Class 10 Maths · Chapter 13
Short answer:
Is chapter me humein grouped data ka mean, median, aur mode nikalna seekhna hai — yani jab data ranges (class intervals) me diya ho, individual values me nahi. Exercise 13.1 me mean ke teen tareeke (direct, assumed mean, step-deviation) cover hote hain, Exercise 13.2 me mode ka formula, aur Exercise 13.3 me median ka formula with cumulative frequency table. Total milakar teeno exercises me karib 16 questions hain jo neeche is spec me solve kiye gaye hain. Note: is current rationalised book me purana graphical representation (ogive) wala portion — less-than curve, more-than curve, aur median graph se nikalna — hata diya gaya hai, isliye yahan sirf numerical (formula-based) tareeke cover honge.
Grouped data me exact value pata nahi hoti, sirf range pata hoti hai — isliye mean/median/mode nikalne ke liye class mark (mid-value) use karte hain, formulas ke through. Jaise agar bola jaye "25 students ne 20-30 range me marks liye", toh humein exact pata nahi kis student ne kitna score kiya — bas itna pata hai ki 20 se 30 ke beech hai. Is situation me humein class mark (midpoint) ko us poore group ka representative maan kar calculation karni padti hai. Grouped data me mean nikalne ke teen tareeke hain — direct method, assumed mean method, aur step deviation method — teeno same answer dete hain, bas calculation ka tareeka alag hai. Numbers bade hon (jaise 100, 500) to step deviation sabse aasan padta hai kyunki usme deviations ko class-width se divide karke chhote numbers bana lete hain.
Chapter 13 Summary — 5 Minute Revision
1. Mean of Grouped Data
Teen methods hain — teeno ka answer same aata hai, bas calculation ka style alag hai.
(a) Direct Method
| Formula | Mean (x̄) = Σfᵢxᵢ / Σfᵢ |
|---|---|
| xᵢ | class mark = (upper limit + lower limit) / 2 |
↔ Table ko side me swipe karein
Har class ka class mark nikalo, frequency se multiply karo, sab jod do, aur total frequency se divide kar do.
(b) Assumed Mean Method
| Formula | Mean (x̄) = a + (Σfᵢdᵢ / Σfᵢ) |
|---|---|
| a | assumed mean — kisi bhi class ka class mark (usually beech wali class jisme calculation aasan ho) |
| dᵢ | = xᵢ − a (deviation) |
↔ Table ko side me swipe karein
Jab class marks bade numbers hon (jaise 45, 65, 85...), toh directly multiply karne ke bajaye ek assumed mean 'a' choose karke deviation nikalna aasan padta hai.
(c) Step Deviation Method
| Formula | Mean (x̄) = a + h × (Σfᵢuᵢ / Σfᵢ) |
|---|---|
| h | class width (uniform hona chahiye) |
| uᵢ | = (xᵢ − a) / h = dᵢ / h |
↔ Table ko side me swipe karein
Assumed mean method jaisa hi hai, bas deviation ko class-width se divide karke aur chhota bana dete hain. Sabse cleanest tareeka bade data ke liye.
2. Mode of Grouped Data
| Formula | Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h |
|---|---|
| Modal class | sabse zyada frequency wali class |
| l | lower limit of modal class |
| f₁ | frequency of modal class |
| f₀ | frequency of class jo modal class se pehle aati hai (preceding) |
| f₂ | frequency of class jo modal class ke baad aati hai (succeeding) |
| h | class width of modal class |
↔ Table ko side me swipe karein
Sabse pehle sabse zyada frequency wali class dhoondo — wahi modal class hai. Agar modal class sabse pehli class ho, to f₀ = 0 le lo.
3. Median of Grouped Data
| Formula | Median = l + [(n/2 − cf) / f] × h |
|---|---|
| n | Σfᵢ (total frequency) |
| Median class | wo class jiski cumulative frequency (cf) sabse pehle n/2 ke barabar ya usse zyada ho jaye |
| l | lower limit of median class |
| cf | cumulative frequency of the class before median class |
| f | frequency of median class (only median class ka, cumulative nahi) |
| h | class width of median class |
↔ Table ko side me swipe karein
Pehle cumulative frequency (cf) column banao. n/2 nikalo. Jis class ki cf pehli baar n/2 se >= ho jaye, wo median class hai.
