RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2: In this exercise, we’ll look at the mean deviation of a discrete frequency distribution and how to compute it using an algorithm. Students who are experiencing difficulty solving arithmetic questions can use the solutions prepared by experts, who have presented the solutions in the most straightforward manner possible for any pupil to understand. Students who are unable to resolve their doubts during class can refer the RD Sharma Class 11 Maths Solutions. Students can easily obtain the RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2 pdf by clicking on the links below.
Download RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.2 Free PDF
RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2
Access answers to RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2- Important Question with Answers
1. Calculate the mean deviation from the median of the following frequency distribution:
|
Heights in inches |
58 |
59 |
60 |
61 |
62 |
63 |
64 |
65 |
66 |
|
No. of students |
15 |
20 |
32 |
35 |
35 |
22 |
20 |
10 |
8 |
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
We know, Median is the Middle term,
So, Median = 61
Let xi =Heights in inches
And, fi = Number of students
|
xi |
fi |
Cumulative Frequency |
|di| = |xi – M| = |xi – 61| |
fi |di| |
|
58 |
15 |
15 |
3 |
45 |
|
59 |
20 |
35 |
2 |
40 |
|
60 |
32 |
67 |
1 |
32 |
|
61 |
35 |
102 |
0 |
0 |
|
62 |
35 |
137 |
1 |
35 |
|
63 |
22 |
159 |
2 |
44 |
|
64 |
20 |
179 |
3 |
60 |
|
65 |
10 |
189 |
4 |
40 |
|
66 |
8 |
197 |
5 |
40 |
|
N = 197 |
Total = 336 |
N=197
\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)
= 1/197 × 336
= 1.70
∴ The mean deviation is 1.70.
2. The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:
|
Number of calls |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
|
Frequency |
14 |
21 |
25 |
43 |
51 |
40 |
39 |
12 |
Compute the mean deviation about the median.
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
We know, Median is the even term, 3+5/2 = 4
So, Median = 8
Let xi =Number of calls
And, fi = Frequency
|
xi |
fi |
Cumulative Frequency |
|di| = |xi – M| = |xi – 61| |
fi |di| |
|
0 |
14 |
14 |
4 |
56 |
|
1 |
21 |
35 |
3 |
63 |
|
2 |
25 |
60 |
2 |
50 |
|
3 |
43 |
103 |
1 |
43 |
|
4 |
51 |
154 |
0 |
0 |
|
5 |
40 |
194 |
1 |
40 |
|
6 |
39 |
233 |
2 |
78 |
|
7 |
12 |
245 |
3 |
36 |
|
Total = 366 |
||||
|
Total = 245 |
N = 245
\(MD = \frac{1}{n}\sum_{i=1}^{n}|d_{i}|\)
= 1/245 × 336
= 1.49
∴ The mean deviation is 1.49.
Statistics Ex 32.2 Q3
Statistics Ex 32.2 Q4(i)
Statistics Ex 32.2 Q4(ii)
Statistics Ex 32.2 Q4(iii)
Statistics Ex 32.2 Q4(iv)
Statistics Ex 32.2 Q5
We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2. If you have any queries related to CBSE Class 11, feel free to ask us in the comment section below.
FAQs on RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.2
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