Get summaries, questions, answers, solutions, extra MCQs, PDF for Measurement of Central Tendencies: NBSE Class 12 Education, which is part of the syllabus for students studying under NBSE (Nagaland Board). These solutions, however, should only be treated as references and can be modified/changed.
Summary
When we have a lot of numbers or information, it can be hard to understand everything at once. A measure of central tendency helps by giving us one single number that represents the center or the typical value of the whole group. Think of it like describing a whole basket of apples by talking about one typical apple from the center. This makes it easier to grasp large amounts of data. It gives a “bird’s eye view,” which means you can see the general picture from above without needing to look at every single detail on the ground. This helps us compare different groups, like seeing which class did better on a test by comparing their average scores.
The most common measure is the mean, which is also known as the average. To find the mean, you add up all the values in a group and then divide that total by the number of values you added. For example, to find the mean of the numbers 2, 3, and 7, you add them together to get 12, and then divide by 3, which gives a mean of 4. The mean is useful because it uses every single number in the data. However, this also means that a very high or very low number can change the mean a lot.
Another measure is the median. The median is the middle value in a list of numbers that has been arranged in order from smallest to largest. If you have the numbers 1, 4, 6, 8, and 9, the median is 6 because it’s right in the middle. If the list has an even number of values, like 2, 4, 6, and 8, there are two middle numbers (4 and 6). In this case, the median is the mean of those two numbers, which would be 5. The median is a good measure to use when there are extreme values, because it is not affected by them.
The third measure is the mode. The mode is the value that appears most often in a set of data. In the list 2, 5, 5, 7, 8, the mode is 5 because it appears more than any other number. A group of data can have one mode, more than one mode, or no mode at all. The mode is especially helpful when you want to find out what is most popular or common, such as the most frequently sold shirt size in a store. Each of these three measures gives a different way to describe the center of the data.
Textual
Very Short Answer Questions
1. Find the mean of the following data:
(a) 9, 7, 11, 13, 2, 4, 5, 5
Answer: The data set is: 9, 7, 11, 13, 2, 4, 5, 5
Number of observations (N) = 8
Sum of observations (ΣX) = 9 + 7 + 11 + 13 + 2 + 4 + 5 + 5 = 56
Mean (X̄) = (Sum of observations) / (Number of observations)
Mean = ΣX / N
Mean = 56 / 8 = 7
The mean of the data is 7.
(b) 16, 18, 19, 21, 23, 23, 27, 29, 29, 35
Answer: The data set is: 16, 18, 19, 21, 23, 23, 27, 29, 29, 35
Number of observations (N) = 10
Sum of observations (ΣX) = 16 + 18 + 19 + 21 + 23 + 23 + 27 + 29 + 29 + 35 = 240
Mean (X̄) = (Sum of observations) / (Number of observations)
Mean = ΣX / N
Mean = 240 / 10 = 24
The mean of the data is 24.
(c) 2.2, 10.2, 14.7, 5.9, 4.9, 11.1, 10.5
Answer: The data set is: 2.2, 10.2, 14.7, 5.9, 4.9, 11.1, 10.5
Number of observations (N) = 7
Sum of observations (ΣX) = 2.2 + 10.2 + 14.7 + 5.9 + 4.9 + 11.1 + 10.5 = 59.5
Mean (X̄) = (Sum of observations) / (Number of observations)
Mean = ΣX / N
Mean = 59.5 / 7 = 8.5
The mean of the data is 8.5.
(d) 1 1/4, 2 1/2, 5 1/2, 3 1/4, 2 1/2
Answer: First, we will convert the mixed fractions into decimals to simplify the calculation.
- 1 1/4 = 1.25
- 2 1/2 = 2.5
- 5 1/2 = 5.5
- 3 1/4 = 3.25
- 2 1/2 = 2.5
The data set in decimal form is: 1.25, 2.5, 5.5, 3.25, 2.5
Number of observations (N) = 5
Sum of observations (ΣX) = 1.25 + 2.5 + 5.5 + 3.25 + 2.5 = 15.0
Mean (X̄) = (Sum of observations) / (Number of observations)
Mean = ΣX / N
Mean = 15.0 / 5 = 3.0
The mean of the data is 3.0.
2. Find the mean of the first ten whole numbers.
Answer: The first ten whole numbers are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Number of observations (N) = 10
Sum of observations (ΣX) = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Mean (X̄) = (Sum of observations) / (Number of observations)
Mean = ΣX / N
Mean = 45 / 10 = 4.5
The mean of the first ten whole numbers is 4.5.
3. Find the mean of the first 5 prime numbers.
Answer: The first 5 prime numbers are: 2, 3, 5, 7, 11
Number of observations (N) = 5
Sum of observations (ΣX) = 2 + 3 + 5 + 7 + 11 = 28
Mean (X̄) = (Sum of observations) / (Number of observations)
Mean = ΣX / N
Mean = 28 / 5 = 5.6
The mean of the first 5 prime numbers is 5.6.