4. Empirical Relationship (Mean, Median, Mode)
| Formula | Mode = 3 × Median − 2 × Mean |
|---|
↔ Table ko side me swipe karein
Ye ek approximate relation hai (moderately skewed distribution ke liye), exact nahi. Agar do value pata ho (jaise mean aur median), to teesri (mode) is formula se estimate ki ja sakti hai.
5. Class Intervals ka Type — Inclusive vs Exclusive
Agar data inclusive form (jaise 1-10, 11-20) me diya ho, to median/mean nikalne se pehle use exclusive form (0.5-10.5, 10.5-20.5) me convert karna padta hai — boundary ko 0.5 se adjust karke. Mode aur mean formula me class mark automatically theek ho jata hai, lekin median formula me continuity zaroori hai.
Poore Class 10 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–Q16)
Q1 (Ex 13.1). Following table shows the number of plants in 20 houses in a locality. Find the mean number of plants per house by the direct method.
Number of plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14 Number of houses 1 2 1 5 6 2 3
↔ Table ko side me swipe karein
| Number of plants | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 |
|---|---|---|---|---|---|---|---|
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
| Class | xᵢ (class mark) | fᵢ | fᵢxᵢ |
|---|---|---|---|
| 0-2 | 1 | 1 | 1 |
| 2-4 | 3 | 2 | 6 |
| 4-6 | 5 | 1 | 5 |
| 6-8 | 7 | 5 | 35 |
| 8-10 | 9 | 6 | 54 |
| 10-12 | 11 | 2 | 22 |
| 12-14 | 13 | 3 | 39 |
| Total | — | Σfᵢ = 20 | Σfᵢxᵢ = 162 |
↔ Table ko side me swipe karein
Mean = Σfᵢxᵢ / Σfᵢ = 162 / 20 = 8.1
Mean number of plants per house = 8.1
Q2 (Ex 13.1). Marks obtained by 50 students in a test are grouped below. Find the mean marks using the assumed mean method.
Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 5 9 15 12 6 3
↔ Table ko side me swipe karein
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| No. of students | 5 | 9 | 15 | 12 | 6 | 3 |
Assumed mean a = 25 (class mark of 20-30).
| Class | xᵢ | fᵢ | dᵢ = xᵢ−a | fᵢdᵢ |
|---|---|---|---|---|
| 0-10 | 5 | 5 | -20 | -100 |
| 10-20 | 15 | 9 | -10 | -90 |
| 20-30 | 25 | 15 | 0 | 0 |
| 30-40 | 35 | 12 | 10 | 120 |
| 40-50 | 45 | 6 | 20 | 120 |
| 50-60 | 55 | 3 | 30 | 90 |
| Total | — | 50 | — | Σfᵢdᵢ = 140 |
↔ Table ko side me swipe karein
Mean = a + Σfᵢdᵢ/Σfᵢ = 25 + 140/50 = 25 + 2.8 = 27.8
Mean marks = 27.8
Q3 (Ex 13.1). Using the same marks data as Q2, find the mean using the step-deviation method and verify it matches Q2.
a = 25, h = 10.
| Class | xᵢ | fᵢ | uᵢ = (xᵢ−a)/h | fᵢuᵢ |
|---|---|---|---|---|
| 0-10 | 5 | 5 | -2 | -10 |
| 10-20 | 15 | 9 | -1 | -9 |
| 20-30 | 25 | 15 | 0 | 0 |
| 30-40 | 35 | 12 | 1 | 12 |
| 40-50 | 45 | 6 | 2 | 12 |
| 50-60 | 55 | 3 | 3 | 9 |
| Total | — | 50 | — | Σfᵢuᵢ = 14 |
↔ Table ko side me swipe karein
Mean = a + h × (Σfᵢuᵢ/Σfᵢ) = 25 + 10 × (14/50) = 25 + 2.8 = 27.8
Mean = 27.8 — same as Q2's assumed mean method. Yahi teeno methods ka fayda hai: answer hamesha same aata hai.
