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Ch 13Statisticsसांख्यिकीStatistics

Measures of dispersion — range, mean deviation about the mean and median (ungrouped and grouped data), variance and standard deviation with the short-cut method — with every NCERT exercise solved.प्रकीर्णन की माप — परिसर, माध्य और माध्यिका के सापेक्ष माध्य विचलन (अवर्गीकृत और वर्गीकृत आँकड़े), प्रसरण और मानक विचलन, लघु विधि सहित — NCERT के सभी प्रश्नों के हल के साथ।Dispersion ke measures — range, mean aur median ke saapeksh mean deviation (ungrouped aur grouped data), variance aur standard deviation, short-cut method ke saath — NCERT ke saare questions ke hal ke saath.

Didn’t understand a paragraph? Tap it (or its अ button) → हिंदी / Hinglish / Simple English.कोई पैराग्राफ़ समझ नहीं आया? उस पर (या अ बटन पर) टैप करें → English / Hinglish।Koi paragraph samajh nahi aaya? Us par (ya अ button par) tap kijiye → हिंदी / English.

👆 Words with a open a diagram or graph — tap any label to read what it means.👆 वाले शब्द पर टैप करें — चित्र या ग्राफ़ खुलेगा (हर लेबल पर टैप = उसका अर्थ)।👆 wale word par tap kijiye — diagram ya graph khulega (har label par tap = uska matlab).
📘 NCERT chapter (English PDF)📗 NCERT हिंदी PDF∑ Formulas✎ NCERT solutions🎯 Quiz

Diagrams in this chapterइस अध्याय के चित्रIs chapter ke diagrams

Same mean, different spread — why we need measures of dispersionमाध्य एक, फैलाव अलग — प्रकीर्णन की मापों की ज़रूरत क्योंSame mean, different spread — why we need measures of dispersionSections 13.2–13.3Mean deviation: average distance from the mean (Ex 13.1 Q1)माध्य विचलन: माध्य से औसत दूरी (प्रश्नावली 13.1 प्रश्न 1)Mean deviation: average distance from the mean (Ex 13.1 Q1)Section 13.4
In this chapter (NCERT sections)इस अध्याय में (NCERT खंड)Is chapter mein (NCERT sections)
  1. 13.2Measures of Dispersionप्रकीर्णन की मापDispersion ke measures
  2. 13.3RangeपरिसरRange
  3. 13.4Mean Deviationमाध्य विचलनMean deviation
  4. 13.5Variance and Standard Deviationप्रसरण और मानक विचलनVariance aur standard deviation
  5. ★Yaad rakho — key formulasयाद रखो — मुख्य सूत्रYaad rakho — key formulas
  6. ⚠Common mistakesआम गलतियाँCommon mistakes
  7. ✎NCERT exercises — solutionsNCERT प्रश्नावली — हलNCERT exercises — solutions
  8. 🎯Quizक्विज़Quiz

13.2Measures of Dispersionप्रकीर्णन की मापDispersion ke measures

Two data sets can have the same mean but very different spread. Dispersion measures how scattered the observations are about a central value. Measures: range, quartile deviation, mean deviation, standard deviation.दो आँकड़ा समुच्चयों का माध्य समान होकर भी फैलाव बहुत अलग हो सकता है। प्रकीर्णन बताता है कि प्रेक्षण किसी केंद्रीय मान के आसपास कितने बिखरे हैं। मापें: परिसर, चतुर्थक विचलन, माध्य विचलन, मानक विचलन।Do data sets ka mean same hokar bhi spread bahut alag ho sakta hai. Dispersion batata hai ki observations kisi central value ke aas-paas kitne bikhre hain. Measures: range, quartile deviation, mean deviation, standard deviation.

Same mean, different spread — why we need measures of dispersionमाध्य एक, फैलाव अलग — प्रकीर्णन की मापों की ज़रूरत क्योंSame mean, different spread — why we need measures of dispersionSections 13.2–13.3 · 4 tappable partsअनुभाग 13.2–13.3 · 4 टैप करने लायक हिस्सेSections 13.2–13.3 · 4 tappable partsOpen

13.3RangeपरिसरRange

\text{Range} = \text{Max} - \text{Min}\text{परिसर} = \text{अधिकतम} - \text{न्यूनतम}\text{Range} = \text{Max} - \text{Min}
uses only two valuesकेवल दो मानों पर आधारितsirf do values par based

13.4Mean Deviationमाध्य विचलनMean deviation

The mean deviation about a central value a is the mean of the absolute deviations |x_i - a|. It is usually found about the mean \bar{x} or the median M.किसी केंद्रीय मान a के सापेक्ष माध्य विचलन, निरपेक्ष विचलनों |x_i - a| का माध्य है। इसे प्रायः माध्य \bar{x} या माध्यिका M के सापेक्ष निकालते हैं।Kisi central value a ke saapeksh mean deviation, absolute deviations |x_i - a| ka mean hai. Ise aksar mean \bar{x} ya median M ke saapeksh nikaalte hain.

