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Statistics revision notes

Every idea in the Statistics chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 13 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Measures of central tendency

Arithmetic mean (grouped, step deviation)
10× asked, last 2026
xˉ=∑fixiN=A+h ∑fiuiN,ui=xi−Ah\bar x=\frac{\sum f_i x_i}{N}=A+h\,\frac{\sum f_i u_i}{N},\qquad u_i=\frac{x_i-A}{h}
Grouped data: xix_i = class mid-point, N=∑fiN=\sum f_i. AA is any assumed mean, hh the common class width.
  • Frequencies nCr{}^nC_r (total 2n2^n): ∑r nCr=n2n−1\sum r\,{}^nC_r=n2^{n-1}, ∑r(r−1) nCr=n(n−1)2n−2\sum r(r-1)\,{}^nC_r=n(n-1)2^{n-2}, ∑nCrar=(1+a)n\sum{}^nC_ra^r=(1+a)^n. Values rising with rr: median at r=n/2r=n/2 (nn even), else the mean of r=(n±1)/2r=(n\pm1)/2.
Median
10× asked, last 2026
sorted data: middle value,M=l+N/2−Cf×h\text{sorted data: middle value},\qquad M=l+\frac{N/2-C}{f}\times h
ll: lower limit of the class where the cf first reaches N/2N/2; CC: cf before that class; ff: its frequency; hh: width.
xxcfO48121620N2=18\frac N2=1836M=10.4M=10.4
Ogive (cumulative frequency, orange) read at N2=18\frac N2=18: M=8+18−1210×4=10.4M=8+\frac{18-12}{10}\times4=10.4. Shaded: the histogram (×2 height)
  • Ungrouped, sorted: odd nn gives the ((n+1)/2)((n+1)/2)th value, even nn the mean of the (n/2)(n/2)th and (n/2+1)(n/2+1)th. Discrete table: the first value whose cf exceeds N/2N/2; if a cf equals N/2N/2, the mean of that value and the next.
  • Not CC including the median class. CC is the cf of the classes before it. Classes 0–4, …, 16–20 with f=3,9,10,8,6f=3,9,10,8,6: N=36N=36, C=12C=12, M=8+(18−12)×4/10=10.4M=8+(18-12)\times4/10=10.4.
  • Quick method. Unknown frequencies: one equation from NN or the mean, one from the median formula. Median 14 in class 12–18 (f=12f=12, C=18C=18): 14=12+(N/2−18)×6/1214=12+(N/2-18)\times6/12, so N=44N=44.
Mode
1× asked, last 2025
Z=l+f1−f02f1−f0−f2×h,Mode≈3 Median−2 MeanZ=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h,\qquad \text{Mode}\approx3\,\text{Median}-2\,\text{Mean}
f1f_1: modal class; f0,f2f_0,f_2: classes before and after. The relation holds only for moderately skewed data.
xxffO4812162010Z=913Z=9\frac13
Modal class 8–12: the crossing lines meet above Z=8+10−920−9−8×4=913Z=8+\frac{10-9}{20-9-8}\times4=9\frac13
  • Ungrouped or discrete table: the mode is the value with the highest frequency. A tie for the highest gives two or more modes (bimodal for two); if all values occur equally often, there is no mode. 2,3,3,4,4,4,5,72,3,3,4,4,4,5,7: mode 4.

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10 more concepts in 5 topics

Mean deviation, Variance and standard deviation, Change of origin and scale and Combined groups and corrected data, plus ideas from Probability distributions. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Mean deviation

  • Mean deviation about a point12×
  • MD of consecutive integers and APs3×

Variance and standard deviation

  • Variance and standard deviation13×
  • Sums about another point9×
  • Frequency tables with unknowns7×
  • First n natural numbers and APs8×

Change of origin and scale

  • Linear change y = ax + b10×

Combined groups and corrected data

  • Combined mean and variance15×
  • Correcting wrong observations15×
  • Two missing observations23×

Also asked here: Probability distributions

  • Key points1

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