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Statistics

Lesson 96 of 113

1. Ye topic hai kya?

Bahut saari readings ko teen numbers mein sametna: mean (balance point), median (beech wali, sort ke baad), mode (sabse zyada baar). Paper: class-interval table, mid-value xᵢ, assumed mean, median class n/2, modal class max f, empirical 3 Median = Mode + 2 Mean (approx), less-than ogive.

Yahan lock

Frequency fᵢ, Σf = n. Grouped: exclusive 10–20, 20–30. Mean teen methods. Median/mode ki l usi class ki lower limit. Inclusive → pehle exclusive.

Yahan kya nahi

Cricket average, “40 chahiye ab kitne runs”, mixture alligation. AP mean of series. Die/card P(E). Coordinate midpoint. Sphere 4πr².

Mean sabko equal weight (frequency se). Median extreme se kam hilta. Mode “sabse common”. Grouped mein class ke andar formula se interpolate — class ka mid median nahi hota automatically.

2. Ungrouped — pehle sort

Raw list: mean (Σx)/n. Median: pehle chhote se bada. Odd n → beech wala. Even n → beech ke do ka mean. Mode = jo value sabse zyada baar; sab unique to mode nahi; do peaks = bimodal.

12 15 18 20 25 median

12, 18, 20, 15, 25 → sort 12, 15, 18, 20, 25. Beech = median 18. Mean bhi 90/5 = 18. Sort ke bina 20 galat.

Discrete frequency table (x alag-alag, class nahi): mean (Σ fᵢxᵢ)/(Σ fᵢ). Median = n/2-th observation (even: n/2 aur n/2+1 ka mean) — cf se dekho kaun sa x. Mode = max f wala x.
Example

x = 1, 2, 3, 4, 5 aur f = 4, 6, 8, 5, 2. Mean, median, mode?

Solution

  1. n

    4+6+8+5+2 = 25.

  2. Mean

    (4×1 + 6×2 + 8×3 + 5×4 + 2×5)/25 = 70/25 = 2.8. Cricket “average 40 chahiye” nahi — yahan Σfx/Σf.

  3. Median

    Odd, 13th value. cf: 4, 10, 18, 23, 25. 13th 11–18 ke beech → x=3.

  4. Mode

    Max f=8 → x=3. AP ka 1,2,3,4,5 agla term nahi.

Answer2.8 ; 3 ; 3
Even list 11, 4, 8, 15: sort 4, 8, 11, 15. Median (8+11)/2 = 9.5. Beech ke “jaise hain” 8 aur 15 se mean 11.5 — galat, sort pehle. Range = max−min = 11 (ye spread, mean nahi).

3. Grouped mean — teen raaste, ek jawab

Class 0–10, 10–20, … exclusive: 10 pehle class ke upper, doosri ke lower — 10 doosri mein. Mid xᵢ = (lower + upper)/2. Width h (yahan 10). Direct / assumed-mean / step-deviation — theoretically same mean.

CI fᵢ cf xᵢ fᵢxᵢ uᵢ=(xᵢ−25)/10 fᵢuᵢ
0–1055525−2−10
10–2081315120−1−8
20–3012252530000
30–409343531519
40–5064045270212
4010303

Yahi table aage median/mode/ogive mein. n = 40. Assumed mean a = 25 (kisi xᵢ pe rakhna easy).

Direct: x̄ = (Σ fᵢxᵢ) / n.
Assumed mean: dᵢ = xᵢ − a, x̄ = a + (Σ fᵢdᵢ)/n.
Step-deviation: uᵢ = (xᵢ − a)/h, x̄ = a + h (Σ fᵢuᵢ)/n. h se divide isliye ke u chhote integers.
Example

Upar wali table. Mean teen tarike se?

Solution

  1. Direct

    1030 / 40 = 25.75.

  2. Step

    a=25, h=10, Σfu = 3. 25 + 10×(3/40) = 25 + 0.75 = 25.75. Same. Σf = 40 bhool ke /5 mat.

  3. Trap

    xᵢ ki jagah class lower (0,10,20…) se mean galat. 2D πr² mix nahi. Age “family average 30” alag chapter.

Answer25.75
Inclusive 1–10, 11–20, 21–30: beech gap. True class = lower−0.5, upper+0.5 → 0.5–10.5, 10.5–20.5, … h ab bhi 10. Median/mode ki l true lower.
Example

CI 5–15, 15–25, 25–35, 35–45. f = 8, f, 10, 6. Mean = 25. f?

