Download Class 11 Mathematics Concept Summaries: CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01
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Arithmetic Mean
Arithmetic Mean
Arithmetic mean is defined as the sum of the values of all observations divided by the number of observations.
Arithmetic Mean in an Individual Series
(i) Direct method: \( \bar{x} = \frac{\Sigma x}{N} \), where \( \Sigma x = x_1 + x_2 + \ldots + x_n \) and N = total number of items.
| Students | A | B | C | D | E | F | G | H | I | J |
|---|---|---|---|---|---|---|---|---|---|---|
| Marks | 85 | 60 | 50 | 75 | 55 | 40 | 55 | 70 | 45 | 65 |
Answer:
| Students | Marks (X) |
|---|---|
| A | 85 |
| B | 60 |
| C | 50 |
| D | 75 |
| E | 55 |
| F | 40 |
| G | 55 |
| H | 70 |
| I | 45 |
| J | 65 |
| N = 10 | Σx = 600 |
Arithmetic mean \( \bar{x} = \frac{\Sigma x}{N} = \frac{600}{10} \) = 60 marks
(ii) Short cut method: Under this method, any figure is assumed as the mean and deviations are calculated from this assumed mean.
\( \bar{x} = A + \frac{\Sigma d}{N} \)
where \( \bar{x} \) = arithmetic mean, A = assumed mean, d = X − A = deviation of the variable from the assumed mean, Σd = Σ(X − A) = sum of the deviations from the assumed mean and N = total number of items.
Answer:
| Students | Marks (X) | d = X − A (A = 50) |
|---|---|---|
| A | 85 | +35 |
| B | 60 | +10 |
| C | 50 (A) | 0 |
| D | 75 | +25 |
| E | 55 | +5 |
| F | 40 | −10 |
| G | 55 | +5 |
| H | 70 | +20 |
| I | 45 | −5 |
| J | 65 | +15 |
| N = 10 | Σd = 100 |
\( \bar{x} = A + \frac{\Sigma d}{N} = 50 + \frac{100}{10} \) = 60 marks
(iii) Step deviation method: In this method, the deviations from the assumed mean are divided by a common factor (C) to get step deviations. Apply the following formula:
\( \bar{x} = A + \frac{\Sigma d'}{N} \times C \)
where d = X − A is the deviation of the variable from the assumed mean, \( d' = \frac{X - A}{C} \) is the step deviation, Σd′ = sum of the step deviations and C = common factor.
Note: The step deviation method can be used only when the deviations from the assumed mean (d) are divisible by a common factor.
Answer:
| Students | Marks (X) | d = X − A, A = 50 | \( d' = \frac{X - A}{C} \), C = 5 |
|---|---|---|---|
| A | 85 | +35 | +7 |
| B | 60 | +10 | +2 |
| C | 50 (A) | 0 | 0 |
| D | 75 | +25 | +5 |
| E | 55 | +5 | +1 |
| F | 40 | −10 | −2 |
| G | 55 | +5 | +1 |
| H | 70 | +20 | +4 |
| I | 45 | −5 | −1 |
| J | 65 | +15 | +3 |
| N = 10 | Σd′ = 20 |
\( \bar{x} = A + \frac{\Sigma d'}{N} \times C = 50 + \frac{20}{10} \times 5 \) = 60 marks
Arithmetic Mean in a Discrete Series
In case of a discrete series (ungrouped frequency distribution), the values of the variable show repetitions, i.e., frequencies are given corresponding to different values of the variable. N = sum total of frequencies = Σf.