Mean deviation: average distance from the mean (Ex 13.1 Q1)माध्य विचलन: माध्य से औसत दूरी (प्रश्नावली 13.1 प्रश्न 1)Mean deviation: average distance from the mean (Ex 13.1 Q1)Section 13.4 · 4 tappable partsअनुभाग 13.4 · 4 टैप करने लायक हिस्सेSection 13.4 · 4 tappable partsOpen
FormulasUngrouped: \text{M.D.}(a) = \frac{1}{n}\sum |x_i - a|. Grouped (discrete or continuous, x_i = class mark): \text{M.D.}(a) = \frac{1}{N}\sum f_i|x_i - a|, N = \sum f_i.सूत्रअवर्गीकृत: \text{M.D.}(a) = \frac{1}{n}\sum |x_i - a|। वर्गीकृत (असतत या सतत, x_i = वर्ग चिह्न): \text{M.D.}(a) = \frac{1}{N}\sum f_i|x_i - a|, N = \sum f_i।FormulasUngrouped: \text{M.D.}(a) = \frac{1}{n}\sum |x_i - a|. Grouped (discrete ya continuous, x_i = class mark): \text{M.D.}(a) = \frac{1}{N}\sum f_i|x_i - a|, N = \sum f_i.

Median of a continuous distribution: find the class whose cumulative frequency first reaches N/2 (the median class), then M = l + \frac{\frac{N}{2} - C}{f} \times h (l = lower limit, C = cumulative frequency before it, f = its frequency, h = class width).सतत बंटन की माध्यिका: वह वर्ग खोजिए जिसकी संचयी बारंबारता पहली बार N/2 तक पहुँचे (माध्यिका वर्ग), फिर M = l + \frac{\frac{N}{2} - C}{f} \times h (l = निम्न सीमा, C = पिछली संचयी बारंबारता, f = उसकी बारंबारता, h = वर्ग माप)।Continuous distribution ka median: woh class dhoondhiye jiski cumulative frequency pehli baar N/2 tak pahunche (median class), phir M = l + \frac{\frac{N}{2} - C}{f} \times h (l = lower limit, C = pichhli cumulative frequency, f = uski frequency, h = class width).

LimitationAbsolute values cannot be handled algebraically, and when variability is very high the median is not representative — so we move on to standard deviation.सीमानिरपेक्ष मानों पर आगे बीजगणितीय क्रियाएँ नहीं हो पातीं, और बहुत अधिक विचरण में माध्यिका प्रतिनिधि नहीं रहती — इसलिए मानक विचलन।LimitationAbsolute values par aage algebra nahi ho pata, aur bahut zyada variability mein median representative nahi rehta — isliye standard deviation.

13.5Variance and Standard Deviationप्रसरण और मानक विचलनVariance aur standard deviation

\sigma^2 = \frac{1}{n}\sum (x_i - \bar{x})^2,\qquad \sigma = \sqrt{\sigma^2}
ungroupedअवर्गीकृतungrouped
\sigma^2 = \frac{1}{N}\sum f_i (x_i - \bar{x})^2
frequency distributionबारंबारता बंटनfrequency distribution
\sigma^2 = \frac{1}{N^2}\left[N\sum f_i x_i^2 - \left(\sum f_i x_i\right)^2\right]
same, easier to computeवही, गणना में आसानwahi, calculate karna aasaan
Short-cut (step-deviation) methodTake an assumed mean A and width h, put y_i = \frac{x_i - A}{h}. Then \bar{x} = A + h\frac{\sum f_i y_i}{N} and \sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right].लघु (पद-विचलन) विधिकल्पित माध्य A और माप h लेकर y_i = \frac{x_i - A}{h} रखिए। तब \bar{x} = A + h\frac{\sum f_i y_i}{N} और \sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]।Short-cut (step-deviation) methodAssumed mean A aur width h lekar y_i = \frac{x_i - A}{h} rakhiye. Tab \bar{x} = A + h\frac{\sum f_i y_i}{N} aur \sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right].
Example 13The variance of 20 observations is 5. Each is multiplied by 2. Find the new variance.उदाहरण 1320 प्रेक्षणों का प्रसरण 5 है। हर प्रेक्षण को 2 से गुणा किया गया। नया प्रसरण ज्ञात कीजिए।Example 1320 observations ka variance 5 hai. Har observation ko 2 se multiply kiya. Naya variance nikaliye.
Solution — step by step (2)हल — चरण दर चरण (2)Hal — step by step (2)
  1. y_i = 2x_i \Rightarrow \bar{y} = 2\bar{x},\ y_i - \bar{y} = 2(x_i - \bar{x})
  2. \sigma_y^2 = \frac{1}{20}\sum 4(x_i - \bar{x})^2 = 4 \times 5 = 20