4. The mean of 8, 11, 6, 14, X and 13 is 66. Find the value of the observation X.
Answer: The given observations are: 8, 11, 6, 14, X, 13
Number of observations (N) = 6
The given mean (X̄) = 66
Using the formula for the mean:
Mean = (Sum of observations) / (Number of observations)
66 = (8 + 11 + 6 + 14 + X + 13) / 6
66 = (52 + X) / 6
To find X, multiply both sides by 6:
66 × 6 = 52 + X
396 = 52 + X
Now, isolate X:
X = 396 – 52
X = 344
The value of the observation X is 344.
Short Answer Questions
1. Find the median of the following data:
(a) 27, 39, 49, 20, 21, 28, 38
Answer: Step 1: Arrange the data in ascending order.
| X |
| 20 |
| 21 |
| 27 |
| 28 |
| 38 |
| 39 |
| 49 |
Step 2: Find the number of observations (N). Here, N = 7, which is an odd number.
Step 3: The position of the median is calculated using the formula for an odd number series:
Median = size of ((N + 1) / 2)th item
Median = size of ((7 + 1) / 2)th item
Median = size of (8 / 2)th item = size of 4th item.
The value of the 4th item in the arranged series is 28.
Thus, the Median is 28.
(b) 10, 19, 54, 80, 15, 16
Answer: Step 1: Arrange the data in ascending order.
10, 15, 16, 19, 54, 80
Step 2: Find the number of observations (N). Here, N = 6, which is an even number.
Step 3: The median is the arithmetic mean of the two middle observations. The positions are:
(N / 2)th item = (6 / 2)th item = 3rd item
((N / 2) + 1)th item = (6 / 2 + 1)th item = 4th item
Step 4: The value of the 3rd item is 16, and the value of the 4th item is 19.
Median = (Value of 3rd item + Value of 4th item) / 2
Median = (16 + 19) / 2 = 35 / 2 = 17.5
Thus, the Median is 17.5.
(c) 47, 41, 52, 43, 56, 35, 49, 55, 42
Answer: Step 1: Arrange the data in ascending order.
| X |
| 35 |
| 41 |
| 42 |
| 43 |
| 47 |
| 49 |
| 52 |
| 55 |
| 56 |
Step 2: Find the number of observations (N). Here, N = 9, which is an odd number.
Step 3: The position of the median is:
Median = size of ((N + 1) / 2)th item
Median = size of ((9 + 1) / 2)th item
Median = size of (10 / 2)th item = size of 5th item.
The value of the 5th item in the arranged series is 47.
Thus, the Median is 47.
(d) 12, 17, 3, 14, 5, 8, 7, 15
Answer:
Step 1: Arrange the data in ascending order.
3, 5, 7, 8, 12, 14, 15, 17
Step 2: Find the number of observations (N). Here, N = 8, which is an even number.
Step 3: The median is the arithmetic mean of the two middle observations. The positions are:
(N / 2)th item = (8 / 2)th item = 4th item
((N / 2) + 1)th item = (8 / 2 + 1)th item = 5th item
Step 4: The value of the 4th item is 8, and the value of the 5th item is 12.
Median = (Value of 4th item + Value of 5th item) / 2
Median = (8 + 12) / 2 = 20 / 2 = 10
Thus, the Median is 10.
2. The following observations are arranged in ascending order. 17, x, 24, x + 7, 35, 36, 46. The median of the data is 25. Find the value of x.
Answer:
Step 1: The given observations are already arranged in ascending order: 17, x, 24, x + 7, 35, 36, 46.
Step 2: The number of observations (N) is 7, which is an odd number.
Step 3: The position of the median is the ((N + 1) / 2)th item.
Position = ((7 + 1) / 2)th item = 4th item.
Step 4: The value of the 4th item in the series is (x + 7).
Step 5: We are given that the median is 25. Therefore, we can set up the equation:
x + 7 = 25
Step 6: Solve for x.
x = 25 – 7
x = 18
Thus, the value of x is 18.
3. Find the mode of the following data:
(a) 12, 8, 4, 8, 1, 8, 9, 11, 9, 10, 12, 8
Answer: By inspecting the data, we count the frequency of each number:
- 8 appears 4 times.
- 9 appears 2 times.
- 12 appears 2 times.
- All other numbers appear once.
The number 8 occurs most frequently. Therefore, the Mode is 8.
(b) 15, 22, 17, 19, 22, 17, 29, 24, 17, 15
Answer: By inspecting the data, we count the frequency of each number:
- 17 appears 3 times.
- 15 appears 2 times.