(i) Direct method: \( \bar{x} = \frac{\Sigma fx}{\Sigma f} \), where Σfx = sum of the products of the variable with the respective frequencies and Σf = total number of items.
| Marks | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|
| No. of students | 8 | 12 | 20 | 10 | 6 | 4 |
Answer:
| Marks (X) | No. of students (f) | fx |
|---|---|---|
| 20 | 8 | 160 |
| 30 | 12 | 360 |
| 40 | 20 | 800 |
| 50 | 10 | 500 |
| 60 | 6 | 360 |
| 70 | 4 | 280 |
| Σf = 60 | Σfx = 2460 |
\( \bar{x} = \frac{\Sigma fx}{\Sigma f} = \frac{2460}{60} \) = 41 marks
(ii) Short cut method: \( \bar{x} = A + \frac{\Sigma fd}{\Sigma f} \), where A = assumed mean, d = X − A is the deviation of the variable from the assumed mean, Σfd = sum of the products of the deviations (d) with the respective frequencies (f) and Σf = total number of items.
Answer:
| Marks | No. of students (f) | d = X − A (A = 40) | fd |
|---|---|---|---|
| 20 | 8 | −20 | −160 |
| 30 | 12 | −10 | −120 |
| 40 (A) | 20 | 0 | 0 |
| 50 | 10 | +10 | +100 |
| 60 | 6 | +20 | +120 |
| 70 | 4 | +30 | +120 |
| Σf = 60 | Σfd = +60 |
\( \bar{x} = A + \frac{\Sigma fd}{\Sigma f} = 40 + \frac{60}{60} \) = 41 marks
(iii) Step deviation method: In this method, the values of the deviations (d) are divided by a common factor (C) to ease the calculation process.
\( \bar{x} = A + \frac{\Sigma fd'}{\Sigma f} \times C \)
where d′ = step deviations (deviations from the assumed mean divided by the common factor) and Σfd′ = sum of the products of the step deviations (d′) with the respective frequencies (f).
Answer:
| Marks X | No. of students f | d = X − A (A = 40) | \( d' = \frac{X - A}{C} \) (C = 10) | fd′ |
|---|---|---|---|---|
| 20 | 8 | −20 | −2 | −16 |
| 30 | 12 | −10 | −1 | −12 |
| 40 (A) | 20 | 0 | 0 | 0 |
| 50 | 10 | +10 | +1 | +10 |
| 60 | 6 | +20 | +2 | +12 |
| 70 | 4 | +30 | +3 | +12 |
| Σf = 60 | Σfd′ = +6 |
\( \bar{x} = A + \frac{\Sigma fd'}{\Sigma f} \times C = 40 + \frac{6}{60} \times 10 \) = 41 marks
Arithmetic Mean in a Continuous Series
In the case of a continuous series (grouped frequency distribution), the values of the variable are grouped in various class intervals along with their respective frequencies. In a continuous series, the mid-points of the various class intervals are used to replace the class intervals. Once it is done, there is no difference between a continuous series and a discrete series.
In a continuous series also, the following three methods are used to calculate the arithmetic mean: (i) direct method, (ii) short cut method, (iii) step deviation method.
| Marks | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
|---|---|---|---|---|---|
| No. of students | 2 | 8 | 6 | 10 | 4 |
Answer:
| Marks X | No. of students f | Mid-value (m) | fm |
|---|---|---|---|
| 0 – 5 | 2 | 2.5 | 5 |
| 5 – 10 | 8 | 7.5 | 60 |
| 10 – 15 | 6 | 12.5 | 75 |
| 15 – 20 | 10 | 17.5 | 175 |
| 20 – 25 | 4 | 22.5 | 90 |
| Σf = 30 | Σfm = 405 |
\( \bar{x} = \frac{\Sigma fm}{\Sigma f} = \frac{405}{30} \) = 13.50 marks
Answer:
| Marks X | No. of students f | Mid-value m | d = m − A (A = 12.5) | fd |
|---|---|---|---|---|
| 0 – 5 | 2 | 2.5 | −10 | −20 |
| 5 – 10 | 8 | 7.5 | −5 | −40 |
| 10 – 15 | 6 | 12.5 (A) | 0 | 0 |
| 15 – 20 | 10 | 17.5 | +5 | +50 |
| 20 – 25 | 4 | 22.5 | +10 | +40 |
| Σf = 30 | Σfd = +30 |