Solution

  1. n aur Σfx

    n = 24+f. xᵢ = 10, 20, 30, 40. Σfx = 80 + 20f + 300 + 240 = 620 + 20f.

  2. Mean

    (620+20f)/(24+f) = 25. 620+20f = 600+25f → 20 = 5f → f = 4. (Quadratic nahi, linear. Probability nahi.)

Answer4

4. Grouped median aur mode

Median class = jisme n/2-th observation padti (cf pehli baar ≥ n/2). Mode class = sabse badi f. Dono ek class hon to bhi formulas alag — mid-point copy mat.

0–10 10–20 20–30 30–40 40–50 f=12

Tallest bar = modal class (max f). Median class n/2 se — yahan bhi 20–30, lekin mode 25.71, median 25.83, mean 25.75: kareeb, equal nahi.

Median: l + [(n/2 − cf)/f] × h. l = median class lower, cf = usse pehle wali cf, f = usi class ki frequency, h = width.
Mode: l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h. f₁ modal, f₀ pehle class, f₂ baad wali. Denominator 0 ho to formula fail (bimodal / flat) — Class-X usually nahi.
Example

Wahi table n=40. Median?

Solution

  1. n/2

    20. cf: 5, 13, 25, 34, 40. Pehli cf ≥ 20 = 25 → class 20–30.

  2. Plug

    l=20, cf=13, f=12, h=10. 20 + (20−13)/12 × 10 = 20 + 70/12 = 20 + 35/6 = 155/6 ≈ 25.83. Trap: 25 mid-point. Trap: cf=25 use karna (woh cumulative, pehle wali 13).

Answer155/6 ≈ 25.83
Example

Wahi table. Mode?

Solution

  1. Modal class

    Max f=12 → 20–30. (Median class coincidentally same — hamesha nahi.)

  2. Plug

    l=20, f₁=12, f₀=8, f₂=9, h=10. 20 + (12−8)/(24−8−9)×10 = 20 + 4/7 × 10 = 20 + 40/7 = 180/7 ≈ 25.71. (f₁−f₂)/(2f₁−f₀−f₂) mix: 3/7×10 galat.

Answer180/7 ≈ 25.71
Empirical Mode ≈ 3 Median − 2 Mean (ya 3 Med = Mode + 2 Mean). Approx, paper pe “relation se nikaalo” tab. Yahan 3×25.83 − 2×25.75 ≈ 26 — table mode 25.71 ke kareeb, barabar claim mat. Variance / σ Class-XI.
Example

Mean = 30, median = 28. Empirical mode?

Solution

  1. 3Med − 2Mean

    3×28 − 2×30 = 84 − 60 = 24. Mean se mode “average” nahi. P(mode) Probability nahi.

Answer24

5. Cumulative frequency aur ogive

Less-than cf: us upper limit se chhote kitne. Table: <10 → 5, <20 → 13, <30 → 25, <40 → 34, <50 → 40. Graph: x = upper limits, y = less-than cf. Smooth curve = ogive.

More-than: x = lower limits. ≥0 → 40, ≥10 → 35, ≥20 → 27, ≥30 → 15, ≥40 → 6, ≥50 → 0. Dono ogives ka katna ≈ median.

n/2 = 20 ≈26 10 20 30 40 50 ogive

Maroon: y = 20 se curve, phir neeche x. Formula wala 25.83; graph ≈26. Histogram bars (upar) ogive nahi. Missing-number grid nahi.

Ogive se median: N/2 horizontal, curve pe point, vertical drop, x-padhao. Mean ogive se nahi nikalta. Mode histogram / formula se.
Paper “less than type ogive” = upper limits. “More than” = lower. Dono ek hi axes pe — intersection median. Statistics graph; Applications tower 30° nahi; Circles tangent nahi.

6. Sawal — basic se pro

Q1 · Basic

10, 12, 17, 21. Mean?

  1. 15
  2. 17
  3. 12
  4. 60

Solution

  1. 60/4

    15. D sum. B median-ish without sort.

Answer15
Q2 · Basic

8, 3, 11, 6, 9. Median?

  1. 8
  2. 11
  3. 6
  4. 7.4

Solution

  1. Sort

    3, 6, 8, 9, 11 → 8. D mean. B unsorted beech.

Answer8
Q3 · Basic

2, 5, 5, 7, 5, 9. Mode?

  1. 5
  2. 7
  3. 5.5
  4. 9

Solution

  1. Sabse baar

    5 teen dafa. C mean/median mix.