In general, adding a constant to every value changes the mean but not the variance; multiplying by a changes the mean to a\bar{x} and the variance to a^2\sigma^2 (SD to |a|\sigma).सामान्यतः हर मान में अचर जोड़ने से माध्य बदलता है, प्रसरण नहीं; a से गुणा करने पर माध्य a\bar{x} और प्रसरण a^2\sigma^2 (मानक विचलन |a|\sigma) हो जाता है।General rule: har value mein constant jodne se mean badalta hai, variance nahi; a se multiply karne par mean a\bar{x} aur variance a^2\sigma^2 (SD |a|\sigma) ho jaata hai.

★ Yaad rakho — key formulas★ याद रखो — मुख्य सूत्र★ Yaad rakho — key formulas

  1. Range = max − min.परिसर = अधिकतम − न्यूनतम।Range = max − min.
  2. \text{M.D.}(a) = \frac{1}{N}\sum f_i|x_i - a|; about the median it is the smallest.\text{M.D.}(a) = \frac{1}{N}\sum f_i|x_i - a|; माध्यिका के सापेक्ष यह सबसे छोटा होता है।\text{M.D.}(a) = \frac{1}{N}\sum f_i|x_i - a|; median ke saapeksh yeh sabse chhota hota hai.
  3. \sigma^2 = \frac{1}{N}\sum f_i x_i^2 - \bar{x}^2 ("mean of squares − square of mean").\sigma^2 = \frac{1}{N}\sum f_i x_i^2 - \bar{x}^2 ("वर्गों का माध्य − माध्य का वर्ग")।\sigma^2 = \frac{1}{N}\sum f_i x_i^2 - \bar{x}^2 ("squares ka mean − mean ka square").
  4. Short-cut: \bar{x} = A + h\bar{y}, \sigma_x = h\,\sigma_y.लघु विधि: \bar{x} = A + h\bar{y}, \sigma_x = h\,\sigma_y।Short-cut: \bar{x} = A + h\bar{y}, \sigma_x = h\,\sigma_y.
  5. First n natural numbers: mean \frac{n + 1}{2}, variance \frac{n^2 - 1}{12}.पहली n प्राकृत संख्याएँ: माध्य \frac{n + 1}{2}, प्रसरण \frac{n^2 - 1}{12}।Pehli n natural numbers: mean \frac{n + 1}{2}, variance \frac{n^2 - 1}{12}.

⚠ Common mistakes⚠ आम गलतियाँ⚠ Common mistakes

  • Forgetting to sort the data before finding the median.माध्यिका निकालने से पहले आँकड़े क्रम में न लगाना।Median nikaalne se pehle data ko order mein na lagana.
  • Dropping the absolute value: Σ(xᵢ − x̄) is always 0, so it cannot measure spread.निरपेक्ष मान छोड़ देना: Σ(xᵢ − x̄) सदा 0 होता है, इससे फैलाव नहीं मपता।Absolute value chhod dena: Σ(xᵢ − x̄) hamesha 0 hota hai, isse spread nahi napta.
  • Using class limits instead of class marks (mid-points) for xᵢ in grouped data.वर्गीकृत आँकड़ों में xᵢ के लिए वर्ग चिह्न (मध्य-बिंदु) की जगह वर्ग सीमाएँ लेना।Grouped data mein xᵢ ke liye class mark (mid-point) ki jagah class limits lena.
  • Not converting inclusive classes (16–20, 21–25) to continuous ones (15.5–20.5) before using the median formula.माध्यिका सूत्र से पहले समावेशी वर्गों (16–20, 21–25) को सतत (15.5–20.5) में न बदलना।Median formula se pehle inclusive classes (16–20, 21–25) ko continuous (15.5–20.5) mein na badalna.
  • In the short-cut method, forgetting the factor h² in the variance (or h in the SD).लघु विधि में प्रसरण में h² (या मानक विचलन में h) का गुणक भूल जाना।Short-cut method mein variance mein h² (ya SD mein h) ka factor bhool jaana.
  • Reporting the variance when the standard deviation was asked (or vice versa).मानक विचलन पूछा हो तब प्रसरण बताना (या उल्टा)।SD poocha ho tab variance batana (ya ulta).

✎ NCERT exercises — solutions✎ NCERT प्रश्नावली — हल✎ NCERT exercises — solutions

All 28 questions of the NCERT exercises. Tap a question to see the solution — try it yourself first!NCERT प्रश्नावली के सभी 28 प्रश्न। हल देखने के लिए प्रश्न पर टैप करें — पहले स्वयं हल करके देखें!NCERT exercises ke saare 28 questions. Hal dekhne ke liye question par tap kijiye — pehle khud try kijiye!