- 22 appears 2 times.
- All other numbers appear once.
The number 17 occurs most frequently. Therefore, the Mode is 17.
(c) 0, 3, 2, 1, 3, 5, 4, 3, 42, 1, 2, 0
Answer: By inspecting the data, we count the frequency of each number:
- 3 appears 3 times.
- 0 appears 2 times.
- 1 appears 2 times.
- 2 appears 2 times.
- All other numbers appear once.
The number 3 occurs most frequently. Therefore, the Mode is 3.
(d) 1, 7, 2, 4, 5, 9, 8, 3
Answer: In this data set, every number appears only once. Since no value occurs more frequently than any other, there is no mode.
4. The weights in kg of 10 students are given below: 39, 43, 36, 38, 46, 51, 33, 44, 44, 43. Find the mode of this data. Is there more than 1 mode? If yes, why?
Answer: Step 1: Find the frequency of each weight by inspection.
- 43 appears 2 times.
- 44 appears 2 times.
- All other weights (33, 36, 38, 39, 46, 51) appear once.
Step 2: The highest frequency of occurrence is 2. The values with this frequency are 43 and 44.
Conclusion:
- The modes of this data are 43 and 44.
- Yes, there is more than one mode.
- This is because two different values, 43 and 44, share the same highest frequency. A series with two modes is called a bi-modal series.
5. Find the mode, median and mean of the following:
(a) 3, 12, 11, 7, 5, 5, 6, 4, 10
Answer:
- Mean:
Sum (ΣX) = 3 + 12 + 11 + 7 + 5 + 5 + 6 + 4 + 10 = 63
Number of items (N) = 9
Mean = ΣX / N = 63 / 9 = 7 - Median:
Arranged data: 3, 4, 5, 5, 6, 7, 10, 11, 12
N = 9 (odd). The median is the ((9+1)/2)th = 5th item.
Median = 6 - Mode:
The number 5 occurs twice, which is more than any other number.
Mode = 5
(b) 16, 19, 10, 24, 19
Answer:
- Mean:
Sum (ΣX) = 16 + 19 + 10 + 24 + 19 = 88
Number of items (N) = 5
Mean = ΣX / N = 88 / 5 = 17.6 - Median:
Arranged data: 10, 16, 19, 19, 24
N = 5 (odd). The median is the ((5+1)/2)th = 3rd item.
Median = 19 - Mode:
The number 19 occurs twice.
Mode = 19
(c) 8, 2, 8, 5, 5, 8
Answer:
- Mean:
Sum (ΣX) = 8 + 2 + 8 + 5 + 5 + 8 = 36
Number of items (N) = 6
Mean = ΣX / N = 36 / 6 = 6 - Median:
Arranged data: 2, 5, 5, 8, 8, 8
N = 6 (even). The median is the average of the 3rd and 4th items.
Median = (5 + 8) / 2 = 13 / 2 = 6.5 - Mode:
The number 8 occurs three times.
Mode = 8
(d) 28, 39, 42, 29, 39, 40, 36, 46, 41, 30
Answer:
- Mean:
Sum (ΣX) = 28+39+42+29+39+40+36+46+41+30 = 370
Number of items (N) = 10
Mean = ΣX / N = 370 / 10 = 37 - Median:
Arranged data: 28, 29, 30, 36, 39, 39, 40, 41, 42, 46
N = 10 (even). The median is the average of the 5th and 6th items.
Median = (39 + 39) / 2 = 39 - Mode:
The number 39 occurs twice.
Mode = 39
(e) 133, 215, 250, 108, 206, 159, 206, 178
Answer:
- Mean:
Sum (ΣX) = 133+215+250+108+206+159+206+178 = 1455
Number of items (N) = 8
Mean = ΣX / N = 1455 / 8 = 181.875 - Median:
Arranged data: 108, 133, 159, 178, 206, 206, 215, 250
N = 8 (even). The median is the average of the 4th and 5th items.
Median = (178 + 206) / 2 = 384 / 2 = 192 - Mode:
The number 206 occurs twice.
Mode = 206
(f) 76, 94, 76, 82, 78, 86, 90
Answer:
- Mean:
Sum (ΣX) = 76+94+76+82+78+86+90 = 582
Number of items (N) = 7
Mean = ΣX / N = 582 / 7 ≈ 83.14 - Median:
Arranged data: 76, 76, 78, 82, 86, 90, 94
N = 7 (odd). The median is the ((7+1)/2)th = 4th item.
Median = 82 - Mode:
The number 76 occurs twice.
Mode = 76
(g) 52, 61, 49, 52, 49, 52, 41, 58
Answer:
- Mean:
Sum (ΣX) = 52+61+49+52+49+52+41+58 = 414
Number of items (N) = 8
Mean = ΣX / N = 414 / 8 = 51.75 - Median:
Arranged data: 41, 49, 49, 52, 52, 52, 58, 61
N = 8 (even). The median is the average of the 4th and 5th items.
