Class 11 Maths · Chapter 13
Short answer: "
Class 11 Maths Chapter 13 Statistics single line mein: yeh chapter data ke \"spread\" (dispersion) ko measure karna sikhata hai — Range, Mean Deviation, Variance aur Standard Deviation — aur akhir mein Coefficient of Variation se do datasets ki consistency compare karna. Neeche di gayi Class 11 Maths NCERT solutions mein Exercise 13.1 (12 questions), Exercise 13.2 (10 questions) aur Miscellaneous Exercise (7 questions) — teeno ke step-by-step solutions hain, saath mein ek quick formulas table jo Class 11 Maths formulas pdf download ki jagah revision ke liye kaafi hai.
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Chapter 13 Statistics Class 11 Maths ka woh chapter hai jo \"average\" ki limitation dikhata hai. Class 10 mein humne Mean, Median, Mode seekha tha — woh sirf data ke center ko batate hain. Lekin do cricket players ka average score same ho sakta hai (jaise dono ka average 40 runs) — phir bhi ek consistent hai aur doosra kabhi 0 kabhi 100 maarta hai. Yehi \"consistency vs inconsistency\" ka gap measures of dispersion — Range, Mean Deviation, Variance aur Standard Deviation — bharte hain.
Yeh chapter Class 11 Maths ke rationalised syllabus (2026-27 session) mein Chapter 13 hai — is se pehle Chapter 12 Limits and Derivatives hai aur is ke baad Chapter 14 Probability aata hai. Agar aap poore session ka plan bana rahe ho to class 11 maths syllabus 2026-27 pdf aur class 11 maths ncert book pdf download se chapter list cross-check kar sakte ho — purane 16-chapter edition mein \"Mathematical Induction\" aur \"Mathematical Reasoning\" jaise chapters the jo ab hata diye gaye hain, isliye numbering match karna zaroori hai. Yeh page us bade class 11 maths ncert solutions series ka hissa hai jisme Chapter 1 Sets, Chapter 3 Trigonometric Functions, Chapter 8 Sequences and Series, Chapter 9 Straight Lines aur Chapter 12 Limits and Derivatives bhi cover kiye jaate hain.
Formula-wise yeh chapter halka lagta hai (bas 4-5 formulas), lekin exam mein sabse zyada marks calculation mistakes mein katte hain — isliye har solution neeche full working ke saath diya gaya hai, taaki board exam mein step-by-step likhne ki practice ho jaaye.
"Chapter 13 Summary — 5 Minute Revision
"Class 11 Maths Chapter 13 Statistics measures of dispersion ke around build hota hai — yani data apne central value (mean/median) se kitna \"bikhra hua\" (spread) hai. Chapter Range se start hota hai (Max − Min, sabse crude measure), phir Mean Deviation (about mean aur about median dono) sikhata hai jo har observation ke absolute deviation ka average leta hai — ungrouped, discrete frequency aur continuous frequency, teeno tarah ke data ke liye.
Iske baad Variance aur Standard Deviation aate hain jo sabse important hain kyunki inhe negative-cancellation ka problem nahi hota (absolute value ki jagah square lete hain). Continuous data ke liye shortcut (step-deviation) method bhi sikhaya jaata hai jisse bade numbers ki calculation aasan ho jaati hai. Chapter ka last topic Analysis of Frequency Distributions hai jahan Coefficient of Variation (C.V.) use karke do alag-alag datasets ki relative variability compare ki jaati hai — kam C.V. wala dataset zyada consistent maana jaata hai.
NCERT ne is chapter mein teen exercises rakhe hain: 13.1 (Mean Deviation — 12 questions), 13.2 (Variance & Standard Deviation — 10 questions) aur Miscellaneous Exercise (mixed application — 7 questions), total 29 questions.
"In-Text Questions — Solutions
Range ki definition aur formula kya hai?
Range kisi data ke sabse bade (Maximum) aur sabse chhote (Minimum) value ka difference hota hai — dispersion ka sabse simple measure.
Range = Xmax − Xmin
Mean deviation about mean aur mean deviation about median mein kya fark hai?
Dono hi "average distance from a central value" measure karte hain, bas central value alag hoti hai — pehle mein har observation ka deviation mean se liya jaata hai, doosre mein median se. Formula same rehta hai — sirf har observation ka absolute deviation nikal ke unka average lena hai.
Variance ka standard deviation se kya relationship hai?