Exercise 13.1प्रश्नावली 13.1Exercise 13.1 · 12 Q

Q9–10: mean deviation about the mean.प्रश्न 9–10: माध्य के सापेक्ष माध्य विचलन।Q9–10: mean ke saapeksh mean deviation.

Q1. 4, 7, 8, 9, 10, 12, 13, 17
  1. \bar{x} = \frac{80}{8} = 10
  2. |x_i - \bar{x}|:\ 6, 3, 2, 1, 0, 2, 3, 7;\ \ \sum = 24
  3. \text{M.D.}(\bar{x}) = \frac{24}{8} = 3
Q2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
  1. \bar{x} = \frac{500}{10} = 50
  2. |x_i - \bar{x}|:\ 12, 20, 2, 10, 8, 5, 13, 4, 4, 6;\ \ \sum = 84
  3. \text{M.D.}(\bar{x}) = \frac{84}{10} = 8.4
Q3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
  1. \text{sorted: } 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18\text{क्रम में: } 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18\text{order mein: } 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
  2. M = \frac{13 + 14}{2} = 13.5
  3. |x_i - 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
  4. \text{M.D.}(M) = \frac{28}{12} \approx 2.33
Q4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
  1. \text{sorted: } 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 \Rightarrow M = \frac{46 + 49}{2} = 47.5\text{क्रम में: } 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 \Rightarrow M = \frac{46 + 49}{2} = 47.5\text{order mein: } 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 \Rightarrow M = \frac{46 + 49}{2} = 47.5
  2. |x_i - M|:\ 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5;\ \sum = 70
  3. \text{M.D.}(M) = \frac{70}{10} = 7
Q5. xᵢ: 5, 10, 15, 20, 25; fᵢ: 7, 4, 6, 3, 5
  1. N = 25,\ \sum f_ix_i = 350,\ \bar{x} = 14
  2. |x_i - 14|:\ 9, 4, 1, 6, 11;\ \ \sum f_i|x_i - \bar{x}| = 63 + 16 + 6 + 18 + 55 = 158
  3. \text{M.D.} = \frac{158}{25} = 6.32
Q6. xᵢ: 10, 30, 50, 70, 90; fᵢ: 4, 24, 28, 16, 8
  1. N = 80,\ \sum f_ix_i = 4000,\ \bar{x} = 50
  2. |x_i - 50|:\ 40, 20, 0, 20, 40;\ \ \sum f_i|x_i - \bar{x}| = 160 + 480 + 0 + 320 + 320 = 1280
  3. \text{M.D.} = \frac{1280}{80} = 16
Q7. xᵢ: 5, 7, 9, 10, 12, 15; fᵢ: 8, 6, 2, 2, 2, 6
  1. N = 26;\ c.f.:\ 8, 14, 16, 18, 20, 26
  2. 13^{th}\ \&\ 14^{th} \text{ values} = 7 \Rightarrow M = 713\text{वाँ व } 14\text{वाँ मान} = 7 \Rightarrow M = 713^{th}\ \&\ 14^{th} \text{ values} = 7 \Rightarrow M = 7
  3. |x_i - 7|:\ 2, 0, 2, 3, 5, 8;\ \sum f_i|x_i - M| = 16 + 0 + 4 + 6 + 10 + 48 = 84
  4. \text{M.D.} = \frac{84}{26} \approx 3.23
Q8. xᵢ: 15, 21, 27, 30, 35; fᵢ: 3, 5, 6, 7, 8
  1. N = 29;\ c.f.:\ 3, 8, 14, 21, 29