Median = (52 + 52) / 2 = 52 - Mode:
The number 52 occurs three times.
Mode = 52
Long Answer Questions
1. If the arithmetic mean of the data given below is 28, find (a) the missing frequency, and (b) the median of the series:
| Profit per retail shop (in ₹) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Number of retail shops | 12 | 18 | 27 | ? | 17 | 6 |
Answer: (a) Finding the missing frequency:
Let the missing frequency for the class interval 30-40 be ‘f’. We can find it using the arithmetic mean formula for a continuous series.
Table for Calculation of Mean
| Profit per retail shop | Mid-value (m) | Number of shops (f) | fm |
| 0-10 | 5 | 12 | 60 |
| 10-20 | 15 | 18 | 270 |
| 20-30 | 25 | 27 | 675 |
| 30-40 | 35 | f | 35f |
| 40-50 | 45 | 17 | 765 |
| 50-60 | 55 | 6 | 330 |
| Total | Σf = 80 + f | Σfm = 2100 + 35f |
Given, Arithmetic Mean (X̄) = 28.
The formula for the mean is: X̄ = Σfm / Σf
Substituting the values from the table:
28 = (2100 + 35f) / (80 + f)
28 * (80 + f) = 2100 + 35f
2240 + 28f = 2100 + 35f
2240 – 2100 = 35f – 28f
140 = 7f
f = 140 / 7
f = 20
Thus, the missing frequency is 20.
(b) Finding the median of the series:
Now that we have the complete frequency distribution, we can calculate the median.
Table for Calculation of Median
| Profit per retail shop | Number of shops (f) | Cumulative Frequency (cf) |
| 0-10 | 12 | 12 |
| 10-20 | 18 | 30 |
| 20-30 | 27 | 57 |
| 30-40 | 20 | 77 |
| 40-50 | 17 | 94 |
| 50-60 | 6 | 100 |
| Total | N = Σf = 100 |
Step 1: Find the median item.
Median item = N / 2 = 100 / 2 = 50th item.
Step 2: Locate the median class.
The 50th item lies in the class interval where the cumulative frequency is just greater than 50. This is the class 20-30, which has a cumulative frequency of 57.
So, the Median Class is 20-30.
Step 3: Apply the median formula.
Median = L + [ (N/2 – c.f.) / f ] * i
Where:
L = Lower limit of the median class = 20
N = Total frequency = 100
c.f. = Cumulative frequency of the class preceding the median class = 30
f = Frequency of the median class = 27
i = Class interval width = 10
Median = 20 + [ (50 – 30) / 27 ] * 10
Median = 20 + [ 20 / 27 ] * 10
Median = 20 + (200 / 27)
Median = 20 + 7.407
Median = 27.407 (approx. 27.41)
2. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.
| Workers | A | B | C | D | E | F | G | H | I | J |
| Daily Income (in ₹) | 120 | 150 | 180 | 200 | 250 | 300 | 220 | 350 | 370 | 260 |
Answer: This is a case of ungrouped data. The arithmetic mean is calculated by summing all the observations and dividing by the number of observations.
Formula: Mean (X̄) = ΣX / N
Step 1: Sum of all observations (ΣX).
ΣX = 120 + 150 + 180 + 200 + 250 + 300 + 220 + 350 + 370 + 260
ΣX = 2400
Step 2: Number of observations (N).
N = 10
Step 3: Calculate the mean.
Mean (X̄) = 2400 / 10 = 240
3. The following information pertains to the daily income of 150 families. Calculate the arithmetic mean.
| Income (in ₹) | Number of families |
| More than 75 | 150 |
| More than 85 | 140 |
| More than 95 | 115 |
| More than 105 | 95 |
| More than 115 | 70 |
| More than 125 | 60 |
| More than 135 | 40 |
| More than 145 | 25 |
Answer: First, we need to convert the ‘more than’ cumulative frequency distribution into a simple frequency distribution with class intervals.
Table for Conversion and Mean Calculation
| Class Interval | Mid-value (m) | Number of families (f) | fm |
| 75-85 | 80 | 150 – 140 = 10 | 800 |
| 85-95 | 90 | 140 – 115 = 25 | 2250 |
| 95-105 | 100 | 115 – 95 = 20 | 2000 |
| 105-115 | 110 | 95 – 70 = 25 | 2750 |
| 115-125 | 120 | 70 – 60 = 10 | 1200 |
| 125-135 | 130 | 60 – 40 = 20 | 2600 |
| 135-145 | 140 | 40 – 25 = 15 | 2100 |
| 145-155 | 150 | 25 | 3750 |
| Total | Σf = 150 | Σfm = 17450 |
Now, we calculate the arithmetic mean using the formula:
Mean (X̄) = Σfm / Σf
Mean (X̄) = 17450 / 150
Mean (X̄) = 116.33
4. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.
| Size of Land Holdings (in acres) | Less than 100 | 100-200 | 200-300 | 300-400 | 400 and above |
| Number of families | 40 | 89 | 148 | 64 | 39 |
Answer: We need to create a table with cumulative frequencies to find the median.