Answer5
Q4 · Basic

Main table (f = 5,8,12,9,6). Mean?

  1. 25.75
  2. 25
  3. 20
  4. 1030

Solution

  1. 1030/40

    25.75. B assumed a. D Σfx.

Answer25.75
Q5 · Medium

Wahi table. Median class?

  1. 20–30
  2. 30–40
  3. 10–20
  4. 0–10

Solution

  1. n/2=20

    cf 13 phir 25. 20–30. B max-f soch ke mode class same yahan, lekin rule n/2 hai. C cf=13 < 20.

Answer20–30
Q6 · Medium

Wahi table. Median?

  1. 155/6
  2. 25
  3. 20
  4. 180/7

Solution

  1. 20+(7/12)×10

    155/6. B mid. D mode.

Answer155/6
Q7 · Medium

Wahi table. Mode?

  1. 180/7
  2. 155/6
  3. 25.75
  4. 12

Solution

  1. 20+40/7

    180/7. D frequency ko mode mat bolo.

Answer180/7
Q8 · Medium

Mean 30, median 28. Empirical mode?

  1. 24
  2. 26
  3. 32
  4. 58

Solution

  1. 3×28−2×30

    24. B (30+28)/2. D 30+28.

Answer24
Q9 · Pro

CI 5–15 … 35–45, f = 8, f, 10, 6, mean 25. Missing f?

  1. 4
  2. 8
  3. 10
  4. 24

Solution

  1. 5f=20

    f=4. D n without f.

Answer4
Q10 · Pro

Ek die, P(6). Is chapter?

  1. Nahi — Probability
  2. Haan, mode 6
  3. Mean of 1–6
  4. Ogive

Solution

  1. P(E)

    Agla chapter. Mean of 1..6 = 3.5 statistics-warmup ho sakta, P(6)=1/6 nahi yahan.

AnswerA

Practice

Pehle khud, phir neeche kholo. Alag table: 10–20 … 40–50, f = 6, 10, 8, 6 (n=30).

12345 678910

P1

4, 7, 9, 12. Median?

  1. 8
  2. 9
  3. 7
  4. 8.5

Solution

  1. (7+9)/2

    8. D mean of all / mix 8.5 nahi. B teesra without even-rule.

Answer8
P2

x=1..5, f=4,6,8,5,2. Median?

  1. 3
  2. 2.8
  3. 8
  4. 13

Solution

  1. 13th

    x=3. B mean. C max f.

Answer3
P3

Practice table n=30, a=25, h=10, Σfu=14. Mean?

  1. 29.67
  2. 25
  3. 14
  4. 39

Solution

  1. 25+10×14/30

    25+14/3 ≈ 29.67. Direct 890/30 same.

Answer29.67
P4

Practice table. Median?

  1. 29
  2. 25
  3. 20
  4. 80/3

Solution

  1. n/2=15, class 20–30

    cf pehle 6, f=10. 20+(15−6)/10×10 = 29. D mode.

Answer29
P5

Practice table. Mode?

  1. 80/3
  2. 29
  3. 10
  4. 25

Solution

  1. 20+(4/6)×10

    20+20/3 = 80/3 ≈ 26.67. C f₁.

Answer80/3
P6

Less-than ogive pe x-axis kya?

  1. Class upper limits
  2. Mid xᵢ
  3. Frequency f
  4. n/2 only

Solution

  1. Upper

    A. More-than pe lower. C histogram height.

AnswerA
P7

Inclusive 1–10, 11–20. True lower of first class?

  1. 0.5
  2. 1
  3. 10.5
  4. 0

Solution

  1. −0.5

    0.5–10.5. C first ka true upper.

Answer0.5
P8

Ogive se median: kaun si line?

  1. y = n/2, phir drop to x
  2. y = max f
  3. x = mean
  4. Sector 90°

Solution

  1. N/2

    A. D Areas. B mode histogram.

AnswerA
P9

3 Median = Mode + 2 Mean. Mean 18, mode 12. Median?

  1. 16
  2. 15
  3. 14
  4. 30

Solution

  1. 3M=12+36=48

    M=16. B (18+12)/2.

Answer16
P10

5 numbers ka mean, “ab 40 average ke liye 6th kitna” — yahan?

  1. Nahi — Average (arithmetic)
  2. Haan, grouped mode
  3. Ogive
  4. Frustum

Solution

  1. Target average

    A. Yahan frequency-class. D Surface areas.

AnswerA

← Index · Surface areas and volumes

Agle topic: Probability.