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The Statistics is an essential branch of mathematics that deals with the collection, organization, analysis, interpretation and presentation of data. In Class 10, Chapter 13 of the NCERT Mathematics textbook covers the various statistical methods such as the mean, median, mode and graphical representation of the data. Exercise 13.1 focuses on the problems that involve calculating measures of the central tendency including mean, median and mode based on the provided datasets.
The Statistics is a field that helps in understanding and interpreting data by using the mathematical concepts. It involves collecting, analyzing and summarizing large sets of information to the make meaningful conclusions. The methods used in statistics allow us to the represent data in an organized way and find patterns or trends that help in decision-making processes.
Number of Plants | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 |
Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.
Step 3: Now we will apply the general formula to calculate the mean
Now, Let's see the detailed solution:
No.of Plants (Class Interval) | No. of Houses (Frequency) (fi) | Class Mark (xi) | fi * xi |
|---|---|---|---|
0-2 | 1 | 1 | 1 |
2-4 | 2 | 3 | 6 |
4-6 | 1 | 5 | 5 |
6-8 | 5 | 7 | 35 |
8-10 | 6 | 9 | 54 |
10-12 | 2 | 11 | 22 |
12-14 | 3 | 13 | 39 |
Sum: ∑ fi = 20 | Sum: ∑ fixi = 162 |
Now, after creating this table we will be able to find the mean very easily -
= 16
= 8.1
Hence, we come to the conclusion that the number of plants per house is 8.1. Since the numeral value of frequency(fi) and the class mark(xi) is small so we use DIRECT METHOD to find the mean number of plants per house.
Daily Wages (in ₹) | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
Number of Workers | 12 | 14 | 8 | 6 | 10 |
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.
ui = (xi – A)/h
=> ui = (xi – 150)/20
Step 3: Now we will apply the Assumed Mean Formula to calculate the mean
Now, Let's see the detailed solution:
Daily wages (Class interval) | Number of workers frequency (fi) | Mid-point (xi) | ui = (xi – 150)/20 | fiui |
|---|---|---|---|---|
100-120 | 12 | 110 | -2 | -24 |
120-140 | 14 | 130 | -1 | -14 |
140-160 | 8 | 150 | 0 | 0 |
160-180 | 6 | 170 | 1 | 6 |
180-200 | 10 | 190 | 2 | 20 |
Total | Sum ∑fi = 50 | Sum ∑fiui = -12 |
So, the formula to find out the mean is:
Mean =
= 150 + (20 × -12/50)
= 150 – 4.8
= 145.20
Thus, mean daily wage of the workers = Rs. 145.20.
Daily pocket allowance (in ₹) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. As a certain frequency is missing and we have an odd number of class intervals hence, we will assume the middle-Class Mark as our Assumed Mean(A).
Step 3: Now we will apply the general formula to calculate the mean
Now, Let's see the detailed solution:
Class interval | Number of children (fi) | Mid-point (xi) | fixi |
|---|---|---|---|
11-13 | 7 | 12 | 84 |
13-15 | 6 | 14 | 84 |
15-17 | 9 | 16 | 144 |
17-19 | 13 | 18 = A | 234 |
19-21 | f | 20 | 20f |
21-23 | 5 | 22 | 110 |
23-25 | 4 | 24 | 96 |
Total | ∑ fi = 44 + f | Sum ∑fixi = 752 + 20f |
The mean formula is
Mean =
= (752 + 20f)/(44 + f)
Now substitute the values and equate to find the missing frequency (f)
⇒ 18 = (752 + 20f)/(44 + f)
⇒ 18(44 + f) = (752 + 20f)
⇒ 792 + 18f = 752 + 20f
⇒ 792 + 18f = 752 + 20f
⇒ 792 – 752 = 20f – 18f
⇒ 40 = 2f
⇒ f = 20
So, the missing frequency, f = 20.
Number of heartbeats per minute | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86 |
Number of Women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 75.5 and class size is h = 3.
di = (xi – A)
=> di = (xi – 75.5)
Step 3: Now we will apply the Assumed Mean Formula to calculate the mean
Now, Let's see the detailed solution:
Class Interval | Number of women (fi) | Mid-point (xi) | di = (xi – 75.5) | fidi |
|---|---|---|---|---|
65-68 | 2 | 66.5 | -9 | -18 |
68-71 | 4 | 69.5 | -6 | -24 |
71-74 | 3 | 72.5 | -3 | -9 |
74-77 | 8 | 75.5 = A | 0 | 0 |
77-80 | 7 | 78.5 | 3 | 21 |
80-83 | 4 | 81.5 | 6 | 24 |
83-86 | 2 | 84.5 | 9 | 18 |
Sum ∑fi = 30 | Sum ∑fiui = 12 |
Mean =
= 75.5 + (12/30)
= 75.5 + 2/5
= 75.5 + 0.4
= 75.9
Therefore, the mean heartbeats per minute for these women is 75.9
Number of Mangoes | 50-52 | 53-55 | 56-58 | 59-61 | 62-64 |
Number of Boxes | 15 | 110 | 135 | 115 | 25 |
Solution:
Step 1: In the above table we find that the class intervals are not continuous and hence to make them a continuous set of data we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals is 1. Then find the Mid Point by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, let us assume the mean value, A = 57 and class size is h = 3.