Q4 (Ex 13.1). The following distribution shows marks of students with one missing frequency f. If the mean is 50, find the missing frequency f and the total number of students.
Marks 0-20 20-40 40-60 60-80 80-100 No. of students 5 f 10 6 3
↔ Table ko side me swipe karein
| Marks | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
|---|---|---|---|---|---|
| No. of students | 5 | f | 10 | 6 | 3 |
a = 50 (class mark of 40-60), h = 20.
| Class | xᵢ | fᵢ | dᵢ = xᵢ−a | fᵢdᵢ |
|---|---|---|---|---|
| 0-20 | 10 | 5 | -40 | -200 |
| 20-40 | 30 | f | -20 | -20f |
| 40-60 | 50 | 10 | 0 | 0 |
| 60-80 | 70 | 6 | 20 | 120 |
| 80-100 | 90 | 3 | 40 | 120 |
| Total | — | 24+f | — | 40 − 20f |
↔ Table ko side me swipe karein
Mean = a + Σfᵢdᵢ/Σfᵢ ⟹ 50 = 50 + (40 − 20f)/(24 + f)
0 = (40 − 20f)/(24 + f) ⟹ 40 − 20f = 0 ⟹ f = 2
Missing frequency f = 2. Total students = 24 + f = 26.
Q5 (Ex 13.1). Weekly transport expenditure (in ₹) of 100 employees is given below. Find the mean expenditure using the step-deviation method.
Expenditure (₹) 100-150 150-200 200-250 250-300 300-350 No. of employees 5 25 32 20 18
↔ Table ko side me swipe karein
| Expenditure (₹) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
|---|---|---|---|---|---|
| No. of employees | 5 | 25 | 32 | 20 | 18 |
a = 225 (class mark of 200-250), h = 50.
| Class | xᵢ | fᵢ | uᵢ | fᵢuᵢ |
|---|---|---|---|---|
| 100-150 | 125 | 5 | -2 | -10 |
| 150-200 | 175 | 25 | -1 | -25 |
| 200-250 | 225 | 32 | 0 | 0 |
| 250-300 | 275 | 20 | 1 | 20 |
| 300-350 | 325 | 18 | 2 | 36 |
| Total | — | 100 | — | Σfᵢuᵢ = 21 |
↔ Table ko side me swipe karein
Mean = a + h × (Σfᵢuᵢ/Σfᵢ) = 225 + 50 × (21/100) = 225 + 10.5 = 235.5
Mean weekly expenditure = ₹235.5
Q6 (Ex 13.2). Ages of patients admitted in a hospital in one day are given below. Find the mode of the data.
Age (years) 5-15 15-25 25-35 35-45 45-55 55-65 No. of patients 6 11 21 23 14 5
↔ Table ko side me swipe karein
| Age (years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |
|---|---|---|---|---|---|---|
| No. of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Highest frequency = 23, so modal class = 35-45.
l = 35, f₁ = 23, f₀ = 21, f₂ = 14, h = 10
Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h
= 35 + [(23 − 21)/(46 − 21 − 14)] × 10 = 35 + (2/11) × 10 = 35 + 1.82
Mode ≈ 36.8 years
Q7 (Ex 13.2). Ages of teachers in schools of a state are given. Find the modal age.
Age 20-25 25-30 30-35 35-40 40-45 45-50 50-55 No. of teachers 10 20 60 32 15 8 5
↔ Table ko side me swipe karein
| Age | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 |
|---|---|---|---|---|---|---|---|
| No. of teachers | 10 | 20 | 60 | 32 | 15 | 8 | 5 |
Highest frequency = 60, so modal class = 30-35.
l = 30, f₁ = 60, f₀ = 20, f₂ = 32, h = 5
Mode = 30 + [(60 − 20)/(120 − 20 − 32)] × 5 = 30 + (40/68) × 5 = 30 + 2.94
Modal age ≈ 32.9 years
Q8 (Ex 13.2). Find the mode of the following distribution where the modal class is the first class interval.