Table for Calculation of Median
| Size of Land Holdings (acres) | Number of families (f) | Cumulative Frequency (cf) |
| 0-100 | 40 | 40 |
| 100-200 | 89 | 129 |
| 200-300 | 148 | 277 |
| 300-400 | 64 | 341 |
| 400-500* | 39 | 380 |
| Total | N = Σf = 380 |
*Assuming the class width of the last interval is the same as the others.
Step 1: Find the median item.
Median item = N / 2 = 380 / 2 = 190th item.
Step 2: Locate the median class.
The 190th item lies in the class interval 200-300, as its cumulative frequency (277) is the first to be greater than 190.
Step 3: Apply the median formula.
Median = L + [ (N/2 – c.f.) / f ] * i
Where:
L = 200
N = 380
c.f. = 129 (cumulative frequency of the preceding class)
f = 148 (frequency of the median class)
i = 100 (class interval width)
Median = 200 + [ (190 – 129) / 148 ] * 100
Median = 200 + [ 61 / 148 ] * 100
Median = 200 + (6100 / 148)
Median = 200 + 41.216
Median = 241.216 (approx. 241.22)
5. The following table gives production yield in kg per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.
| Production yield (kg per hectare) | 50-53 | 53-56 | 56-59 | 59-62 | 62-65 | 65-68 | 68-71 | 71-74 | 74-77 |
| Number of farms | 3 | 8 | 14 | 30 | 36 | 28 | 16 | 10 | 5 |
Answer:
(i) Calculation of Mean
| Class Interval | Mid-value (m) | Frequency (f) | fm |
| 50-53 | 51.5 | 3 | 154.5 |
| 53-56 | 54.5 | 8 | 436.0 |
| 56-59 | 57.5 | 14 | 805.0 |
| 59-62 | 60.5 | 30 | 1815.0 |
| 62-65 | 63.5 | 36 | 2286.0 |
| 65-68 | 66.5 | 28 | 1862.0 |
| 68-71 | 69.5 | 16 | 1112.0 |
| 71-74 | 72.5 | 10 | 725.0 |
| 74-77 | 75.5 | 5 | 377.5 |
| Total | Σf = 150 | Σfm = 9573.0 |
Mean (X̄) = Σfm / Σf = 9573 / 150 = 63.82
(ii) Calculation of Median
| Class Interval | Frequency (f) | Cumulative Frequency (cf) |
| 50-53 | 3 | 3 |
| 53-56 | 8 | 11 |
| 56-59 | 14 | 25 |
| 59-62 | 30 | 55 |
| 62-65 | 36 | 91 |
| 65-68 | 28 | 119 |
| 68-71 | 16 | 135 |
| 71-74 | 10 | 145 |
| 74-77 | 5 | 150 |
| Total | N = 150 |
Median item = N / 2 = 150 / 2 = 75th item.
This lies in the class interval 62-65.
Median = L + [ (N/2 – c.f.) / f ] * i
Median = 62 + [ (75 – 55) / 36 ] * 3
Median = 62 + [ 20 / 36 ] * 3
Median = 62 + (60 / 36) = 62 + 1.67 = 63.67
(iii) Calculation of Mode
By inspection, the highest frequency is 36, which corresponds to the class interval 62-65. This is the modal class.
Mode = l1 + [ (f1 – f0) / (2f1 – f0 – f2) ] * i
Where:
l1 = 62
f1 (frequency of modal class) = 36
f0 (frequency of preceding class) = 30
f2 (frequency of succeeding class) = 28
i = 3
Mode = 62 + [ (36 – 30) / (2*36 – 30 – 28) ] * 3
Mode = 62 + [ 6 / (72 – 58) ] * 3
Mode = 62 + [ 6 / 14 ] * 3
Mode = 62 + (18 / 14) = 62 + 1.2857 = 63.29
6. Compute the mean of the following frequency distribution using step deviation method.
| Class-interval | 0-11 | 11-22 | 22-33 | 33-44 | 44-55 | 55-66 |
| Frequency | 9 | 17 | 28 | 26 | 15 | 8 |
Answer:
Table for Mean Calculation by Step-Deviation Method
Let Assumed Mean (A) = 38.5. Common factor (C) = 11.