Standard deviation (σ) variance (σ²) ka positive square root hota hai.
σ = √(σ²) , ya σ² = σ × σ
Variance ka unit original data ke square mein hota hai (jaise cm²), isliye SD zyada practical hai kyunki uska unit data jaisa hi hota hai (cm).
Coefficient of Variation (C.V.) kis kaam aata hai?
C.V. do datasets ki relative variability compare karne ke liye use hota hai, especially jab unke means alag-alag hon ya units alag hon.
C.V. = (σ / x̄) × 100
Jis dataset ka C.V. kam ho, woh zyada consistent (less variable) maana jaata hai.
Discrete frequency distribution ke liye mean deviation about mean ka formula likho.
MD(x̄) = Σ fi|xi − x̄| / N , jahan N = Σfi
Yani har value ka apne mean se absolute deviation lo, uski frequency se multiply karo, phir sabka sum total frequency N se divide kar do.
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q29)
Exercise 13.1, Q1. Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17.
n = 8, Σxi = 4+7+8+9+10+12+13+17 = 80
x̄ = 80/8 = 10
Absolute deviations |xi − x̄|: 6, 3, 2, 1, 0, 2, 3, 7 → sum = 24
MD(x̄) = 24/8 = 3
Exercise 13.1, Q2. Find the mean deviation about the mean for the data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
n = 10, Σxi = 500
x̄ = 500/10 = 50
Absolute deviations: 12, 20, 2, 10, 8, 5, 13, 4, 4, 6 → sum = 84
MD(x̄) = 84/10 = 8.4
Exercise 13.1, Q3. Find the mean deviation about the median for the data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
n = 12 (even). Ascending order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Median = (6th term + 7th term)/2 = (13+14)/2 = 13.5
Absolute deviations |xi − M|: 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5 → sum = 28
MD(M) = 28/12 = 7/3 ≈ 2.33
Exercise 13.1, Q4. Find the mean deviation about the mean for the data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
n = 10, Σxi = 500
x̄ = 500/10 = 50
Absolute deviations: 14, 22, 4, 8, 10, 5, 3, 4, 1, 1 → sum = 72
MD(x̄) = 72/10 = 7.2
Exercise 13.1, Q5. Find the mean deviation about the median for the data: 34, 66, 30, 38, 44, 50, 40, 60, 42, 51.
n = 10 (even). Ascending order: 30, 34, 38, 40, 42, 44, 50, 51, 60, 66
Median = (5th + 6th term)/2 = (42+44)/2 = 43
Absolute deviations: 13, 9, 5, 3, 1, 1, 7, 8, 17, 23 → sum = 87
MD(M) = 87/10 = 8.7
Exercise 13.1, Q6. Find the mean deviation about the mean for the discrete frequency distribution: xi: 5, 10, 15, 20, 25 with fi: 7, 4, 6, 3, 5.
N = Σfi = 7+4+6+3+5 = 25
Σfixi = 35+40+90+60+125 = 350 → x̄ = 350/25 = 14
|xi−x̄|: 9, 4, 1, 6, 11; fi|xi−x̄|: 63, 16, 6, 18, 55 → sum = 158
MD(x̄) = 158/25 = 6.32
Exercise 13.1, Q7. Find the mean deviation about the median for the discrete frequency distribution: xi: 10, 20, 30, 40, 50 with fi: 4, 3, 7, 4, 2.
N = 20. Cumulative frequency: 4, 7, 14, 18, 20
N/2 = 10 → first cf ≥ 10 is 14, corresponding xi = 30, so Median = 30.
|xi−30|: 20, 10, 0, 10, 20; fi×dev: 80, 30, 0, 40, 40 → sum = 190
MD(M) = 190/20 = 9.5
Exercise 13.1, Q8. Find the mean deviation about the mean for the continuous frequency distribution: Class 0-10, 10-20, 20-30, 30-40, 40-50 with fi: 5, 8, 15, 16, 6.
Mid-values xi: 5, 15, 25, 35, 45; N = 50
Σfixi = 25+120+375+560+270 = 1350 → x̄ = 1350/50 = 27
|xi−x̄|: 22, 12, 2, 8, 18; fi×dev: 110, 96, 30, 128, 108 → sum = 472
MD(x̄) = 472/50 = 9.44
Exercise 13.1, Q9. Find the mean deviation about the median for the same distribution as Q8: Class 0-10 to 40-50 with fi: 5, 8, 15, 16, 6.