Class 0-10 10-20 20-30 30-40 Frequency 30 20 15 10
↔ Table ko side me swipe karein
| Class | 0-10 | 10-20 | 20-30 | 30-40 |
|---|---|---|---|---|
| Frequency | 30 | 20 | 15 | 10 |
Highest frequency = 30 → modal class = 0-10, jo pehli class hai, isliye f₀ = 0 (koi preceding class nahi hai).
l = 0, f₁ = 30, f₀ = 0, f₂ = 20, h = 10
Mode = 0 + [(30 − 0)/(60 − 0 − 20)] × 10 = (30/40) × 10 = 7.5
Mode = 7.5
Q9 (Ex 13.2). Literacy rate (%) of villages is grouped below. Find the modal literacy rate.
Literacy rate (%) 35-40 40-45 45-50 50-55 55-60 60-65 65-70 No. of villages 3 5 9 12 8 4 2
↔ Table ko side me swipe karein
| Literacy rate (%) | 35-40 | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 |
|---|---|---|---|---|---|---|---|
| No. of villages | 3 | 5 | 9 | 12 | 8 | 4 | 2 |
Highest frequency = 12 → modal class = 50-55.
l = 50, f₁ = 12, f₀ = 9, f₂ = 8, h = 5
Mode = 50 + [(12 − 9)/(24 − 9 − 8)] × 5 = 50 + (3/7) × 5 = 50 + 2.14
Modal literacy rate ≈ 52.1%
Q10 (Ex 13.2). For a distribution, the mean is 26.8 and the median is 27. Estimate the mode using the empirical relationship.
Mode = 3 × Median − 2 × Mean = 3(27) − 2(26.8) = 81 − 53.6 = 27.4
Estimated mode = 27.4
Q11 (Ex 13.3). The weights (in kg) of 30 students are grouped below. Find the median weight.
Weight (kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75 No. of students 2 3 8 6 6 3 2
↔ Table ko side me swipe karein
| Weight (kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |
|---|---|---|---|---|---|---|---|
| No. of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
| Class | fᵢ | cf |
|---|---|---|
| 40-45 | 2 | 2 |
| 45-50 | 3 | 5 |
| 50-55 | 8 | 13 |
| 55-60 | 6 | 19 |
| 60-65 | 6 | 25 |
| 65-70 | 3 | 28 |
| 70-75 | 2 | 30 |
↔ Table ko side me swipe karein
n = 30, n/2 = 15. Pehli cf jo 15 se ≥ ho: 19 (class 55-60) → median class = 55-60.
l = 55, cf (before) = 13, f = 6, h = 5
Median = l + [(n/2 − cf)/f] × h = 55 + [(15 − 13)/6] × 5 = 55 + (2/6) × 5 = 55 + 1.67
Median weight ≈ 56.7 kg
Q12 (Ex 13.3). Cumulative frequency (less-than type) of marks of 70 students is given. Find the median marks.
Marks less than 10 20 30 40 50 60 No. of students 5 15 30 50 65 70
↔ Table ko side me swipe karein
| Marks less than | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|
| No. of students | 5 | 15 | 30 | 50 | 65 | 70 |
Pehle isse class-wise frequency table me convert karo (differencing).
| Class | fᵢ | cf |
|---|---|---|
| 0-10 | 5 | 5 |
| 10-20 | 10 | 15 |
| 20-30 | 15 | 30 |
| 30-40 | 20 | 50 |
| 40-50 | 15 | 65 |
| 50-60 | 5 | 70 |
↔ Table ko side me swipe karein
n = 70, n/2 = 35. Pehli cf ≥ 35 hai 50 (class 30-40) → median class = 30-40.
l = 30, cf (before) = 30, f = 20, h = 10
Median = 30 + [(35 − 30)/20] × 10 = 30 + 2.5 = 32.5
Median marks = 32.5
Q13 (Ex 13.3). In a distribution with total frequency 60 and median 28.5, one frequency f₁ (in 10-20) and f₂ (in 40-50) are missing. Find f₁ and f₂.
Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 f₁ 20 15 f₂
↔ Table ko side me swipe karein
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Frequency | 5 | f₁ | 20 | 15 | f₂ |
Total: 5 + f₁ + 20 + 15 + f₂ = 60 ⟹ f₁ + f₂ = 20.
Median = 28.5 falls in class 20-30, so median class = 20-30. l = 20, f = 20, h = 10, n/2 = 30.
cf before median class = 5 + f₁
28.5 = 20 + [(30 − (5 + f₁))/20] × 10
8.5 = (25 − f₁) × 10/20 = (25 − f₁)/2 ⟹ 25 − f₁ = 17 ⟹ f₁ = 8
f₂ = 20 − f₁ = 20 − 8 = 12
Check cf: 5, 13, 33, 48, 60 → n/2 = 30 falls in 20-30 ✓. Median = 20 + [(30−13)/20]×10 = 28.5 ✓
f₁ = 8, f₂ = 12
Q14 (Ex 13.3). Using the marks data of Q2/Q3 (Ex 13.1: 0-10 to 50-60, frequencies 5,9,15,12,6,3), find the median and mode, then verify the empirical relationship 3×Median = Mode + 2×Mean.
| Class | fᵢ | cf |
|---|---|---|
| 0-10 | 5 | 5 |
| 10-20 | 9 | 14 |
| 20-30 | 15 | 29 |
| 30-40 | 12 | 41 |
| 40-50 | 6 | 47 |
| 50-60 | 3 | 50 |
↔ Table ko side me swipe karein
Median: n = 50, n/2 = 25. Pehli cf ≥ 25 hai 29 (class 20-30) → median class = 20-30.
l = 20, cf(before) = 14, f = 15, h = 10
Median = 20 + [(25 − 14)/15] × 10 = 20 + 7.33 = 27.33
Mode: highest frequency = 15 → modal class = 20-30 (same class).
l = 20, f₁ = 15, f₀ = 9, f₂ = 12, h = 10
Mode = 20 + [(15 − 9)/(30 − 9 − 12)] × 10 = 20 + (6/9) × 10 = 20 + 6.67 = 26.67
Verify: Mean (from Q2) = 27.8
3 × Median − 2 × Mean = 3(27.33) − 2(27.8) = 82.0 − 55.6 = 26.4
Ye 26.4, mode 26.67 ke bahut close hai (empirical relation approximate hoti hai, exact match nahi). Median ≈ 27.33, Mode ≈ 26.67.
Q15 (Ex 13.3). Marks of 30 students are given in inclusive class intervals. Find the median (remember to convert to exclusive/continuous form first).
Marks (inclusive) 1-10 11-20 21-30 31-40 41-50 No. of students 3 7 12 5 3
↔ Table ko side me swipe karein
| Marks (inclusive) | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 |
|---|---|---|---|---|---|
| No. of students | 3 | 7 | 12 | 5 | 3 |
Inclusive classes ko continuous banane ke liye har boundary me 0.5 adjust karo (upper limit me +0.5, lower limit me −0.5):
| Class (continuous) | fᵢ | cf |
|---|---|---|
| 0.5-10.5 | 3 | 3 |
| 10.5-20.5 | 7 | 10 |
| 20.5-30.5 | 12 | 22 |
| 30.5-40.5 | 5 | 27 |
| 40.5-50.5 | 3 | 30 |
↔ Table ko side me swipe karein
n = 30, n/2 = 15. Pehli cf ≥ 15 hai 22 (class 20.5-30.5) → median class = 20.5-30.5.
l = 20.5, cf(before) = 10, f = 12, h = 10
Median = 20.5 + [(15 − 10)/12] × 10 = 20.5 + 4.17 = 24.67
Median marks ≈ 24.67
Q16 (Ex 13.3). "More than" type cumulative frequency of ages is given below. Convert it into a frequency distribution and find the median.
