Errors — absolute, relative, percentage; systematic vs random: Visual walkthrough

1.1.5 · D2 · Physics › Measurement, Vectors & Kinematics › Errors — absolute, relative, percentage; systematic vs rando

Hum ek running story use karenge: tumne ek swinging pendulum ko paanch baar time kiya aur paanch numbers seconds mein likhe.

$2.63, 2.56, 2.42, 2.71, 2.80$

Step 1 — Raw readings ko number line par rakh do

YEH PEHLE KYU? Kyunki error analysis poori tarah is baat ke baare mein hai ki yeh dots kitne spread out hain aur unka centre kahan baithta hai. Spread ke baare mein tab tak reason nahi kar sakte jab tak tum use dekh nahi lete. Neeche ki picture poori problem hai, honestly draw ki gayi.

PICTURE: paanch red dots scattered hain. Kuch low hain (woh $2.42$ wala), kuch high (woh $2.80$ wala). Inme se koi "truth" nahi hai — true period ek hidden point hai jise hum guess karne ki koshish kar rahe hain.

Figure — Errors — absolute, relative, percentage; systematic vs random

Step 2 — Balance point dhundho: the mean

AVERAGE KYU, aur koi biggest ya middle wala kyun nahi? Random wobbles readings ko true value ke upar aur neeche roughly equally push karti hain. Agar highs aur lows balanced hain, toh woh point jo tug-of-war ke beech mein baithta hai woh least-biased single guess hai. Saari readings ko sum karna aur equally share karna ("kitne mein divide karo") exactly woh balance point hai.

Har symbol ko dekhte hain jaise woh appear karta hai:

$a_{mean} = share equally among n \frac{1}{n} add up all readings a_{1} \dots a_{n} i = 1 \sum n a_{i}$

$n$ ::: kitni readings li tumne. Yahan $n = 5$ .
$i = 1 \sum n a_{i}$ ::: bada "S" (Greek sigma) ek shorthand hai "sab add karo" ka: $i = 1$ se shuru, $i = n$ par ruko, aur har $a_{i}$ total karo. Yahan yeh hai $2.63 + 2.56 + 2.42 + 2.71 + 2.80 = 13.12$ .
$\frac{1}{n}$ ::: us total ko $n$ equal shares mein baanto — iska matlab "average" hota hai.

$a_{mean} = \frac{13.12}{5} = 2.624 s$

PICTURE: seesaw exactly $2.624$ ke neeche balance karta hai. Woh black triangle hamaara best single guess hai.

Recall Balance point average kyun hota hai? (peek)

Question ::: Seesaw arithmetic mean par kyun balance karta hai, kisi aur jagah kyun nahi? Answer ::: Kyunki mean woh ek point hai jahan uske left ke dots ka total "pull" exactly uske right ke dots ke total pull ke equal hota hai — usse signed distances ka sum zero hota hai. Yahi balance point ki definition hai.

Step 3 — Har dot ne balance point ko kitna miss kiya, measure karo

"ABSOLUTE" word kyun? Ek reading left (too small, negative gap) ya right (too big, positive gap) par baaith sakti hai. Hum sirf kitni door hai care karte hain, kaunsi taraf nahi — 6 thousandths ki miss, 6 thousandths ki miss hai chahe woh left ya right gayi ho. Isliye hum sign throw away kar dete hain. Do vertical bars $∣ ∣$ exactly yahi matlab hai: "size rakho, minus drop karo."

$Δ a_{i} = balance point a_{mean} - this reading a_{i}$

$Δ$ ::: Greek letter delta, physics mein har jagah use hota hai "ek chhoti change / gap / amount of" mean karne ke liye. $Δ a_{i}$ padhte hain "reading $i$ ka gap."
$a_{mean} - a_{i}$ ::: raw gap, jo $+$ ya $-$ ho sakta hai.
$∣ \dots ∣$ ::: sign strip karo, length rakho.

Paanchon compute karna (un-rounded $2.624$ use karke):

$∣2.624 - 2.63∣ = 0.006, 0.064, 0.204, 0.086, 0.176$

PICTURE: har red bracket ek dot se black balance line tak span karta hai. Sabse lamba bracket ( $0.204$ , woh $2.42$ reading) hamaari worst miss hai; sabse chhota ( $0.006$ ) almost bullseye hai.

Step 4 — Misses ko average karo: mean absolute error

MISSES AVERAGE KYUN? Ek miss fluke ho sakta hai. Typical miss tumhe honestly batata hai ki koi bhi single reading kitna wander karna tend karti hai. Woh typical wander exactly wahi hai jo "hamaari uncertainty" ka matlab hona chahiye.

