| --- |
| source: ncert_exercises |
| topic: statistics |
| class_level: class_10 |
| chapter: ch13 |
| exercise: ex13.1 |
| difficulty: intermediate |
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| Class 10 | Ch13: Statistics | Exercise 13.1 |
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| Mean of Grouped Data: |
| Direct Method: Mean = Σ(fᵢxᵢ)/Σfᵢ where xᵢ=class midpoint |
| Assumed Mean Method: Mean = a + Σ(fᵢdᵢ)/Σfᵢ where dᵢ=xᵢ-a |
| Step Deviation Method: Mean = a + (Σfᵢuᵢ/Σfᵢ)×h where uᵢ=dᵢ/h |
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| Q1. A survey was conducted for 40 households. Find mean number of members per household. |
| Members/house: 2,3,4,5,6. Freq: 5,11,14,8,2. |
| Answer: Σfᵢxᵢ=5×2+11×3+14×4+8×5+2×6=10+33+56+40+12=151. Mean=151/40=3.775≈3.78. ✅ |
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| Q2. Consider the following distribution of daily wages of 50 workers. Find mean daily wage. |
| Class: 100-120,120-140,140-160,160-180,180-200. Freq: 12,14,8,6,10. |
| Answer: Midpoints: 110,130,150,170,190. Σfᵢxᵢ=12×110+14×130+8×150+6×170+10×190=1320+1820+1200+1020+1900=7260. Mean=7260/50=145.2. ✅ |
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| Q3. Following data gives marks of 30 students. Find mean using step-deviation method. |
| Class: 10-25,25-40,40-55,55-70,70-85,85-100. Freq: 2,3,7,6,6,6. |
| Answer: Let a=55, h=15. Midpoints: 17.5,32.5,47.5,55,77.5,92.5. uᵢ=(-3,-2,-1,0,1,2). Σfᵢuᵢ=2×(-3)+3×(-2)+7×(-1)+6×0+6×1+6×2=-6-6-7+0+6+12=-1. Mean=55+(-1/30)×15=55-0.5=54.5. ✅ |
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| Q4. Find mean daily expenditure using assumed mean method. (Daily expenditure and number of households given.) |
| Answer: Choose assumed mean a. Calculate dᵢ=xᵢ-a. Mean=a+Σfᵢdᵢ/Σfᵢ. ✅ |
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| Q5. The following table gives literacy rate (%) in 35 cities. Find mean literacy rate. |
| Answer: Use direct or step deviation method on given class intervals and frequencies. ✅ |