N = 50, cf: 5, 13, 28, 44, 50. N/2 = 25 → median class = 20-30 (cf = 28 first ≥ 25).
Median = l + [(N/2 − cf)/f] × h = 20 + [(25−13)/15] × 10 = 28
Mid-values: 5, 15, 25, 35, 45; |xi−28|: 23, 13, 3, 7, 17; fi×dev: 115, 104, 45, 112, 102 → sum = 478
MD(M) = 478/50 = 9.56
Exercise 13.1, Q10. Find the mean deviation about the mean for the data: 5, 10, 15, 20, 25, 30, 35.
n = 7, Σxi = 140
x̄ = 140/7 = 20
Absolute deviations: 15, 10, 5, 0, 5, 10, 15 → sum = 60
MD(x̄) = 60/7 ≈ 8.57
Exercise 13.1, Q11. Find the mean deviation about the mean for the discrete frequency distribution: xi: 2, 5, 8, 11, 14 with fi: 3, 8, 10, 6, 3.
N = 30, Σfixi = 6+40+80+66+42 = 234 → x̄ = 234/30 = 7.8
|xi−7.8|: 5.8, 2.8, 0.2, 3.2, 6.2; fi×dev: 17.4, 22.4, 2, 19.2, 18.6 → sum = 79.6
MD(x̄) = 79.6/30 ≈ 2.65
Exercise 13.1, Q12. Find the mean deviation about the mean for the continuous frequency distribution: Class 10-20, 20-30, 30-40, 40-50, 50-60 with fi: 4, 6, 10, 8, 2.
Mid-values: 15, 25, 35, 45, 55; N = 30
Σfixi = 60+150+350+360+110 = 1030 → x̄ = 1030/30 ≈ 34.33
|xi−x̄|: 19.33, 9.33, 0.67, 10.67, 20.67; fi×dev ≈ 77.32, 55.98, 6.7, 85.36, 41.34 → sum ≈ 266.7
MD(x̄) ≈ 266.7/30 ≈ 8.89
Exercise 13.2, Q1. Find the variance and standard deviation of the data: 6, 7, 10, 12, 13, 4, 8, 12.
n = 8, Σxi = 72
x̄ = 72/8 = 9
Deviations (xi−x̄): −3, −2, 1, 3, 4, −5, −1, 3; squares: 9, 4, 1, 9, 16, 25, 1, 9 → sum = 74
Variance σ² = 74/8 = 9.25 ; SD σ = √9.25 ≈ 3.04
Exercise 13.2, Q2. Find the variance and standard deviation of the first n natural numbers.
Mean x̄ = (n+1)/2, and Σi² = n(n+1)(2n+1)/6
σ² = Σi²/n − x̄² = (n+1)(2n+1)/6 − (n+1)²/4
= (n+1)[2(2n+1) − 3(n+1)]/12 = (n+1)(n−1)/12 = (n²−1)/12
SD = √[(n²−1)/12]
Exercise 13.2, Q3. Find the variance and standard deviation of the first 10 multiples of 3.
Data: 3, 6, 9, ..., 30 = 3 × (1, 2, ..., 10). Multiplying data by a constant k multiplies variance by k².
Variance of (1..10) = (10²−1)/12 = 99/12 = 8.25
Variance of given data = 3² × 8.25 = 9 × 8.25 = 74.25
SD = √74.25 ≈ 8.62
Exercise 13.2, Q4. Find the variance and standard deviation for the discrete frequency distribution: xi: 6, 10, 14, 18, 24, 28, 30 with fi: 2, 4, 7, 12, 8, 4, 3.
N = 40; Σfixi = 12+40+98+216+192+112+90 = 760 → x̄ = 760/40 = 19
Σfixi² = 72+400+1372+3888+4608+3136+2700 = 16176
Variance σ² = 16176/40 − 19² = 404.4 − 361 = 43.4
SD σ = √43.4 ≈ 6.59
Exercise 13.2, Q5. Find the variance and standard deviation for the discrete frequency distribution: xi: 92, 93, 97, 98, 102, 104, 109 with fi: 3, 2, 3, 2, 6, 3, 3.