$Δ a_{mean} = equally share karo \frac{1}{n} saari misses add karo i = 1 \sum n Δ a_{i}$

Step 2 jaise hi machinery — sigma add karta hai, $\frac{1}{n}$ share karta hai — lekin ab yeh misses khaata hai, readings nahi.

$Δ a_{mean} = \frac{0.006 + 0.064 + 0.204 + 0.086 + 0.176}{5} = \frac{0.536}{5} = 0.1072 \approx 0.11 s$

Intuition Yeh spread measure kyun, standard deviation kyun nahi?

Yeh kehne ke teen common tarike hain ki "readings kitni spread out hain," aur yeh jaanna zaroori hai ki tum ek choose kar rahe ho:

Mean absolute deviation $Δ a_{mean} = \frac{1}{n} \sum ∣Δ a_{i} ∣$ — misses directly average karo. Haath se compute karna aur explain karna sabse easy; yeh school/JEE convention hai aur wahi hai jo parent note use karta hai.
Standard deviation $σ = \frac{1}{n} \sum (Δ a_{i})^{2}$ — squared misses average karo, phir square-root. Badi misses ko zyada weight karta hai aur statistics prefer karta hai. Dekho Mean and Standard Deviation.
Standard error of the mean $σ / n$ — mean itself kitna uncertain hai (jaise jaise zyada readings lete ho, shrinke karta hai).

Hum deliberately MAD use karte hain kyunki yeh walkthrough parent note se match karta hai aur koi squares ya square-roots ki zaroorat nahi — har step ek simple picture rehta hai jo ek 12-saal ka baccha bhi draw kar sake. Same data ke liye $σ \approx 0.13$ s, thoda bada; conclusion ("roughly tenth of a second wobble") dono mein same hai.

PICTURE: paanchon bracket-lengths bars ke roop mein stack ki gayi hain; dashed red line unki average height hai — woh ek number $0.1072$ jo "how far off, typically" ke liye stand karta hai.

Step 5 — Ek baar round karo, phir honesty fence banao

SIRF YAHAN KYU ROUND KARO, aur matching decimals kyun? Kyunki rounding lossy hai — har round thodi si information throw away karta hai. Pehle round karo aur Steps 3–4 mein slivers pile up hote hain (yeh rounding-error propagation hai). Ek baar round karo, last moment mein, aur downstream mein corrupt karne ke liye kuch nahi hoga. Matching decimals tumhe ek value claim karne se rokta hai jo uske apne error bar se zyada precise ho — $2.624 \pm 0.11$ likhna pretend karta hai ki tum teesra decimal jaante ho jo $\pm 0.11$ kehta hai ki tum nahi jaante.

EK SINGLE NUMBER KYU NAHI? Kyunki ek exact number claim karna jhooth hoga — tum true value ko infinite precision tak nahi jaante. Band honest statement hai: "kahin yahaan." Notice karo ki $2.42$ wala far-out reading chhod ke har raw dot andar hai; band ka kaam readings ko typically contain karna hai, sabko nahi.

PICTURE: red band balance point ko bracket karta hai. Yeh picture chapter ka poora point visually dikhata hai.

Step 6 — Compare karne layak banao: relative aur percentage error

DIVIDE KYUN? Division units cancel kar deta hai (seconds ÷ seconds = pure number) aur har measurement ko ek common ruler par rescale karta hai: fraction of itself. Ab ek timing error aur ek mass error head-to-head compare ho sakti hain — kuch jo absolute errors, alag alag units mein atak ke, kabhi nahi kar saktihen.

$δ_{r} = \frac{Δ a _{mean} miss (s)}{size (s) a _{mean}} = \frac{0.11}{2.62} = 0.042 (pure number, koi units nahi)$

$δ$ ::: Greek delta (lower-case), ek conventional label "error of" ke liye. Subscript ek label hai, variable nahi — yeh naam deta hai ki kaunsi tarah ki error hai.
subscript $r$ in $δ_{r}$ ::: "relative" (fractional) ke liye stand karta hai.
subscript $%$ in $δ_{%}$ (neeche) ::: ek mnemonic label jo literally padhta hai "error percentage ke roop mein express ki gayi." Kuch books $δ_{p}$ likhti hain ya sirf " $%$ " number ke saath append kar deti hain; symbol sirf ek naam hai, isliye jo bhi mile use "percentage error" padho. Hum $δ_{%}$ use karte hain parent note se match karne ke liye.

Ise percent ke roop mein dress karo 100 se multiply karke — kyunki insaan " $4.2%$ " ko " $0.042$ " se tez padhte hain:

$δ_{%} = 0.042 \times 100% = 4.2%$

PICTURE: ek bar jo tiny red $\approx 4%$ slice ko measured value ke poore $100%$ ke against dikhata hai — ek fraction jo tum dekh sako.

