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RBSE Solutions for Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Ex 17.4

March 6, 2019 by Phani 1 Comment

RBSE Solutions for Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Ex 17.4 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Exercise 17.4.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 10
Subject Maths
Chapter Chapter 17
Chapter Name केन्द्रीय प्रवृत्ति के माप
Exercise Exercise 17.4
Number of Questions Solved 7
Category RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Ex 17.4

निम्न बारम्बारता बंटन का माध्य, कल्पित माध्य की सहायता से ज्ञात कीजिए–(प्रश्न 1 से 4)

प्रश्न 1.

X 800 820 860 900 920 980 1000
f 7 14 19 25 20 10 5

हल:
यहाँ कल्पित (RBSESolutions.com) माध्य a = 900 है।

X f d = (x – a) f × d
800 7 – 100 – 700
820 14 – 80 – 1120
860 19 – 40 – 760
900 25 00 000
920 20 20 400
980 10 80 800
1000 5 100 500
∑f = 100 ∑fx = – 880

अतः समान्तर माध्य
\(\begin{aligned} \overline{\mathrm{X}} &=\mathrm{a}+\frac{\Sigma \mathrm{fd}}{\Sigma \mathrm{f}} \\ &=900+\frac{-880}{100} \\ &=900-8.8 \end{aligned}\)
समान्तर माध्य \(\overline{\mathrm{X}}=891.2\)
अतः अभीष्ट समान्तर माध्य = 891.2 उत्तर

RBSE Solutions

प्रश्न 2.

भरा किग्रा में 61 62 63 64 65
मजदूरों की संख्या 5 8 14 16 10 7

हल:
यहाँ कल्पित (RBSESolutions.com) माध्य 62 है।

X f d = (x – 62) fd
60 5 – 2 – 10
61 8 – 1 – 8
62 14 0 00
63 16 1 16
64 10 2 20
65 7 3 21
∑f = 60 ∑fd = 39

अतः समान्तर माध्य
\(\begin{aligned} \overline{\mathrm{X}} &=\mathrm{a}+\frac{\Sigma \mathrm{fd}}{\Sigma \mathrm{f}} \\ &=62+\frac{39}{60} \end{aligned}\)
= 62.65 किग्रा. उत्तर

प्रश्न 3.

खर्च (रुपयों में ) 100-150 150-200 200-250 250-300
मजदूरों की संख्या 24 40 33 28
खर्च (रुपयों में) 300-350 350-400 400-450 450-500
मजदूरों की संख्या 30 22 16 7

हल:
माना कल्पित (RBSESolutions.com) माध्य a = 325 है।
RBSE Solutions for Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Ex 17.4 1
समान्तर माध्य
\(\begin{aligned} \overline{\mathrm{X}} &=\mathrm{a}+\frac{\Sigma \mathrm{fd}}{\Sigma \mathrm{f}} \\ &=325-\frac{11750}{200} \\ &=325-58.75 \end{aligned}\)
समान्तर माध्य = 266.25 उत्तर

प्रश्न 4.

पानी पर खर्च (रुपयों में) 15-20 20-25 25-30 30-35 35-40 40-45
मकानों की संख्या 7 5 7 8 9 11
पानी पर खर्च (रुपयों में) 45-50 50-55 55-60 60-65 65-70
मकानों की संख्या 7 5 4 4 3

हल:
माना कल्पित (RBSESolutions.com) माध्य 42.5 है।
RBSE Solutions for Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Ex 17.4 2
समान्तर माध्य
\(\begin{aligned} \overline{\mathrm{X}} &=\mathrm{a}+\frac{\Sigma \mathrm{fd}}{\Sigma \mathrm{f}} \\ &=42.5+\frac{-205}{70} \\ &=42.5-2.93 \end{aligned}\)
समान्तर माध्य = 39.57 उत्तर

RBSE Solutions

प्रश्न 5.
कल्पित माध्य 25 मानकर निम्न (RBSESolutions.com) बंटने का माध्य ज्ञात ‘कीजिए–

वर्ग 0-10 10-20 20-30 30-40 40-50
f 6 10 13 7 4

हल:
यहाँ a = 25 दिया हाय है।

वर्ग मध्यमान (x) (f) d = (x – a) fd
0-10 5 6 – 20 – 120
10-20 15 10 – 10 – 100
20-30 25 13 0 0
30-40 35 7 10 70
40-50 45 4 20 80
∑f = 40 ∑fd = – 70

समान्तर माध्य
\(\begin{array}{l}{\overline{\mathrm{X}}=\mathrm{a}+\frac{\Sigma \mathrm{fd}}{\Sigma \mathrm{f}}} \\ {\overline{\mathrm{X}}=25+\left(\frac{-70}{40}\right)} \\ {\overline{\mathrm{X}}=25-1.75} \\ {\mathrm{X}=23.25}\end{array}\)
अतः अभीष्ट समान्तर माध्य = 23.25 उत्तर

प्रश्न 6.
अग्रलिखित सारणी में एक शहर में एक विशेष वर्ष में एक रोग से पीड़ित रोगियों का आयु बंटन दिया गया है। (RBSESolutions.com) प्रति रोगी औसत आयु (वर्षो में) ज्ञात कीजिए।

आयु (वर्षों में) रोगियों की संख्या
5-14 6
15-24 11
25-34 21
35-44 23
45-54 14
55-64 5

हल:
माना कल्पित माध्य (a) = 29.5
RBSE Solutions for Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Ex 17.4 3
समान्तर माध्य
\(\begin{aligned}(\overline{\mathrm{X}}) &=\mathrm{a}+\frac{\Sigma \mathrm{fd}}{\Sigma \mathrm{f}} \\ &=29.5+\frac{430}{80}=29.5+5.375 \\ &=34.875 \end{aligned}\)
अतः समान्तर माध्य = 34.875 वर्ष उत्तर

प्रश्न 7.
निम्नलिखित बारम्बारता बंटन से माध्ये ज्ञात (RBSESolutions.com) कीजिए|–

वर्ग अन्तराल 40-50 50-60 60-70 70-80 80-90 90-100
बारम्बारता 10 25 28 12 10 15

हल:
माना कल्पित माध्य (a) = 65
RBSE Solutions for Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Ex 17.4 4
समान्तर माध्य
\(\begin{array}{l}{\overline{\mathrm{X}}=\mathrm{a}+\frac{\Sigma \mathrm{fd}}{\Sigma \mathrm{f}}} \\ {\overline{\mathrm{X}}=65+\frac{320}{100}}\end{array}\)
समान्तर माध्य = 65 + 3.2 = 68.2
अतः अभीष्ट समान्तर माध्य = 68.2 उत्तर

RBSE Solutions

We hope the RBSE Solutions for Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Ex 17.4 help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 17 केन्द्रीय प्रवृत्ति के माप Exercise 17.4, drop a comment below and we will get back to you at the earliest.

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