N = 22; Σfixi = 276+186+291+196+612+312+327 = 2200 → x̄ = 2200/22 = 100
Σfixi² = 25392+17298+28227+19208+62424+32448+35643 = 220640
Variance σ² = 220640/22 − 100² = 10029.09 − 10000 = 29.09
SD σ = √29.09 ≈ 5.39
Exercise 13.2, Q6. Find the variance and standard deviation using the direct method for the continuous frequency distribution: Class 0-10, 10-20, 20-30, 30-40, 40-50 with fi: 5, 8, 15, 16, 6.
Mid-values: 5, 15, 25, 35, 45; N = 50, x̄ = 27 (from Ex 13.1 Q8)
Σfixi² = 125+1800+9375+19600+12150 = 43050
Variance σ² = 43050/50 − 27² = 861 − 729 = 132
SD σ = √132 ≈ 11.49
Exercise 13.2, Q7. Find the variance and standard deviation using the step-deviation (shortcut) method: Class 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100 with fi: 3, 7, 12, 15, 8, 3, 2.
Mid-values: 35, 45, 55, 65, 75, 85, 95. Assumed mean A = 65, h = 10, yi = (xi−65)/10 = −3,−2,−1,0,1,2,3
Σfiyi = −9−14−12+0+8+6+6 = −15
Σfiyi² = 27+28+12+0+8+12+18 = 105
x̄ = A + h(Σfiyi/N) = 65 + 10(−15/50) = 62
σ² = h²[Σfiyi²/N − (Σfiyi/N)²] = 100[2.1 − 0.09] = 201
SD σ = √201 ≈ 14.18
Exercise 13.2, Q8. Wages of workers in Firm A have mean ₹5253 and variance 100; in Firm B mean is also ₹5253 but variance is 121. Which firm shows greater variability in wages?
Both means are equal, so we compare SD directly. σA = √100 = 10, σB = √121 = 11
C.V.A = (10/5253) × 100 ≈ 0.19% ; C.V.B = (11/5253) × 100 ≈ 0.21%
Since C.V.B > C.V.A, Firm B shows greater variability in wages.
Exercise 13.2, Q9. Find the variance and standard deviation using the step-deviation method: Class 0-30, 30-60, 60-90, 90-120, 120-150, 150-180, 180-210 with fi: 2, 3, 5, 10, 3, 5, 2.
Mid-values: 15, 45, 75, 105, 135, 165, 195. A = 105, h = 30, yi = −3,−2,−1,0,1,2,3
Σfiyi = −6−6−5+0+3+10+6 = 2 ; Σfiyi² = 18+12+5+0+3+20+18 = 76
x̄ = 105 + 30(2/30) = 107
σ² = 900[76/30 − (2/30)²] = 900[2.533 − 0.0044] ≈ 2276
SD σ = √2276 ≈ 47.71
Exercise 13.2, Q10. Average plant height in a District A sample is 50 cm with SD 6 cm; in District B sample, average is also 50 cm with SD 8 cm. Which district shows more consistent plant heights?
Means are equal, so compare C.V.
C.V.A = (6/50) × 100 = 12% ; C.V.B = (8/50) × 100 = 16%
Since C.V.A < C.V.B, District A shows more consistent (less variable) plant heights.
Miscellaneous Exercise, Q1. Find the mean deviation about the mean for the continuous frequency distribution: Class 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 with fi: 6, 8, 14, 16, 4, 2.
Mid-values: 5, 15, 25, 35, 45, 55; N = 50
Σfixi = 30+120+350+560+180+110 = 1350 → x̄ = 27
|xi−27|: 22,12,2,8,18,28; fi×dev: 132,96,28,128,72,56 → sum = 512
MD(x̄) = 512/50 = 10.24
Miscellaneous Exercise, Q2. The mean and standard deviation of 100 observations were found to be 40 and 5.1 respectively. Later it was found that one observation, 50, was misread as 40. Find the correct mean and standard deviation.
Wrong Σx = 100 × 40 = 4000; correct Σx = 4000 − 40 + 50 = 4010
Correct mean = 4010/100 = 40.1
Wrong Σx² : since σ² = Σx²/n − x̄² → 26.01 = Σx²(wrong)/100 − 1600 → Σx²(wrong) = 162601
Correct Σx² = 162601 − 40² + 50² = 163501
Correct variance = 163501/100 − (40.1)² = 1635.01 − 1608.01 = 27
Correct SD = √27 ≈ 5.20
Miscellaneous Exercise, Q3. The standard deviation of 20 observations is 5. If each observation is multiplied by 3, find the new standard deviation.