| Class-interval | Frequency (f) | Mid-value (m) | d = m – A | d’ = d/C | fd’ |
| 0-11 | 9 | 5.5 | -33 | -3 | -27 |
| 11-22 | 17 | 16.5 | -22 | -2 | -34 |
| 22-33 | 28 | 27.5 | -11 | -1 | -28 |
| 33-44 | 26 | 38.5 | 0 | 0 | 0 |
| 44-55 | 15 | 49.5 | 11 | 1 | 15 |
| 55-66 | 8 | 60.5 | 22 | 2 | 16 |
| Total | Σf = 103 | Σfd’ = -58 |
Formula: Mean (X̄) = A + (Σfd’ / Σf) * C
Mean (X̄) = 38.5 + (-58 / 103) * 11
Mean (X̄) = 38.5 – (638 / 103)
Mean (X̄) = 38.5 – 6.194
Mean (X̄) = 32.306
(Ans. The mean of the frequency distribution is 32.306)
7. Calculate the mean from the following data:
| Class-interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 2 | 18 | 30 | 40 | 35 | 20 | 6 | 3 |
Answer:
Table for Mean Calculation by Direct Method
| Class-interval | Mid-value (m) | Frequency (f) | fm |
| 0-10 | 5 | 2 | 10 |
| 10-20 | 15 | 18 | 270 |
| 20-30 | 25 | 30 | 750 |
| 30-40 | 35 | 40 | 1400 |
| 40-50 | 45 | 35 | 1575 |
| 50-60 | 55 | 20 | 1100 |
| 60-70 | 65 | 6 | 390 |
| 70-80 | 75 | 3 | 225 |
| Total | Σf = 154 | Σfm = 5720 |
Mean (X̄) = Σfm / Σf
Mean (X̄) = 5720 / 154
Mean (X̄) = 37.142 (approx. 37.15)
8. Calculate Mean, Median and Mode from the following data.
| Class-Interval | 150-159 | 140-149 | 130-139 | 120-129 | 110-119 | 100-109 | 90-99 | 80-89 | 70-79 | 60-69 | 50-59 |
| Frequency | 2 | 2 | 4 | 1 | 5 | 5 | 12 | 10 | 12 | 10 | 1 |
Answer: First, we must convert the inclusive series into an exclusive series and arrange it in ascending order. The correction factor is (60-59)/2 = 0.5.
Master Table for Calculations
| Original Class | Exclusive Class | Mid-value (m) | Frequency (f) | fm | Cumulative Frequency (cf) |
| 50-59 | 49.5-59.5 | 54.5 | 1 | 54.5 | 1 |
| 60-69 | 59.5-69.5 | 64.5 | 10 | 645.0 | 11 |
| 70-79 | 69.5-79.5 | 74.5 | 12 | 894.0 | 23 |
| 80-89 | 79.5-89.5 | 84.5 | 10 | 845.0 | 33 |
| 90-99 | 89.5-99.5 | 94.5 | 12 | 1134.0 | 45 |
| 100-109 | 99.5-109.5 | 104.5 | 5 | 522.5 | 50 |
| 110-119 | 109.5-119.5 | 114.5 | 5 | 572.5 | 55 |
| 120-129 | 119.5-129.5 | 124.5 | 1 | 124.5 | 56 |
| 130-139 | 129.5-139.5 | 134.5 | 4 | 538.0 | 60 |
| 140-149 | 139.5-149.5 | 144.5 | 2 | 289.0 | 62 |
| 150-159 | 149.5-159.5 | 154.5 | 2 | 309.0 | 64 |
| Total | Σf = 64 | Σfm = 5928.0 |
(i) Mean
Mean (X̄) = Σfm / Σf = 5928 / 64 = 92.625
(ii) Median
Median item = N / 2 = 64 / 2 = 32nd item.
This lies in the class 80-89 (exclusive 79.5-89.5), as its cf (33) is just greater than 32.
Median = L + [ (N/2 – c.f.) / f ] * i
Median = 79.5 + [ (32 – 23) / 10 ] * 10
Median = 79.5 + (9 / 10) * 10
Median = 79.5 + 9 = 88.5
(iii) Mode
The distribution has two classes (70-79 and 90-99) with the maximum frequency of 12. This is a bimodal distribution. To find the single mode, we use the empirical relationship: Mode = 3 Median – 2 Mean.
