निम्न बारम्बारता बंटन का माध्य ज्ञात कीजिए ( प्रश्न 1-4):
प्रश्न 1.
| X: | 3 | 5 | 8 | 11 |
| f: | 2 | 4 | 5 | 3 |
हल:
समान्तर माध्य के लिये सारणी
| X | f | fx |
| 3 | 2 | 3 × 2 = 6 |
| 5 | 4 | 5 × 4 = 20 |
| 8 | 5 | 8 × 5 = 40 |
| 11 | 3 | 11 × 3 = 33 |
| ∑f = 14 | ∑fx = 99 |
समान्तर माध्य
प्रश्न 2.
| X | 2 | 5 | 7 | 9 | 11 |
| f | 1 | 5 | 4 | 7 | 3 |
हल:
समान्तर माध्य के लिये सारणी
| X | f | fx |
| 2 | 1 | 2 |
| 5 | 5 | 25 |
| 7 | 4 | 28 |
| 9 | 7 | 63 |
| 11 | 3 | 33 |
| ∑f = 20 | ∑fx = 151 |
अतः अभीष्ट समान्तर माध्य = 7.55 उत्तर
प्रश्न 3.
| X | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
| f | 30 | 60 | 20 | 40 | 10 | 50 |
हल:
समान्तर माध्य के लिये सारणी
| X | f | fx |
| 0.1 | 30 | 0.1 × 30 = 3.0 |
| 0.2 | 60 | 0.2 × 60 = 12.0 |
| 0.3 | 20 | 0.3 × 20 = 6.0 |
| 0.4 | 40 | 0.4 × 40 = 16.0 |
| 0.5 | 10 | 0.5 × 10 = 5.0 |
| 0.6 | 50 | 0.6 × 50 = 30.0 |
| ∑f = 210 | ∑fx = 72.0 |
समान्तर माध्य
अतः अभीष्ट समान्तर माध्य = 0.34 (लगभग) उत्तर
प्रश्न 4.
| X | 0.1 | 0.3 | 0.5 | 0.7 | 0.89 |
| f | 7 | 8 | 10 | 15 | 10 |
हल:
समान्तर माथ्य के लिये सारणी
| X | f | fx |
| 0.1 | 7 | 0.1 × 7 = 0.7 |
| 0.3 | 8 | 0.3 × 8 = 2.4 |
| 0.5 | 10 | 0.5 × 10 = 5.0 |
| 0.7 | 15 | 0.7 × 15 = 10.0 |
| 0.89 | 10 | 0.89 × 10 = 8.0 |
| ∑f = 50 | ∑fx = 27.5 |
समान्तर माध्य
अतः अभीष्ट समान्तर माध्य = 0.55 उत्तर
प्रश्न 5.
एक सौ परिवारों में बच्चों की संख्या निम्न प्रकार है–
| बच्चों की संख्या | 1 | 2 | 3 | 4 | 5 | 6 |
| परिवारों संख्या | 45 | 25 | 19 | 8 | 2 | 1 |
इनका समान्तर माध्य ज्ञात कीजिए।
हल:
समान्तर माध्य के लिये सारणी
| X | f | fx |
| 1 | 45 | 45 |
| 2 | 25 | 50 |
| 3 | 19 | 57 |
| 4 | 8 | 32 |
| 5 | 2 | 10 |
| 6 | 1 | 06 |
| ∑f = 100 | ∑fx = 200 |
समान्तर माध्य
अतः सौ परिवार में बच्चों का समान्तर माध्य = 2 उत्तर
प्रश्न 6.
एक कक्षा में छात्रों के भार निम्न सारणी में दिये गए हैं—
| भार किग्रा में | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 |
| छात्रों की संख्या | 1 | 2 | 6 | 7 | 4 | 2 | 3 | 2 | 3 |
इनका समान्तर माध्य ज्ञात कीजिए।
हल:
समान्तर माध्य के लिये सारणी
| X | f | fx |
| 20 | 1 | 20 |
| 21 | 2 | 42 |
| 22 | 6 | 132 |
| 23 | 7 | 161 |
| 24 | 4 | 96 |
| 25 | 2 | 50 |
| 26 | 3 | 78 |
| 27 | 2 | 54 |
| 28 | 3 | 84 |
| ∑f = 30 | ∑fx = 717 |
अतः समान्तर माध्य
अतः छात्रों का औसत भार = 23.9 किग्रा.। उत्तर
प्रश्न 7.
यदि निम्न बंटन का माध्य 7.5 हो, तो P का मान ज्ञात कीजिए।
| X | 3 | 5 | 7 | 9 | 11 | 13 |
| f | 6 | 8 | 15 | P | 8 | 4 |
हल:
समान्तर माध्य के लिये सारणी
| X | f | fx |
| 3 | 6 | 18 |
| 5 | 8 | 40 |
| 7 | 15 | 105 |
| 9 | P | 9P |
| 11 | 8 | 88 |
| 13 | 4 | 52 |
| ∑f = (41 + P) | ∑fx = (303 + 9P) |
अतः P का मान = 3 उत्तर
प्रश्न 8.
यदि निम्न बारम्बारता बंटन का माध्य 1.46 हो तो अज्ञात बारम्बारताएं ज्ञात कीजिए।
| X | 0 | 1 | 2 | 3 | 4 | 5 | योग |
| f | 46 | • • • | • • • | 25 | 10 | 5 | 200 |
हल:
समान्तर माध्य के लिये सारणी
| X | f | fx |
| 0 | 46 | 0 |
| 1 | L1 | L1 |
| 2 | L2 | 2L2 |
| 3 | 25 | 75 |
| 4 | 10 | 40 |
| 5 | 5 | 25 |
| योग | ∑f = 200 | ∑fx = 140 + L1 + 2L2 |
यहाँ 86 + L1 + L2 = 200 (दिया है)
L1 + L2 = 200 – 86 = 114 ………………………(i)
प्रश्नानुसार, समान्तर माध्य = 1.46
अतः 140 + L1 + 2L2 = 1.46 x 200 = 292
या L1 + 2L2 = 292 – 140 = 152 ………………………..(ii)
समी. (i) व (ii) से।
L2 = 152 – 114 = 38
L1 = 114 – 38 = 76
अत: L1 = 76 – L2 = 38
अतः अज्ञात बारम्बारतायें 76 व 38 है। उत्तर
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