Multiplying every observation by a constant k multiplies the SD by |k| (variance by k²).
New SD = 3 × 5 = 15
Miscellaneous Exercise, Q4. Two sets of observations have n1=10, mean1=15, SD1=3 and n2=15, mean2=25, SD2=4. Find the combined mean and combined standard deviation.
Combined mean x̄ = (n₁x̄₁+n₂x̄₂)/(n₁+n₂) = (150+375)/25 = 21
d₁ = x̄₁ − x̄ = 15−21 = −6, d₂ = x̄₂ − x̄ = 25−21 = 4
Combined σ² = [n₁(σ₁²+d₁²) + n₂(σ₂²+d₂²)]/(n₁+n₂) = [10(9+36)+15(16+16)]/25 = 930/25 = 37.2
Combined SD = √37.2 ≈ 6.10
Miscellaneous Exercise, Q5. Find the mean deviation about the median for the continuous frequency distribution: Class 0-10, 10-20, 20-30, 30-40, 40-50 with fi: 6, 8, 14, 16, 6.
N = 50, cf: 6, 14, 28, 44, 50. N/2 = 25 → median class = 20-30 (cf = 28 first ≥ 25).
Median = 20 + [(25−14)/14] × 10 ≈ 27.86
Mid-values: 5,15,25,35,45; |xi−27.86|: 22.86,12.86,2.86,7.14,17.14; fi×dev ≈ 137.1,102.9,40,114.3,102.9 → sum ≈ 497.1
MD(M) ≈ 497.1/50 ≈ 9.94
Miscellaneous Exercise, Q6. For a set of 8 observations, Σ(xi−25) = 8 and Σ(xi−25)² = 240. Find the mean and standard deviation.
Mean = 25 + Σ(xi−25)/n = 25 + 8/8 = 26
Variance = Σ(xi−25)²/n − [Σ(xi−25)/n]² = 240/8 − 1² = 30 − 1 = 29
SD = √29 ≈ 5.39
Miscellaneous Exercise, Q7. Two brands of tyres, A and B, were tested on samples of 100 tyres each. Brand A: mean life = 40 (in '000 km), SD = 6. Brand B: mean life = 40, SD = 8. Which brand gives more consistent performance?
Means are equal, so compare C.V.
C.V.A = (6/40) × 100 = 15% ; C.V.B = (8/40) × 100 = 20%
Since C.V.A < C.V.B, Brand A gives more consistent performance.
Important Equations — Ek Nazar Me
"| Concept | Formula |
|---|---|
| Range | Range = Xmax − Xmin |
| Mean Deviation about mean (ungrouped) | MD(x̄) = Σ|xi − x̄| / n |
| Mean Deviation about median (ungrouped) | MD(M) = Σ|xi − M| / n |
| Mean Deviation (discrete/continuous freq.) | MD = Σfi|xi − A| / N , where N = Σfi, A = mean or median |
| Median of continuous distribution | M = l + [(N/2 − cf)/f] × h |
| Variance (ungrouped) | σ² = Σ(xi − x̄)² / n |
| Variance (discrete/continuous freq.) | σ² = Σfi(xi − x̄)² / N = Σfixi²/N − x̄² |
| Standard Deviation | σ = √(σ²) |
| Step-deviation (shortcut) method | yi = (xi − A)/h ; σ² = h²[Σfiyi²/N − (Σfiyi/N)²] |
| Combined SD of two groups | σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)] / (n₁+n₂), di = x̄i − combined mean |
| Coefficient of Variation | C.V. = (σ / x̄) × 100 |
↔ Table ko side me swipe karein
"Common Mistakes — Yahan Marks Kat te Hain
- Mean deviation nikaalte waqt deviations ka absolute value lena bhool jaana — signed deviations ka sum hamesha zero aata hai, isliye |xi − x̄| lena zaroori hai, sign hataye bina answer galat aayega.
- Grouped/continuous data mein n ki jagah galti se N (= Σfi) use na karna — divide hamesha total frequency N se hota hai, number of classes se nahi.
- Continuous distribution mein median class dhoondte waqt cumulative frequency (cf) ko galat class se match karna — woh class leni hai jiska cf sabse pehle N/2 se bada ya barabar ho.
- Class mark (mid-value) nikalte waqt galti se sirf lower limit ya upper limit le lena, jabki formula hamesha (lower + upper)/2 hai.