  2. 15^{th} \text{ value} = 30 \Rightarrow M = 3015\text{वाँ मान} = 30 \Rightarrow M = 3015^{th} \text{ value} = 30 \Rightarrow M = 30
  3. |x_i - 30|:\ 15, 9, 3, 0, 5;\ \sum f_i|x_i - M| = 45 + 45 + 18 + 0 + 40 = 148
  4. \text{M.D.} = \frac{148}{29} \approx 5.1
Q9. Income/day (₹): 0–100, 100–200, …, 700–800; persons: 4, 8, 9, 10, 7, 5, 4, 3दैनिक आय (₹): 0–100, 100–200, …, 700–800; व्यक्ति: 4, 8, 9, 10, 7, 5, 4, 3Income/day (₹): 0–100, 100–200, …, 700–800; log: 4, 8, 9, 10, 7, 5, 4, 3
  1. x_i = 50, 150, \dots, 750;\ N = 50,\ \sum f_ix_i = 17900,\ \bar{x} = 358
  2. |x_i - 358|:\ 308, 208, 108, 8, 92, 192, 292, 392
  3. \sum f_i|x_i - \bar{x}| = 7896 \Rightarrow \text{M.D.} = \frac{7896}{50} = 157.92
Q10. Height (cm): 95–105, 105–115, …, 145–155; boys: 9, 13, 26, 30, 12, 10ऊँचाई (cm): 95–105, 105–115, …, 145–155; लड़के: 9, 13, 26, 30, 12, 10Height (cm): 95–105, 105–115, …, 145–155; ladke: 9, 13, 26, 30, 12, 10
  1. x_i = 100, 110, \dots, 150;\ N = 100,\ \sum f_ix_i = 12530,\ \bar{x} = 125.3
  2. |x_i - 125.3|:\ 25.3, 15.3, 5.3, 4.7, 14.7, 24.7
  3. \sum f_i|x_i - \bar{x}| = 1128.8 \Rightarrow \text{M.D.} = 11.288 \approx 11.28
Q11. Mean deviation about the median — Marks: 0–10, …, 50–60; girls: 6, 8, 14, 16, 4, 2माध्यिका के सापेक्ष माध्य विचलन — अंक: 0–10, …, 50–60; लड़कियाँ: 6, 8, 14, 16, 4, 2Median ke saapeksh mean deviation — Marks: 0–10, …, 50–60; ladkiyan: 6, 8, 14, 16, 4, 2
  1. N = 50,\ c.f.:\ 6, 14, 28, 44, 48, 50;\ \frac{N}{2} = 25 \Rightarrow 20\text{–}30
  2. M = 20 + \frac{25 - 14}{14} \times 10 = 27.86
  3. |x_i - M|\ (x_i = 5, 15, \dots, 55):\ 22.86, 12.86, 2.86, 7.14, 17.14, 27.14
  4. \sum f_i|x_i - M| = 517.1 \Rightarrow \text{M.D.} = \frac{517.1}{50} \approx 10.34
Q12. Mean deviation about the median age — Age: 16–20, 21–25, …, 51–55; persons: 5, 6, 12, 14, 26, 12, 16, 9माध्यिका आयु के सापेक्ष माध्य विचलन — आयु: 16–20, 21–25, …, 51–55; व्यक्ति: 5, 6, 12, 14, 26, 12, 16, 9Median age ke saapeksh mean deviation — Age: 16–20, 21–25, …, 51–55; log: 5, 6, 12, 14, 26, 12, 16, 9
  1. \text{continuous: } 15.5\text{–}20.5, \dots, 50.5\text{–}55.5;\ N = 100\text{सतत: } 15.5\text{–}20.5, \dots, 50.5\text{–}55.5;\ N = 100\text{continuous: } 15.5\text{–}20.5, \dots, 50.5\text{–}55.5;\ N = 100
  2. c.f.:\ 5, 11, 23, 37, 63, \dots;\ \frac{N}{2} = 50 \Rightarrow 35.5\text{–}40.5
  3. M = 35.5 + \frac{50 - 37}{26} \times 5 = 38
  4. |x_i - 38|\ (x_i = 18, 23, \dots, 53):\ 20, 15, 10, 5, 0, 5, 10, 15
  5. \sum f_i|x_i - M| = 735 \Rightarrow \text{M.D.} = 7.35