Age more than (years) 0 10 20 30 40 50 No. of people 35 32 27 18 10 3
↔ Table ko side me swipe karein
| Age more than (years) | 0 | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|---|
| No. of people | 35 | 32 | 27 | 18 | 10 | 3 |
Frequency of each class = (more than lower limit) − (more than next limit).
| Class | fᵢ | cf |
|---|---|---|
| 0-10 | 35−32=3 | 3 |
| 10-20 | 32−27=5 | 8 |
| 20-30 | 27−18=9 | 17 |
| 30-40 | 18−10=8 | 25 |
| 40-50 | 10−3=7 | 32 |
| 50-60 | 3−0=3 | 35 |
↔ Table ko side me swipe karein
n = 35, n/2 = 17.5. Pehli cf ≥ 17.5 hai 25 (class 30-40) → median class = 30-40.
l = 30, cf(before) = 17, f = 8, h = 10
Median = 30 + [(17.5 − 17)/8] × 10 = 30 + 0.625 = 30.625
Median age ≈ 30.6 years
Important Equations — Ek Nazar Me
| Concept | Formula | Symbols |
|---|---|---|
| Mean (direct method) | x̄ = Σfᵢxᵢ / Σfᵢ | xᵢ = class mark, fᵢ = frequency |
| Mean (assumed mean method) | x̄ = a + Σfᵢdᵢ / Σfᵢ | a = assumed mean, dᵢ = xᵢ − a |
| Mean (step-deviation method) | x̄ = a + h × (Σfᵢuᵢ / Σfᵢ) | h = class width, uᵢ = (xᵢ − a)/h |
| Class mark | xᵢ = (upper limit + lower limit) / 2 | — |
| Mode | Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h | l = lower limit of modal class, f₁ = freq of modal class, f₀ = freq of preceding class, f₂ = freq of succeeding class, h = class width |
| Median | Median = l + [(n/2 − cf) / f] × h | l = lower limit of median class, n = Σfᵢ, cf = cumulative frequency before median class, f = frequency of median class, h = class width |
| Empirical relationship | Mode = 3 × Median − 2 × Mean | approximate relation for moderately skewed data |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Class mark ki jagah upper/lower limit use karna. Mean nikalte waqt xᵢ hamesha class ka mid-value hona chahiye — (upper+lower)/2 — na ki sirf upper limit ya lower limit. Ye sabse common calculation error hai.
- Median class ke liye galat cumulative frequency le lena. Median formula me 'cf' hamesha median class se pehle wali class ki cumulative frequency honi chahiye, khud median class ki nahi. Student aksar galti se median class ki hi cf use kar lete hain.
- Mode formula me f₁−f₀ aur f₀−f₂ me confuse ho jana. Sahi formula hai (f₁−f₀)/(2f₁−f₀−f₂) — sign ulta karne se (jaise f₀−f₁ likh dena) poora answer negative ya galat aa jata hai. Formula ko yaad karte waqt 'f₁' hamesha pehle, phir '2f₁' denominator me double hota hai — yahi yaad rakhne ka tareeka hai.
- h (class width) ko class mark samajh lena. Mode aur median dono formulas me 'h' matlab class ki width (jaise 10-20 ki class ka h=10), na ki us class ka midpoint. In dono ko mix karna badi galti hai.
- n/2 ki jagah sirf n use kar dena median class dhoondte waqt. Median class find karne ke liye cumulative frequency ko n/2 se compare karna hai, total n se nahi — is chhoti si galti se poori median class hi galat pakad li jaati hai.
- Modal class ko sirf 'sabse zyada number wali class' samajh kar last row ya kisi bhi row utha lena bina poora table check kiye. Modal class hamesha wo class hai jiski frequency sabse zyada ho — poore data set me compare karke dekhna zaroori hai, na ki table ke last ya first row ko directly le lena.
Board-Style Important Questions
- 1 mark: Grouped data ke mean ko step-deviation method se nikalne ke formula me 'uᵢ' kis cheez ko represent karta hai?
- 2 marks: Ek modal class ka lower limit 25, f₁=18, f₀=10, f₂=12, aur class width 5 hai. Mode nikalo.
- 3 marks: Kisi grouped data ka mean 42 aur mode 46 diya gaya hai. Empirical relationship use karke median nikalo.
- 3 marks: Diye gaye grouped frequency distribution (5 classes) ka median nikalo, cumulative frequency table banate hue poora tareeka dikhao.
