WHAT: We want the "fair share". If total is S and we split it equally among n people:
xˉ=nS=n∑xi
WHY this is the balance point: The mean is the value m that makes the deviations cancel:
∑(xi−m)=0⇒∑xi−nm=0⇒m=n∑xiWhy this step? Setting total deviation to zero is the definition of a balance point — pushes above equal pulls below.
We can't see individual values, so we find where the middle person lands using cumulative frequency (cf).
Derivation (linear interpolation), WHY every piece:
We need the position of the 2n-th item. Before the median class, cf items are already used up. We still need 2n−cf more items. Assume the f items in the median class are spread evenly across its width h. So each item occupies fh of width. To reach (2n−cf) items into the class we move:
(2n−cf)×fh
past the lower boundary l. Add to l → done.
The modal class = class with highest frequency f1. But the true peak may lie left or right of centre depending on the neighbouring classes.
Derivation intuition, WHY: Look at how much the peak rises above the class before it, (f1−f0), vs the total "excess" over both neighbours, (f1−f0)+(f1−f2)=2f1−f0−f2. The fraction tells us how far into the modal class the true peak sits. If the class before is heavier, the mode leans left; if the class after is heavier, it leans right. Multiply by width h, add to l.
The point where total deviation ∑(xi−m)=0; pulls above and below cancel.
Grouped mean direct formula?
xˉ=∑fi∑fixi using class midpoints.
Step-deviation mean formula and what is ui?
xˉ=a+h∑fi∑fiui, where ui=hxi−a.
Grouped median formula?
l+(fn/2−cf)h.
In the median formula, what is cf?
Cumulative frequency of the class before the median class.
How do you identify the median class?
First class whose cumulative frequency ≥n/2.
Grouped mode formula?
l+(2f1−f0−f2f1−f0)h.
What are f0,f1,f2 in mode?
f1=modal class freq, f0=preceding class freq, f2=following class freq.
Empirical relation between the three?
Mode ≈3Median−2Mean.
Which measure is most affected by extreme values?
The mean (median and mode ignore outliers).
Median of even n data?
Average of the (n/2)-th and (n/2+1)-th ordered values.
Why use assumed-mean method?
Same mean, but smaller deviation numbers make arithmetic easier.
Recall Feynman: explain to a 12-year-old
Imagine a bunch of kids holding different numbers of candies.
Mean: dump ALL candies in one pile and share equally — that equal share is the mean.
Median: line the kids up from fewest to most candies; the kid standing exactly in the middle — that's the median.
Mode: the candy-count that the MOST kids happen to have — that's the mode.
When candies are given in groups ("5 to 10 candies"), we can't see each kid, so we pretend everyone in a group has the middle amount, and we slide along the crowded group to guess where the middle kid or the peak really is. That sliding is the little formula with l, f, h.
Central tendency ka matlab hai: poore data ko ek hi number se represent karna. Teen main measures hote hain — mean (average, yaani sab jod ke count se divide), median (bilkul beech wala value jab data sort kiya ho), aur mode (jo value sabse zyada baar aaye). Har ek alag kahani batata hai: mean ko extreme values (outliers) kheench lete hain, median unhe ignore karta hai, aur mode sirf frequency dekhta hai.
Jab data grouped hota hai (jaise 20–30, 30–40 wale classes), tab humein individual values nahi dikhti. Isliye hum maan lete hain ki har class ke sab values uske midpoint ke barabar hain — isse mean nikalte hain: xˉ=∑fi∑fixi. Numbers chhote karne ke liye step-deviation trick use karo (ui=hxi−a) — answer wahi aata hai, bas arithmetic aasaan ho jaata hai.
Median ke liye pehle n/2 nikaalo, phir cumulative frequency dekh ke wo class dhoondo jahan cf pehli baar n/2 tak pahunchti hai — wahi median class. Formula l+fn/2−cf×h basically linear interpolation hai: us class ke andar kitne items aur chahiye, unhe evenly spread maan ke slide karte hain. Mode me sabse zyada frequency wali class lo aur dono padosi classes (f0,f2) se compare karke peak ka exact position nikaalo.
Yaad rakho — hamesha class boundaries use karo (limits nahi), median ke liye data sort karo, aur mode ke liye individual frequency dekho, cumulative nahi. Exam me yeh choti si galtiyan hi marks kha jaati hain!