Exercise 13.2प्रश्नावली 13.2Exercise 13.2 · 10 Q

Q7–8: mean and variance of the frequency distribution.प्रश्न 7–8: बारंबारता बंटन का माध्य और प्रसरण।Q7–8: frequency distribution ka mean aur variance.

Q1. 6, 7, 10, 12, 13, 4, 8, 12
  1. \bar{x} = \frac{72}{8} = 9
  2. (x_i - 9)^2:\ 9, 4, 1, 9, 16, 25, 1, 9;\ \ \sum = 74
  3. \sigma^2 = \frac{74}{8} = 9.25
Q2. First n natural numbersपहली n प्राकृत संख्याएँPehli n natural numbers
  1. \bar{x} = \frac{n(n + 1)/2}{n} = \frac{n + 1}{2}
  2. \sigma^2 = \frac{1}{n}\cdot\frac{n(n + 1)(2n + 1)}{6} - \left(\frac{n + 1}{2}\right)^2 = \frac{(n + 1)(n - 1)}{12} = \frac{n^2 - 1}{12}
Q3. First 10 multiples of 33 के पहले 10 गुणज3 ke pehle 10 multiples
  1. 3, 6, \dots, 30:\ \bar{x} = \frac{165}{10} = 16.5
  2. \sigma^2 = 9 \times \frac{10^2 - 1}{12} = 9 \times 8.25 = 74.25
Q4. xᵢ: 6, 10, 14, 18, 24, 28, 30; fᵢ: 2, 4, 7, 12, 8, 4, 3
  1. N = 40,\ \sum f_ix_i = 760,\ \bar{x} = 19
  2. \sum f_i(x_i - 19)^2 = 338 + 324 + 175 + 12 + 200 + 324 + 363 = 1736
  3. \sigma^2 = \frac{1736}{40} = 43.4
Q5. xᵢ: 92, 93, 97, 98, 102, 104, 109; fᵢ: 3, 2, 3, 2, 6, 3, 3
  1. N = 22,\ \sum f_ix_i = 2200,\ \bar{x} = 100
  2. \sum f_i(x_i - 100)^2 = 192 + 98 + 27 + 8 + 24 + 48 + 243 = 640
  3. \sigma^2 = \frac{640}{22} \approx 29.09
Q6. Mean and SD by short-cut — xᵢ: 60, 61, …, 68; fᵢ: 2, 1, 12, 29, 25, 12, 10, 4, 5लघु विधि से माध्य और मानक विचलन — xᵢ: 60, 61, …, 68; fᵢ: 2, 1, 12, 29, 25, 12, 10, 4, 5Short-cut se mean aur SD — xᵢ: 60, 61, …, 68; fᵢ: 2, 1, 12, 29, 25, 12, 10, 4, 5
  1. A = 64,\ y_i = x_i - 64:\ -4, \dots, 4;\ N = 100
  2. \sum f_iy_i = 0,\ \sum f_iy_i^2 = 286
  3. \bar{x} = 64 + 0 = 64,\quad \sigma^2 = \frac{286}{100} - 0 = 2.86 \Rightarrow \sigma \approx 1.69
Q7. Classes 0–30, 30–60, …, 180–210; f: 2, 3, 5, 10, 3, 5, 2वर्ग 0–30, 30–60, …, 180–210; f: 2, 3, 5, 10, 3, 5, 2Classes 0–30, 30–60, …, 180–210; f: 2, 3, 5, 10, 3, 5, 2
  1. x_i = 15, 45, \dots, 195;\ N = 30,\ \sum f_ix_i = 3210,\ \bar{x} = 107
  2. \sum f_i(x_i - 107)^2 = 68280
  3. \sigma^2 = \frac{68280}{30} = 2276
Q8. Classes 0–10, …, 40–50; f: 5, 8, 15, 16, 6वर्ग 0–10, …, 40–50; f: 5, 8, 15, 16, 6Classes 0–10, …, 40–50; f: 5, 8, 15, 16, 6
  1. x_i = 5, 15, 25, 35, 45;\ N = 50,\ \sum f_ix_i = 1350,\ \bar{x} = 27
  2. \sum f_i(x_i - 27)^2 = 2420 + 1152 + 60 + 1024 + 1944 = 6600
  3. \sigma^2 = \frac{6600}{50} = 132
Q9. Mean, variance and SD by short-cut — Height 70–75, …, 110–115; children: 3, 4, 7, 7, 15, 9, 6, 6, 3लघु विधि से माध्य, प्रसरण, मानक विचलन — ऊँचाई 70–75, …, 110–115; बच्चे: 3, 4, 7, 7, 15, 9, 6, 6, 3Short-cut se mean, variance, SD — Height 70–75, …, 110–115; bachche: 3, 4, 7, 7, 15, 9, 6, 6, 3
  1. A = 92.5,\ h = 5,\ y_i = -4, \dots, 4;\ N = 60
  2. \sum f_iy_i = 6,\ \sum f_iy_i^2 = 254
  3. \bar{x} = 92.5 + 5 \cdot \frac{6}{60} = 93
  4. \sigma^2 = \frac{25}{3600}\left[60 \cdot 254 - 6^2\right] = 105.58,\quad \sigma \approx 10.27
Q10. Diameters 33–36, 37–40, 41–44, 45–48, 49–52 (mm); circles: 15, 17, 21, 22, 25. Find SD and mean diameter.व्यास 33–36, 37–40, 41–44, 45–48, 49–52 (mm); वृत्त: 15, 17, 21, 22, 25। मानक विचलन और माध्य व्यास ज्ञात कीजिए।Diameters 33–36, 37–40, 41–44, 45–48, 49–52 (mm); circles: 15, 17, 21, 22, 25. SD aur mean diameter nikaliye.
  1. x_i = 34.5, 38.5, 42.5, 46.5, 50.5;\ A = 42.5,\ h = 4;\ N = 100
  2. \sum f_iy_i = 25,\ \sum f_iy_i^2 = 199
  3. \bar{x} = 42.5 + 4 \cdot \frac{25}{100} = 43.5
  4. \sigma^2 = 16\left[\frac{199}{100} - \left(\frac{25}{100}\right)^2\right] = 30.84 \Rightarrow \sigma \approx 5.55