- 5 marks: Ek grouped data table diya hai jisme ek frequency missing hai aur median value bhi di gayi hai. Missing frequency find karo, poora working dikhate hue.
Quick Quiz — Score Check Karein
Q1. Grouped data me 'class mark (xᵢ)' kaise nikalte hain?
Q2. Ex 13.1 Q1: 20 houses me plants ki number of houses distribution di gayi hai (classes 0-2 se 12-14, Σfᵢ=20, Σfᵢxᵢ=162). Direct method se mean number of plants per house kya hoga?
Q3. Assumed mean method me 'dᵢ' kise represent karta hai?
Q4. Step-deviation method me 'uᵢ' formula kya hai?
Q5. 50 students ke marks data (classes 0-10 se 50-60, frequencies 5,9,15,12,6,3) me assumed mean a=25 lekar assumed mean method se mean kya aaya (Q2)?
Q6. Mode ka formula grouped data ke liye kya hai?
Q7. Hospital patients ki ages data me (classes 5-15 se 55-65, frequencies 6,11,21,23,14,5) modal class kaunsi hai, aur uska l, f₁, f₀, f₂ kya hain?
Q8. Agar modal class distribution ki sabse pehli class ho (koi preceding class na ho), to mode formula me f₀ ki value kya loge?
Q9. Median formula me 'cf' kis cheez ko represent karta hai?
Q10. Empirical relationship ke according, agar kisi distribution ka mean = 26.8 aur median = 27 diya ho, to estimated mode kya hoga?
Aksar Poochhe Jaane Wale Sawaal
Grouped data me exact mean kyun nahi nikal sakte?
Kyunki humein sirf ye pata hai ki kitne values kis range (class interval) me aati hain, individual values pata nahi. Isliye har class ke saare data points ko us class ke class mark (midpoint) ke barabar maan liya jata hai — ye ek approximation hai, exact value nahi.
Teeno mean methods (direct, assumed mean, step-deviation) me se kaunsa use karna chahiye?
Teeno ka final answer same hi aata hai. Chhote, simple numbers ho to direct method theek hai. Class marks bade hon (jaise 45, 65, 85) to assumed mean se calculation aasan ho jaati hai. Aur agar class width bhi uniform ho, to step-deviation method sabse fast padta hai kyunki numbers sabse chhote ho jaate hain.
Median class aur modal class same class ho sakti hai kya?
Haan, bilkul ho sakti hai — jaisa Q14 me dikhaya gaya hai. Modal class wo hai jiski frequency sabse zyada ho, aur median class wo jiski cumulative frequency n/2 ko pehli baar cross kare. Kai baar dono conditions ek hi class satisfy karti hain, lekin ye hamesha guaranteed nahi hai.
Agar modal class hi sabse pehli class ho to f₀ kya lein?
f₀ = 0 le lo, kyunki modal class se pehle koi class hi nahi hai (jaisa Q8 me dikhaya). Isi tarah agar modal class sabse aakhri class ho, to f₂ = 0 le lo.
Empirical relationship (Mode = 3Median − 2Mean) hamesha exact match karti hai kya?
Nahi, ye ek approximate relation hai jo moderately skewed distributions ke liye kaam karti hai. Jaisa Q14 me dikha — calculated mode (26.67) aur empirical formula se aaya value (26.4) close hain lekin exactly same nahi. Isse estimate ke roop me use karo, exact proof ke roop me nahi.
Inclusive class intervals (jaise 1-10, 11-20) ko continuous kyun banana padta hai?
Kyunki median formula continuous (exclusive) class intervals maanti hai jahan ek class ki upper limit doosri ki lower limit ke barabar ho. Inclusive form me 10 aur 11 ke beech gap hota hai, isliye har limit ko 0.5 se adjust karke continuous banate hain — jaisa Q15 me dikhaya gaya.
Class 10 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
Class 10 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
NCERT Kaksha ek swatantra shaikshik platform hai aur NCERT ya CBSE se aadhikarik roop se sambaddh (officially affiliated) nahi hai. Kisi galti ki jaankari dene ke liye contact kijiye.