\( \bar{x} = A + \frac{\Sigma fd}{\Sigma f} = 12.5 + \frac{30}{30} \) = 13.50 marks
Step deviation method: When the class intervals for all the classes in a continuous series are of the same magnitude (width), the short cut method can be further simplified by the step deviation method: \( \bar{x} = A + \frac{\Sigma fd'}{\Sigma f} \times C \)
Answer:
| Marks X | No. of students f | Mid-value m | d = m − A (A = 12.5) | \( d' = \frac{m - A}{C} \) (C = 5) | fd′ |
|---|---|---|---|---|---|
| 0 – 5 | 2 | 2.5 | −10 | −2 | −4 |
| 5 – 10 | 8 | 7.5 | −5 | −1 | −8 |
| 10 – 15 | 6 | 12.5 (A) | 0 | 0 | 0 |
| 15 – 20 | 10 | 17.5 | +5 | +1 | +10 |
| 20 – 25 | 4 | 22.5 | +10 | +2 | +8 |
| Σf = 30 | Σfd′ = +6 |
Average marks \( \bar{x} = A + \frac{\Sigma fd'}{\Sigma f} \times C = 12.5 + \frac{6}{30} \times 5 \) = 13.50 marks
Median
Median
Median may be defined as the middle value in the data set when its elements are arranged in a sequential order, that is, in either ascending or descending order of magnitude.
Median in an Individual Series
Median (Me) = size of \( \left(\frac{N + 1}{2}\right) \)th item
(i) In case of an odd number of items: Median = middle item of the distribution.
(ii) In case of an even number of items: Median = average of the two middle items.
Example 10. Marks of five students are 42, 40, 60, 55 and 50. Calculate the median.
Answer:
Step 1: Marks arranged in ascending order: 40, 42, 50, 55, 60.
Step 2: Me = size of \( \left(\frac{N + 1}{2}\right) \)th item = size of \( \left(\frac{5 + 1}{2}\right) \)th item = size of 3rd item
Median = 50
Median in a Discrete Series
In a discrete series, the values of the variable are given along with their frequencies. Cumulative frequencies are found and Me = size of \( \left(\frac{N + 1}{2}\right) \)th item.
| Size (X) | Frequency (f) |
|---|---|
| 3 | 2 |
| 4 | 1 |
| 5 | 3 |
| 6 | 7 |
| 7 | 4 |
Answer:
| Size (X) | Frequency (f) | Cumulative frequency |
|---|---|---|
| 3 | 2 | 2 |
| 4 | 1 | 3 |
| 5 | 3 | 6 |
| 6 | 7 | 13 |
| 7 | 4 | 17 |
N = Σf = 17
Me = size of \( \left(\frac{N + 1}{2}\right) \)th item = size of \( \left(\frac{17 + 1}{2}\right) \)th item = size of 9th item
The 9th item falls in the cumulative frequency 13, which corresponds to size 6.
∴ Me = 6
Median in a Continuous Series
Steps to calculate the median:
Step 1: Arrange the data in ascending or descending order.
Step 2: Calculate the cumulative frequencies (c.f.).
Step 3: Find the median item as: Me = size of \( \left(\frac{N}{2}\right) \)th item.
Step 4: Find the c.f. which is either equal to or just greater than this.
Step 5: Find the class corresponding to the c.f. equal to \( \frac{N}{2} \) or just greater than this. This class is called the median class.
Step 6: \( Me = l_1 + \frac{\frac{N}{2} - c.f.}{f} \times i \)
where \( l_1 \) = lower limit of the median class, c.f. = cumulative frequency of the class preceding the median class, f = frequency of the median class and i = width of the median class.
| X | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
|---|---|---|---|---|---|
| f | 3 | 4 | 2 | 7 | 9 |
Answer:
| X | f | c.f. |
|---|---|---|
| 0 – 10 | 3 | 3 |
| 10 – 20 | 4 | 7 |
| 20 – 30 | 2 | 9 |
| 30 – 40 | 7 | 16 |
| 40 – 50 | 9 | 25 |
N = Σf = 25
Median item = \( \frac{N}{2} = \frac{25}{2} \) = 12.5th item. The 12.5th item lies in the group 30 – 40.