Miscellaneous Exercise on Chapter 13अध्याय 13 पर विविध प्रश्नावलीMiscellaneous Exercise on Chapter 13 · 6 Q

Q1. Mean and variance of 8 observations are 9 and 9.25. Six of them are 6, 7, 10, 12, 12, 13. Find the other two.8 प्रेक्षणों का माध्य और प्रसरण 9 और 9.25 हैं। इनमें छह 6, 7, 10, 12, 12, 13 हैं। शेष दो ज्ञात कीजिए।8 observations ka mean aur variance 9 aur 9.25 hain. Inmein chhe 6, 7, 10, 12, 12, 13 hain. Baaki do nikaliye.
  1. x + y = 72 - 60 = 12
  2. \sum x_i^2 = 8(9.25 + 81) = 722 \Rightarrow x^2 + y^2 = 722 - 642 = 80
  3. xy = \frac{144 - 80}{2} = 32 \Rightarrow \{x, y\} = \{4, 8\}

✔ 4 and 84 और 84 aur 8

Q2. Mean and variance of 7 observations are 8 and 16. Five are 2, 4, 10, 12, 14. Find the other two.7 प्रेक्षणों का माध्य और प्रसरण 8 और 16 हैं। पाँच प्रेक्षण 2, 4, 10, 12, 14 हैं। शेष दो ज्ञात कीजिए।7 observations ka mean aur variance 8 aur 16 hain. Paanch 2, 4, 10, 12, 14 hain. Baaki do nikaliye.
  1. x + y = 56 - 42 = 14
  2. \sum x_i^2 = 7(16 + 64) = 560 \Rightarrow x^2 + y^2 = 560 - 460 = 100
  3. xy = \frac{196 - 100}{2} = 48 \Rightarrow \{x, y\} = \{6, 8\}

✔ 6 and 86 और 86 aur 8

Q3. Mean and SD of six observations are 8 and 4. Each is multiplied by 3. New mean and SD?छह प्रेक्षणों का माध्य और मानक विचलन 8 और 4 हैं। हर प्रेक्षण को 3 से गुणा किया। नया माध्य और मानक विचलन?Chhe observations ka mean aur SD 8 aur 4 hain. Har observation ko 3 se multiply kiya. Naya mean aur SD?
  1. \bar{y} = 3 \times 8 = 24,\quad \sigma_y = 3 \times 4 = 12
Q4. If x̄ and σ² are the mean and variance of x₁, …, xₙ, prove that ax₁, …, axₙ have mean ax̄ and variance a²σ² (a ≠ 0).यदि x₁, …, xₙ का माध्य x̄ और प्रसरण σ² है, तो सिद्ध कीजिए कि ax₁, …, axₙ का माध्य ax̄ और प्रसरण a²σ² है (a ≠ 0)।Agar x₁, …, xₙ ka mean x̄ aur variance σ² hai, to prove kijiye ki ax₁, …, axₙ ka mean ax̄ aur variance a²σ² hai (a ≠ 0).
  1. \bar{y} = \frac{1}{n}\sum ax_i = a\bar{x}
  2. \sigma_y^2 = \frac{1}{n}\sum (ax_i - a\bar{x})^2 = a^2 \cdot \frac{1}{n}\sum (x_i - \bar{x})^2 = a^2\sigma^2
Q5. Mean and SD of 20 observations are 10 and 2. An observation 8 was wrong. Find the correct mean and SD if (i) it is omitted (ii) it is replaced by 12.20 प्रेक्षणों का माध्य और मानक विचलन 10 और 2 हैं। एक प्रेक्षण 8 गलत था। सही माध्य और मानक विचलन ज्ञात कीजिए यदि (i) इसे हटा दें (ii) इसकी जगह 12 रखें।20 observations ka mean aur SD 10 aur 2 hain. Ek observation 8 galat tha. Sahi mean aur SD nikaliye agar (i) ise hata dein (ii) iski jagah 12 rakhein.
  1. \sum x_i = 200,\quad \sum x_i^2 = 20(4 + 100) = 2080
  2. (i)\ n = 19:\ \bar{x} = \frac{192}{19} \approx 10.1,\ \sigma^2 = \frac{2016}{19} - \left(\frac{192}{19}\right)^2 \approx 3.99,\ \sigma \approx 1.997
  3. (ii)\ \sum x_i = 204,\ \sum x_i^2 = 2160:\ \bar{x} = 10.2,\ \sigma^2 = 108 - 104.04 = 3.96,\ \sigma \approx 1.990

✔ (i) 10.1, ≈ 1.99 (ii) 10.2, ≈ 1.99. (The NCERT answer key prints 1.98 for (ii) by cutting off √3.96 = 1.98997… instead of rounding.)(i) 10.1, ≈ 1.99 (ii) 10.2, ≈ 1.99। (NCERT उत्तरमाला (ii) में √3.96 = 1.98997… को पूर्णांकित करने के बजाय काटकर 1.98 छापती है।)(i) 10.1, ≈ 1.99 (ii) 10.2, ≈ 1.99. (NCERT answer key (ii) mein √3.96 = 1.98997… ko round karne ki jagah kaatkar 1.98 chhaapti hai.)