Mode = 3 * (88.5) – 2 * (92.625)
Mode = 265.5 – 185.25
Mode = 80.25
9. Data of exports of a certain firms for the year 2018 – 2019 are mentioned in the following table: Compare average value of exports for these firms using Assumed Mean Method.
| Firms | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Value of Exports (₹ Cr) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 |
Answer: Table for Mean Calculation by Assumed Mean Method
Let Assumed Mean (A) = 50.
| Firms | Value of Exports (X) | d = X – A |
| 1 | 10 | -40 |
| 2 | 20 | -30 |
| 3 | 30 | -20 |
| 4 | 40 | -10 |
| 5 | 50 | 0 |
| 6 | 60 | 10 |
| 7 | 70 | 20 |
| 8 | 80 | 30 |
| 9 | 90 | 40 |
| Total | N = 9 | Σd = 0 |
Formula: Mean (X̄) = A + (Σd / N)
Mean (X̄) = 50 + (0 / 9)
Mean (X̄) = 50 + 0 = 50
10. From the following data of the marks obtained by 60 students of a class. Calculate the arithmetic mean by (i) Direct Method, (ii) Assumed Mean Method, and (iii) Step-Deviation Method.
| Marks | 20 | 30 | 40 | 50 | 60 | 70 |
| No. of Students | 8 | 12 | 20 | 10 | 6 | 4 |
Answer:
(i) Direct Method
| 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 |
| Total | Σf = 60 | Σfx = 2460 |
Mean (X̄) = Σfx / Σf = 2460 / 60 = 41
(ii) Assumed Mean Method
Let Assumed Mean (A) = 40.
| Marks (x) | Frequency (f) | d = x – A | fd |
| 20 | 8 | -20 | -160 |
| 30 | 12 | -10 | -120 |
| 40 | 20 | 0 | 0 |
| 50 | 10 | 10 | 100 |
| 60 | 6 | 20 | 120 |
| 70 | 4 | 30 | 120 |
| Total | Σf = 60 | Σfd = 60 |
Mean (X̄) = A + (Σfd / Σf) = 40 + (60 / 60) = 40 + 1 = 41
(iii) Step-Deviation Method
Let Assumed Mean (A) = 40, Common factor (C) = 10.
| Marks (x) | Frequency (f) | d = x – A | d’ = d/C | fd’ |
| 20 | 8 | -20 | -2 | -16 |
| 30 | 12 | -10 | -1 | -12 |
| 40 | 20 | 0 | 0 | 0 |
| 50 | 10 | 10 | 1 | 10 |
| 60 | 6 | 20 | 2 | 12 |
| 70 | 4 | 30 | 3 | 12 |
| Total | Σf = 60 | Σfd’ = 6 |
Mean (X̄) = A + (Σfd’ / Σf) * C = 40 + (6 / 60) * 10 = 40 + (1/10) * 10 = 40 + 1 = 41
11. For the two frequency distributions given below, the mean calculated from the first was 25.4 and that from the second was 32.5. Find the values of x and y.
| Class-Interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Distribution I | 20 | 15 | 10 | x | y |
| Distribution II | 4 | 8 | 4 | 2x | Y |
Assuming ‘Y’ in Distribution II is a typo for ‘y’.
Answer:
For Distribution I (Mean = 25.4):
Mid-values (m): 15, 25, 35, 45, 55
Σf₁ = 20 + 15 + 10 + x + y = 45 + x + y
Σfm₁ = (2015) + (1525) + (1035) + (x45) + (y*55) = 300 + 375 + 350 + 45x + 55y = 1025 + 45x + 55y
Mean = Σfm₁ / Σf₁
25.4 = (1025 + 45x + 55y) / (45 + x + y)
25.4(45 + x + y) = 1025 + 45x + 55y
1143 + 25.4x + 25.4y = 1025 + 45x + 55y
1143 – 1025 = 45x – 25.4x + 55y – 25.4y
118 = 19.6x + 29.6y — (Equation 1)
For Distribution II (Mean = 32.5):
Σf₂ = 4 + 8 + 4 + 2x + y = 16 + 2x + y
Σfm₂ = (415) + (825) + (435) + (2x45) + (y*55) = 60 + 200 + 140 + 90x + 55y = 400 + 90x + 55y
Mean = Σfm₂ / Σf₂
32.5 = (400 + 90x + 55y) / (16 + 2x + y)
32.5(16 + 2x + y) = 400 + 90x + 55y
520 + 65x + 32.5y = 400 + 90x + 55y
520 – 400 = 90x – 65x + 55y – 32.5y
120 = 25x + 22.5y — (Equation 2)
Now we solve the system of equations:
- 19.6x + 29.6y = 118
- 25x + 22.5y = 120
From (2), 25x = 120 – 22.5y => x = (120 – 22.5y) / 25 = 4.8 – 0.9y
Substitute x into (1):
19.6(4.8 – 0.9y) + 29.6y = 118
94.08 – 17.64y + 29.6y = 118
11.96y = 118 – 94.08
11.96y = 23.92
y = 23.92 / 11.96 = 2
Now find x:
x = 4.8 – 0.9y = 4.8 – 0.9(2) = 4.8 – 1.8 = 3
12. Determine the modal value in the following series:
| Value | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 | 26 | 28 | 30 | 32 |
| Frequency | 7 | 15 | 21 | 38 | 34 | 34 | 11 | 19 | 10 | 38 | 5 | 2 |
Answer:
The mode is the value that occurs with the highest frequency.