\( l_1 \) = 30, c.f. = 9, f = 7, i = 10
By applying the formula, \( Me = l_1 + \frac{\frac{N}{2} - c.f.}{f} \times i = 30 + \frac{12.5 - 9}{7} \times 10 = 30 + 5 = 35 \)
Median = 35
Mode
Mode
Mode is the value occurring most frequently in a set of observations and around which the other items of the set cluster most densely.
Mode in an Individual Series
(A) Mode by way of observation:
Step 1: Arrange the data in ascending or descending order.
Step 2: The item which occurs most often in the series is the ‘mode’.
Example 13. From the heights of 15 students, calculate the value of mode.
Height (in inches): 52, 50, 66, 70, 66, 72, 71, 66, 60, 67, 69, 67, 48, 60, 65.
Answer:
By arranging the series in ascending order we get:
48, 50, 52, 60, 60, 65, 66, 66, 66, 67, 67, 69, 70, 71, 72.
By observation, the height 66 inches occurs most often, therefore the mode (Z) is 66 inches.
(B) Mode by converting an individual series into a discrete series: The data are written as a frequency table and the value with the maximum frequency is the mode.
Answer:
| Height (in inches) | Frequency |
|---|---|
| 48 | 1 |
| 50 | 1 |
| 52 | 1 |
| 60 | 2 |
| 65 | 1 |
| 66 | 3 |
| 67 | 2 |
| 69 | 1 |
| 70 | 1 |
| 71 | 1 |
| 72 | 1 |
The height of 66 inches has the maximum frequency (3), therefore the modal height (Z) is 66.
Mode = 66 inches
Mode in a Discrete Series
There are two methods to determine the mode in a discrete series:
(i) Mode by observation
(ii) Mode by grouping method
| Daily wages (in Rs.) | 100 | 110 | 120 | 130 | 140 | 150 |
|---|---|---|---|---|---|---|
| No. of persons | 2 | 4 | 8 | 10 | 5 | 4 |
Answer:
By inspection, we can see that 130 occurs most frequently (10 times) in the series.
Hence the mode is Rs. 130.
Steps of the grouping method:
Step 1: Prepare a grouping table with six columns. Column I contains the given frequencies.
Step 2: In Column II, add the frequencies in two’s starting from the first; in Column III, add them in two’s leaving the first frequency.
Step 3: In Column IV, add the frequencies in three’s starting from the first; in Column V, in three’s leaving the first; and in Column VI, in three’s leaving the first two.
Step 4: Prepare an analysis table: for each column, mark (✓) the values of the variable that make up the maximum total of that column. The value with the largest number of marks is the mode.
Answer:
Grouping table
| Wages (in Rs.) | Frequency | In two’s | In three’s | |||
|---|---|---|---|---|---|---|
| Column I | Column II | Column III | Column IV | Column V | Column VI | |
| 100 | 2 | 2+4 = 6 | 2+4+8 = 14 | |||
| 110 | 4 | 4+8 = 12 | 4+8+10 = 22 | |||
| 120 | 8 | 8+10 = 18 | 8+10+5 = 23 | |||
| 130 | 10 | 10+5 = 15 | 10+5+4 = 19 | |||
| 140 | 5 | 5+4 = 9 | ||||
| 150 | 4 | |||||
In the analysis table, we mark (✓) the values having the maximum frequencies in each column of the grouping table.
Analysis table
| Column No. | 100 | 110 | 120 | 130 | 140 | 150 |
|---|---|---|---|---|---|---|
| I | ✓ | |||||
| II | ✓ | ✓ | ||||
| III | ✓ | ✓ | ||||
| IV | ✓ | ✓ | ✓ | |||
| V | ✓ | ✓ | ✓ | |||
| VI | ✓ | ✓ | ✓ | |||
| Total | 1 | 3 | 6 | 3 | 1 |
Since the value 130 has occurred the maximum number of times, i.e. 6, the modal wage is Rs. 130.