Q6. Mean and SD of 100 observations are 20 and 3. Three wrong values 21, 21, 18 are omitted. Find the new mean and SD.100 प्रेक्षणों का माध्य और मानक विचलन 20 और 3 हैं। तीन गलत मान 21, 21, 18 हटा दिए गए। नया माध्य और मानक विचलन ज्ञात कीजिए।100 observations ka mean aur SD 20 aur 3 hain. Teen galat values 21, 21, 18 hata di gayi. Naya mean aur SD nikaliye.
  1. \sum x_i = 2000,\quad \sum x_i^2 = 100(9 + 400) = 40900
  2. n = 97:\ \sum x_i = 2000 - 60 = 1940 \Rightarrow \bar{x} = 20
  3. \sum x_i^2 = 40900 - 1206 = 39694;\ \sigma^2 = \frac{39694}{97} - 400 \approx 9.216 \Rightarrow \sigma \approx 3.036

🎯 Quiz — 12 questions🎯 क्विज़ — 12 प्रश्न🎯 Quiz — 12 questions

Tap an option. The answer and a short reason appear at once.किसी विकल्प पर टैप करें। उत्तर और छोटा कारण तुरंत दिखेगा।Kisi option par tap kijiye. Answer aur chhota reason turant dikhega.

1. The range of 7, 3, 12, 9, 5 is7, 3, 12, 9, 5 का परिसर है7, 3, 12, 9, 5 ka range hai

2. Mean deviation about the mean of 2, 4, 6, 8, 10 is2, 4, 6, 8, 10 का माध्य के सापेक्ष माध्य विचलन है2, 4, 6, 8, 10 ka mean ke saapeksh mean deviation hai

3. Σ(xᵢ − x̄) for any data equalsकिसी भी आँकड़े के लिए Σ(xᵢ − x̄) बराबर हैKisi bhi data ke liye Σ(xᵢ − x̄) barabar hai

4. The median of 3, 8, 5, 10, 7, 4 is3, 8, 5, 10, 7, 4 की माध्यिका है3, 8, 5, 10, 7, 4 ka median hai

5. The variance of 1, 2, 3, 4, 5 is1, 2, 3, 4, 5 का प्रसरण है1, 2, 3, 4, 5 ka variance hai

6. Standard deviation isमानक विचलन हैStandard deviation hai

7. If each observation is increased by 5, the varianceयदि हर प्रेक्षण में 5 जोड़ दें, तो प्रसरणAgar har observation mein 5 jod dein, to variance

8. The SD of some data is 3. If each value is multiplied by 4, the new SD isकिसी आँकड़े का मानक विचलन 3 है। हर मान को 4 से गुणा करने पर नया मानक विचलन हैKisi data ka SD 3 hai. Har value ko 4 se multiply karne par naya SD hai

9. For grouped data, xᵢ in the formulas is theवर्गीकृत आँकड़ों में सूत्रों का xᵢ हैGrouped data mein formulas ka xᵢ hai

10. σ² equalsσ² बराबर हैσ² barabar hai

11. Mean = 10 and Σxᵢ² = 1100 for 10 observations. The variance is10 प्रेक्षणों के लिए माध्य = 10 और Σxᵢ² = 1100। प्रसरण है10 observations ke liye mean = 10 aur Σxᵢ² = 1100. Variance hai

12. In the short-cut method with yᵢ = (xᵢ − A)/h, σₓ equalsyᵢ = (xᵢ − A)/h वाली लघु विधि में σₓ बराबर हैyᵢ = (xᵢ − A)/h wale short-cut method mein σₓ barabar hai

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These are Simple Learning Hub's own study notes and solutions, written to follow the NCERT Class 11 Mathematics textbook (Reprint 2026-27) section by section; read them with the official NCERT chapter. Diagrams are original SLH drawings.ये Simple Learning Hub के अपने स्टडी नोट्स हैं, जो NCERT कक्षा 11 गणित की किताब (2026-27) के हर अनुभाग के साथ लिखे गए हैं; इन्हें NCERT के आधिकारिक अध्याय के साथ पढ़िए। डायग्राम SLH के अपने बनाए चित्र हैं।Ye Simple Learning Hub ke apne study notes hain, jo NCERT Class 11 Maths ki kitaab (2026-27) ke har section ke saath likhe gaye hain; inhe NCERT ke official chapter ke saath padhiye. Diagrams SLH ki apni drawings hain.