mean number of seats occupied over the flights.
Answer: 109.92

Question. The following distribution gives the daily income of 50 workers of a factory:
Daily income (in \( Rs \)): \( 100-120, 120-140, 140-160, 160-180, 180-200 \)
Number of workers: \( 12, 14, 8, 6, 10 \)
Write the above distribution as ‘less than type’ cumulative frequency distribution. 
Answer:
Daily income (in \( Rs \)) | Cumulative frequency
Less than 120 | 12
Less than 140 | 26
Less than 160 | 34
Less than 180 | 40
Less than 200 | 50

Question. Find the mode of the following frequency distribution. 
Class: \( 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70 \)
Frequency: \( 8, 10, 10, 16, 12, 6, 7 \)
Answer: 36

Long Answer Type Questions

Question. Find the mean of the following frequency distribution using assumed mean method.
Class: \( 2-8, 8-14, 14-20, 20-26, 26-32 \)
Frequency: \( 6, 3, 12, 11, 8 \)
Answer: 18.8