In this series, the highest frequency is 38.
This frequency occurs for two different values: 16 and 28.
Therefore, the series is bimodal.
(Ans. The modal values are 16 and 28)
13. The median and mode of the following wage distribution are known to be ₹33.5 and ₹34 respectively. Three frequency values from the table are however missing. Find the missing values.
| Wages in ₹ | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
| Frequencies | 10 | 10 | ? | ? | ? | 6 | 4 | 230 |
Answer:
Let the missing frequencies for 20-30, 30-40, and 40-50 be f₁, f₂, and f₃ respectively.
Total Frequency N = 230.
So, 10 + 10 + f₁ + f₂ + f₃ + 6 + 4 = 230
30 + f₁ + f₂ + f₃ = 230
f₁ + f₂ + f₃ = 200 — (Equation 1)
Using Mode Information:
Mode = 34, which lies in the 30-40 class. So, 30-40 is the modal class.
Mode = l1 + [ (f₂ – f₁) / (2f₂ – f₁ – f₃) ] * i
34 = 30 + [ (f₂ – f₁) / (2f₂ – f₁ – f₃) ] * 10
4 = 10 * (f₂ – f₁) / (2f₂ – f₁ – f₃)
0.4 = (f₂ – f₁) / (2f₂ – f₁ – f₃)
0.4(2f₂ – f₁ – f₃) = f₂ – f₁
0.8f₂ – 0.4f₁ – 0.4f₃ = f₂ – f₁
0.6f₁ – 0.2f₂ – 0.4f₃ = 0
Multiply by 5: 3f₁ – f₂ – 2f₃ = 0 — (Equation 2)
Using Median Information:
Median = 33.5, which also lies in the 30-40 class. This is the median class.
N/2 = 230/2 = 115.
Cumulative frequency (c.f.) of the class preceding the median class = 10 + 10 + f₁ = 20 + f₁.
Median = L + [ (N/2 – c.f.) / f ] * i
33.5 = 30 + [ (115 – (20 + f₁)) / f₂ ] * 10
3.5 = [ (95 – f₁) / f₂ ] * 10
0.35 = (95 – f₁) / f₂
0.35f₂ = 95 – f₁
f₁ + 0.35f₂ = 95 — (Equation 3)
Solving the three equations:
From (3): f₁ = 95 – 0.35f₂
From (1): f₃ = 200 – f₁ – f₂ = 200 – (95 – 0.35f₂) – f₂ = 105 – 0.65f₂
Substitute f₁ and f₃ into (2):
3(95 – 0.35f₂) – f₂ – 2(105 – 0.65f₂) = 0
285 – 1.05f₂ – f₂ – 210 + 1.3f₂ = 0
75 – 0.75f₂ = 0
0.75f₂ = 75
f₂ = 100
Now find f₁ and f₃:
f₁ = 95 – 0.35(100) = 95 – 35 = 60
f₃ = 105 – 0.65(100) = 105 – 65 = 40
The missing frequencies are 60, 100, and 40.
Additional
Extra Questions and Answers
1. What are measures of central tendency?
Answer: Measures of central tendency are single values that give some idea of the trend of a group of data taken as a whole. A score that indicates where the centre of the distribution tends to be located is called the measure of central tendency, which is a central value of the items in the series.
2. What is the purpose of summarising a set of data by a single number?
Answer: The purpose of summarising a set of data by a single number, known as a measure of central tendency, is to get a bird’s eye view, or a general overview, of a huge mass of statistical data. These measures are devices that make the human mind capable of grasping the true significance of large amounts of facts and measurements. They set aside unnecessary details and present a concise picture of the complex information being studied.
3. What does a score that indicates the centre of a distribution tend to be called?
Answer: A score that indicates where the centre of the distribution tends to be located is called the measure of central tendency.
4. Name the three most common measures of central tendency.
Answer: The three most common measures of central tendency are:
- Mean
- Median
- Mode
5. What is a measure of central tendency according to the provided quote?
Answer: A measure of central tendency is a typical value around which other figures congregate.
44. What are the characteristics of a good average?
Answer: The characteristics of a good average are:
- An average should be rigidly defined. If it is, there will be no instability in its value, and it would always be a definite figure.
- An average should be based on all the items or observations of a series.
- An average should be easily comprehended. It should not be abstract.
- An average should be calculable with reasonable ease and rapidity.
- The effects of fluctuations of a sample on an average should be as little as possible.
- An ideal average should be able to be located graphically so that one can know the average just by looking at the graph.
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