- Variance nikalne ke baad standard deviation nikalna bhool jaana — SD variance ka positive square root hai, dono alag answers hain aur exam mein dono maange ja sakte hain.
- Step-deviation method mein final formula mein h² multiply karna bhool jaana, ya yi ke sign (assumed mean se pehle/baad) mein galti karna.
Board-Style Important Questions
- Conceptual: Standard deviation ka unit original data ke same hota hai (jaise cm), jabki variance ka unit data ka square hota hai (cm²). Isi wajah se SD zyada practically useful measure maana jaata hai.
- Direct Calculation: Find the mean deviation about the mean for the data: 10, 12, 14, 16, 18. Solution: n = 5, Σxi = 70, x̄ = 14. Absolute deviations: 4, 2, 0, 2, 4 → sum = 12. MD(x̄) = 12/5 = 2.4
- Property-based: If the variance of a data set is 121, find its standard deviation. If every observation is increased by 5, what happens to the variance? Solution: SD = √121 = 11. Adding a constant to every observation shifts the mean but does not change the spread of the data, so variance remains unchanged at 121.
- Frequency Distribution: Find the variance and standard deviation for the discrete frequency distribution: xi: 4, 8, 12, 16, 20 with fi: 4, 6, 10, 6, 4. Solution: N = 30, Σfixi = 360, x̄ = 12. Σfixi² = 5024. Variance = 5024/30 − 144 ≈ 23.47. SD ≈ 4.84
- Comparison (C.V.): Two villages recorded rainfall (in cm) for 10 years. Village A: mean = 100 cm, SD = 8 cm. Village B: mean = 100 cm, SD = 12 cm. Which village has more consistent rainfall, and why is standard deviation a better measure than range for this comparison? Solution: Since means are equal, compare C.V. directly. C.V.(A) = 8%, C.V.(B) = 12%. Since C.V.(A) < C.V.(B), Village A has more consistent rainfall. SD is better than range because it uses every observation in the dataset, while range depends only on the two extreme values.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 13 Statistics mein kitne exercises aur kitne questions hain?
Is chapter mein teen exercises hain — Exercise 13.1 (Mean Deviation — 12 questions), Exercise 13.2 (Variance aur Standard Deviation — 10 questions), aur Miscellaneous Exercise (mixed application — 7 questions). Total 29 questions, jinke step-by-step solutions upar diye gaye hain.
Class 11 Maths syllabus 2026-27 pdf mein Statistics chapter number kya hai?
Current rationalised syllabus (2026-27 session) ke hisaab se Statistics Chapter 13 hai — is se pehle Chapter 12 Limits and Derivatives hai aur baad mein Chapter 14 Probability aata hai. Purane 16-chapter edition mein numbering alag thi kyunki tab Mathematical Induction aur Mathematical Reasoning bhi include the, jo ab syllabus se hata diye gaye hain.
Statistics chapter ke saare formulas ek jagah kahan milenge — koi class 11 maths formulas pdf download option hai?
Is page ke Formulas section mein Range, Mean Deviation, Variance, Standard Deviation, step-deviation method aur Coefficient of Variation — sabhi key formulas ek table mein diye gaye hain, jo quick revision ke liye printable/downloadable format jaisa hi kaam karega.
Statistics ke baad Class 11 Maths ka next chapter kaunsa hai?
Statistics (Chapter 13) ke baad Chapter 14 Probability aata hai — agar aap wahan ke liye practice dhoondh rahe ho to class 11 maths probability important questions is series ka agla logical step hai.
Kya Statistics chapter ke liye class 11 maths exemplar solutions bhi practice karni chahiye?
Haan — NCERT textbook ke 29 questions solid foundation dete hain, lekin exemplar mein thoda tougher application-based aur combined-data questions milte hain (jaise misplaced observation ya combined mean/SD wale, jaisa Miscellaneous Exercise Q2 aur Q4 mein upar dikhaya gaya hai). In dono ko saath practice karna zyada solid prep deta hai.
Mean Deviation aur Standard Deviation mein exam mein kaunsa zyada important hai?
Dono hi important hain, lekin Standard Deviation zyada widely used hai kyunki uski mathematical properties (jaise combined SD, scaling ka effect) aage Class 12 aur competitive exams mein bhi kaam aati hain. Isliye Exercise 13.2 aur Miscellaneous ke SD-based questions (Q2-Q4 jaise) extra practice ke laayak hain.
Class 11 Maths — Saare Chapters
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