Step 7 — Edge cases jinpar kabhi mat trip karo

Case A — saari readings identical. Maan lo tumne $2.62, 2.62, 2.62$ measure kiya. Har gap $Δ a_{i} = 0$ , toh $Δ a_{mean} = 0$ . Kya iska matlab zero error hai? Nahi. Iska matlab hai ki tumhari random scatter tumhare instrument ki resolution ke neeche hai. Real floor tab tumhare device ka least count hai — dekho Least Count and Vernier Calipers. Ek stopwatch jo "0 spread" read kare woh bhi apni tick size se better nahi ho sakta.

Case B — balance point zero par baithta hai. Relative error $δ_{r} = Δ a_{mean} / a_{mean}$ blow up karta hai agar $a_{mean} = 0$ : tum kuch nahi se divide kar rahe hoge. Physically yeh warn karta hai ki "relative error" us quantity ke liye meaningless hai jiska best estimate zero hai — $0$ ke around $\pm 0.1$ ka matlab " $\infty%$ off" nahi hai, iska matlab sirf yeh hai ki true value either sign ho sakti hai.

Case C — ek systematic offset mean ke andar chhupta hai. Agar tumhara stopwatch hamesha $0.3$ s late start karta hai, har reading $+ 0.3$ zyada hogi. Averaging (Step 2) woh $+ 0.3$ rakhta hai — balance point shifted hai, band galat jagah centre hua hai. Upar ki pictures ne pure random scatter assume kiya. Systematic error in formulas mein hai hi nahi; ise alag se dhundha aur subtract kiya jaana chahiye. Yahi exactly wajah hai kyun precise ≠ accurate.

PICTURE: do number lines. Top = random-only (band true value ke aas paas hai, acha). Bottom = systematic shift (tight band, lekin poora band true value se slide ho gaya hai — confidently galat).

Ek picture mein summary

Recall Feynman retelling — poora walk ek 12-saal ke bacche ko explain karo

Tumne ek swing ko paanch baar time kiya aur paanch thode different numbers mile, kyunki stopwatch par tumhara thumb magic nahi hai. Pehle tum un paanch ka middle dhundhte ho — seesaw balance point — aur ise apna best guess bolte ho ( $2.624$ s, aur woh extra digit abhi rakhte ho taaki agli steps kharab na ho). Phir har koshish ke liye poochte ho "main middle se kitna miss kiya?" aur ignore karte ho ki tum pehle the ya baad mein — bas miss ka size. Un misses ko average karte ho "main usually kitna off hota hoon" paane ke liye ( $0.107$ s). Sirf end mein tidy up karte ho: wobble $0.11$ ho jaata hai, toh answer match karne ke liye round hota hai — "about 2.62 seconds, give or take 0.11." Woh "give or take," 2.62 ke muqable mein, 100 mein se roughly 4 hai — 4% wobble, chhota. Do warnings: beech mein round mat karo ya chhoti chops pile up hongi; aur agar tumhara stopwatch hamesha same amount se lag karta hai, zyada tries lena tumhe rescue nahi karega — har number same tarah shift hua hai, toh middle bhi shift hua hai. Averaging shaky-hand mistakes fix karta hai, ruler-broken mistakes kabhi nahi.

Recall Quick self-test

Poora pipeline order mein ::: mean (guard digits rakho) → har ek ki absolute error → mean absolute deviation → ek baar round karo → $a_{mean} \pm Δ a_{mean}$ likho → relative ke liye divide karo → ×100 for percent. $Δ a_{i}$ mein sign strip kyun karte hain? ::: Hum miss ka size chahte hain, uski direction nahi; left-misses aur right-misses equally buri hain. Bilkul end mein hi round kyun karte hain? ::: Rounding lossy hai; pehle rounding karna chhoti chops ko har baad ke step mein propagate karne deta hai aur final band corrupt karta hai. Yahan mean absolute deviation standard deviation ki jagah kyun use karte hain? ::: Isme koi squares ya square-roots nahi chahiye, isliye har step ek simple picture rehta hai; yeh parent-note convention se match karta hai. Standard deviation statistically preferred alternative hai. Agar saari readings identical hain, kya error zero hai? ::: Nahi — random scatter zero hai, lekin least count aur koi bhi systematic offset abhi bhi tumhe limit karta hai. Jab $a_{mean} = 0$ ho toh relative error kyun compute nahi ho sakti? ::: Tum zero se divide kar rahe hoge; "fraction of the value" ek zero value ke liye undefined hai.

Yeh bhi dekho: Mean and Standard Deviation · Combination of Errors · Significant Figures and Rounding · Dimensional Analysis.