RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3: The mean deviation of a grouped or continuous frequency distribution is the focus of this exercise. Experts at Kopykitab have solved the exercise-by-exercise problems using shortcut approaches to help students do well on their board exams. These RD Sharma Class 11 Maths Solutions assist students in resolving any questions they may have concerning the concepts covered in this topic. Students can download the RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3 in pdf format based on their needs and begin practising on their own schedule.
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Download RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.3 Free PDF
RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3
Access answers to RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3- Important Question with Answers
1. Compute the mean deviation from the median of the following distribution:
Class |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
Frequency |
5 |
10 |
20 |
5 |
10 |
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
Median is the middle term of the X_{i},
Here, the middle term is 25
So, Median = 25
Class Interval |
x_{i} |
f_{i} |
Cumulative Frequency |
|d_{i}| = |x_{i} – M| |
f_{i} |d_{i}| |
0-10 |
5 |
5 |
5 |
20 |
100 |
10-20 |
15 |
10 |
15 |
10 |
100 |
20-30 |
25 |
20 |
35 |
0 |
0 |
30-40 |
35 |
5 |
91 |
10 |
50 |
40-50 |
45 |
10 |
101 |
20 |
200 |
Total = 50 |
Total = 450 |
\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)
= 1/50 × 450
= 9
∴ The mean deviation is 9
2. Find the mean deviation from the mean for the following data:
i
Classes |
0-100 |
100-200 |
200-300 |
300-400 |
400-500 |
500-600 |
600-700 |
700-800 |
Frequencies |
4 |
8 |
9 |
10 |
7 |
5 |
4 |
3 |
Solution:
To find the mean deviation from the mean, firstly let us calculate the mean.
By using the formula,
= 17900/50
= 358
Class Interval |
x_{i} |
f_{i} |
Cumulative Frequency |
|d_{i}| = |x_{i} – M| |
f_{i} |d_{i}| |
0-100 |
50 |
4 |
200 |
308 |
1232 |
100-200 |
150 |
8 |
1200 |
208 |
1664 |
200-300 |
250 |
9 |
2250 |
108 |
972 |
300-400 |
350 |
10 |
3500 |
8 |
80 |
400-500 |
450 |
7 |
3150 |
92 |
644 |
500-600 |
550 |
5 |
2750 |
192 |
960 |
600-700 |
650 |
4 |
2600 |
292 |
1168 |
700-800 |
750 |
3 |
2250 |
392 |
1176 |
Total = 50 |
Total = 17900 |
Total = 7896 |
N = 50
\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)
= 1/50 × 7896
= 157.92
∴ The mean deviation is 157.92
ii
Classes |
95-105 |
105-115 |
115-125 |
125-135 |
135-145 |
145-155 |
Frequencies |
9 |
13 |
16 |
26 |
30 |
12 |
Solution:
To find the mean deviation from the mean, firstly let us calculate the mean.
By using the formula,
= 13630/106
= 128.58
Class Interval |
x_{i} |
f_{i} |
Cumulative Frequency |
|d_{i}| = |x_{i} – M| |
f_{i} |d_{i}| |
95-105 |
100 |
9 |
900 |
28.58 |
257.22 |
105-115 |
110 |
13 |
1430 |
18.58 |
241.54 |
115-125 |
120 |
16 |
1920 |
8.58 |
137.28 |
125-135 |
130 |
26 |
3380 |
1.42 |
36.92 |
135-145 |
140 |
30 |
4200 |
11.42 |
342.6 |
145-155 |
150 |
12 |
1800 |
21.42 |
257.04 |
N = 106 |
Total = 13630 |
Total = 1272.6 |
N = 106
\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)
= 1/106 × 1272.6
= 12.005
∴ The mean deviation is 12.005
Statistics Ex 32.3 Q2(iii)
Statistics Ex 32.3 Q3
Statistics Ex 32.3 Q4
Statistics Ex 32.3 Q5
Statistics Ex 32.3 Q6
Statistics Ex 32.3 Q7
Statistics Ex 32.3 Q8
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FAQs on RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.3
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There are a total of 6 questions in RD Sharma Class 11 Solutions Chapter 32 Exercise 32.3.
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