Step 3: Since the frequency values are big, hence we are using the STEP-DEVIATION METHOD.
Now, Lets see the detailed solution:
Class Interval | Number of boxes (fi) | Mid-point (xi) | di = xi – A | ui=(xi – A)/h | fiui |
|---|---|---|---|---|---|
49.5-52.5 | 15 | 51 | -6 | -2 | -30 |
52.5-55.5 | 110 | 54 | -3 | -1 | -110 |
55.5-58.5 | 135 | 57 =A | 0 | 0 | 0 |
58.5-61.5 | 115 | 60 | 3 | 1 | 115 |
61.5-64.5 | 25 | 63 | 6 | 2 | 50 |
Sum ∑fi = 400 | Sum ∑fiui = 25 |
Mean =
= 57 + 3 * (25/400)
= 57 + 0.1875
= 57.19
Therefore, the mean number of mangoes kept in a packing box is 57.19
Daily Expenditure (in ₹) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
Number of Households | 4 | 5 | 12 | 2 | 2 |
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 225 and class size is h = 50.
di = (xi – A)
=> di = (xi – 225)
ui = (xi – A)/h
=> ui = (xi – 225)/50
Step 3: Now we will apply the Step Deviation Formula to calculate the mean
Now, Let's see the detailed solution:
Class Interval | Number of households (fi) | Mid-point (xi) | di = xi – A | ui = di/50 | fiui |
|---|---|---|---|---|---|
100-150 | 4 | 125 | -100 | -2 | -8 |
150-200 | 5 | 175 | -50 | -1 | -5 |
200-250 | 12 | 225 = A | 0 | 0 | 0 |
250-300 | 2 | 275 | 50 | 1 | 2 |
300-350 | 2 | 325 | 100 | 2 | 4 |
Sum ∑fi = 25 | Sum ∑fiui = -7 |
Mean =
= 225 + 50 (-7/25)
= 225 - 14
= 211
Therefore, the mean daily expenditure on food is ₹211
Concentration of SO2 (in ppm) | Frequency |
|---|---|
0.00-0.04 | 4 |
0.04-0.08 | 9 |
0.08-0.12 | 9 |
0.12-0.16 | 2 |
0.16-0.20 | 4 |
0.20-0.24 | 2 |
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.
Step 3: Now we will apply the general formula to calculate the mean
Now, Let's see the detailed solution:
Concentration of SO2 (in ppm) | Frequency (fi) | Mid-point (xi) | fixi |
|---|---|---|---|
0.00-0.04 | 4 | 0.02 | 0.08 |
0.04-0.08 | 9 | 0.06 | 0.54 |
0.08-0.12 | 9 | 0.10 | 0.90 |
0.12-0.16 | 2 | 0.14 | 0.28 |
0.16-0.20 | 4 | 0.18 | 0.72 |
0.20-0.24 | 2 | 0.22 | 0.44 |
Sum ∑fi = 30 | Sum ∑fixi = 2.96 |
The formula to find out the mean is
Mean =
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO2 in the air is 0.099 ppm.
Number of Days | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |
Number of Students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.
Step 3: Now we will apply the general formula to calculate the mean
Now, Let's see the detailed solution:
Class Interval | Frequency (fi) | Mid-point (xi) | fixi |
|---|---|---|---|
0-6 | 11 | 3 | 33 |
6-10 | 10 | 8 | 80 |
10-14 | 7 | 12 | 84 |
14-20 | 4 | 17 | 68 |
20-28 | 4 | 24 | 96 |
28-38 | 3 | 33 | 99 |
38-40 | 1 | 39 | 39 |
Sum ∑fi = 40 | Sum ∑fixi = 499 |
The mean formula is,
Mean =
= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.
Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |
Number of Cities | 3 | 10 | 11 | 8 | 3 |
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class size is h = 10.
di = (xi – A)
=> di = (xi – 70)
ui = (xi – A)/h
=> ui = (xi – 70)/10
Step 3: Now we will apply the Step Deviation Formula to calculate the mean
Now, Let's see the detailed solution:
Class Interval | Frequency (fi) | Class Mark(xi) | di = xi – a | ui = di/h | fiui |
|---|---|---|---|---|---|
45-55 | 3 | 50 | -20 | -2 | -6 |
55-65 | 10 | 60 | -10 | -1 | -10 |
65-75 | 11 | 70 = A | 0 | 0 | 0 |
75-85 | 8 | 80 | 10 | 1 | 8 |
85-95 | 3 | 90 | 20 | 2 | 6 |
Sum ∑fi = 35 | Sum ∑fiui = -2 |
So,
Mean =
= 70 + (-2/35) × 10
= 69.42
Therefore, the mean literacy rate = 69.42%.
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Exercise 13.1 of Chapter 13 is focused on building the foundational knowledge of the calculating measures of the central tendency such as mean, median and mode. These calculations allow students to the better understand how data can be summarized to the represent the overall characteristics of the dataset. Mastering these concepts will aid in higher-level statistical studies and help in the practical situations where data analysis is required.