Mode = Rs. 130
Mode in a Continuous Series
In a continuous series, the mode lies in a particular class or group, which is called the modal class. The following two methods are used in determining the mode:
(i) Observation method:
Step 1: Determine the modal class, i.e., the class with the highest frequency.
Step 2: Determine the exact value of the mode by the following formula:
\( Mo = l_1 + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times i \)
where Mo = mode, \( l_1 \) = lower limit of the modal class, \( f_1 \) = 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 and i = class interval (width) of the modal class.
| Class interval | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
|---|---|---|---|---|---|
| Frequency | 2 | 4 | 15 | 6 | 7 |
Answer:
| Class interval | Frequency |
|---|---|
| 0 – 5 | 2 |
| 5 – 10 | 4 (\( f_0 \)) |
| 10 – 15 | 15 (\( f_1 \)) modal class |
| 15 – 20 | 6 (\( f_2 \)) |
| 20 – 25 | 7 |
The class 10 – 15 has the highest frequency, so it is the modal class.
Mode (Z) = \( l_1 + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times i \), where \( l_1 \) = 10, \( f_1 \) = 15, \( f_0 \) = 4, \( f_2 \) = 6, i = 5
\( Z = 10 + \frac{15 - 4}{2 \times 15 - 4 - 6} \times 5 = 10 + \frac{11}{20} \times 5 \) = 12.75
Mode = 12.75
(ii) Grouping method: When the frequencies are irregular or the maximum frequency is close to other frequencies, the modal class is found with the help of a grouping table and an analysis table, and then the formula is applied.
| Size | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 | 80 – 90 |
|---|---|---|---|---|---|---|---|---|
| Frequency | 4 | 10 | 25 | 15 | 23 | 22 | 12 | 3 |
Answer:
Grouping table
| Size X | Frequency | In two’s | In three’s | |||
|---|---|---|---|---|---|---|
| Column I | Column II | Column III | Column IV | Column V | Column VI | |
| 10 – 20 | 4 | 4+10 = 14 | 4+10+25 = 39 | |||
| 20 – 30 | 10 | 10+25 = 35 | 10+25+15 = 50 | |||
| 30 – 40 | 25 | 25+15 = 40 | 25+15+23 = 63 | |||
| 40 – 50 | 15 | 15+23 = 38 | 15+23+22 = 60 | |||
| 50 – 60 | 23 | 23+22 = 45 | 23+22+12 = 57 | |||
| 60 – 70 | 22 | 22+12 = 34 | 22+12+3 = 37 | |||
| 70 – 80 | 12 | 12+3 = 15 | ||||
| 80 – 90 | 3 | |||||
Analysis table
| Column No. | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 | 80 – 90 |
|---|---|---|---|---|---|---|---|---|
| I | ✓ | |||||||
| II | ✓ | ✓ | ||||||
| III | ✓ | ✓ | ||||||
| IV | ✓ | ✓ | ✓ | |||||
| V | ✓ | ✓ | ✓ | |||||
| VI | ✓ | ✓ | ✓ | |||||
| Total | 2 | 3 | 5 | 3 | 1 |
It is clear that the modal class is 50 – 60 (5 marks) and the frequency of this class is 23.
Using the formula, Mode (Z) = \( l_1 + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times i \), where \( l_1 \) = 50, \( f_1 \) = 23, \( f_0 \) = 15, \( f_2 \) = 22, i = 10
\( Z = 50 + \frac{23 - 15}{2 \times 23 - 15 - 22} \times 10 = 50 + \frac{8}{9} \times 10 \) = 58.89
Mode = 58.89
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Chapter 13 Statistics Concepts and Summary for Class 11 Mathematics
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