Summarize the data without hiding its shape: center, spread, transformations and grouped estimates
JEE Main · JEE Advanced80 min study guide10 indexed sections
BY THE END, YOU CAN
Compute and interpret mean, median, mode and measures of dispersion.
Distinguish population-style variance from sample variance and retain correct units.
Derive transformation, combined-group and correction formulas from sums.
Recognize grouped-data estimates and state the assumptions behind interpolation.
BEFORE YOU START
Finite sums, frequencies, squares and square roots.
Basic algebra and ordered data lists.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A summary describes a specified collection
Descriptive statistics compress a data collection into quantities such as its center and spread. Before calculating, identify the observational unit, the variable, its units and the number of observations. A list of measurements and a frequency table may represent the same data, but a table of class intervals usually contains less information than the original list.
A numerical summary does not replace the distribution. Two data sets can have the same mean and different spreads, or the same mean and variance with different shapes. A sensible interpretation states what the summary captures and what it leaves unresolved, especially when outliers or multiple clusters are present.
For observations x₁,…,xₙ with n≥1, the arithmetic mean is x̄=(Σxᵢ)/n. With distinct values xⱼ occurring at frequencies fⱼ, use n=Σfⱼ and x̄=(Σfⱼxⱼ)/n. Frequencies count repeated observations; treating each distinct value as equally frequent changes the data.
A weighted mean Σwᵢxᵢ/Σwᵢ uses nonnegative weights with positive total weight. It lies between the smallest and largest values receiving positive weight. If the weights represent group sizes, it is the combined mean. If they represent something else, explain that meaning rather than automatically calling the result an ordinary unweighted average.
02 / UNDERSTAND & APPLY
2. Different centers answer different questions
The median is determined after sorting the observations. For odd n it is the middle observation, at position (n+1)/2. For even n the usual convention is the average of the two central observations, at positions n/2 and n/2+1. The median need not itself occur in the data when n is even.
The mode is a most frequent value. There may be more than one mode. When every value occurs equally often, conventions vary between calling every value modal and reporting no unique mode; state the convention rather than implying that every data set has exactly one. A modal category can be meaningful even when numerical means are not.
The mean responds to the magnitude of every observation. Changing one value by d changes the mean by d/n. The median depends on ordering and middle positions, so an already extreme value can move further outward without changing it. This resistance does not mean the median is unaffected by every possible data change.
For a strongly skewed distribution, mean and median can differ substantially. There is no universal exact equation tying mean, median and mode together. Any empirical approximation between them is a contextual rule of thumb, not an algebraic identity that may be imposed on arbitrary data.
ORIGINAL WORKED EXAMPLE
One data set, three centers
Find mean, median and mode of 2,4,4,6,9. Then replace 9 by 29 and compare.
The original list is already sorted. Its sum is 25, so the mean is 25/5=5. The third observation is 4, giving median 4.
The value 4 occurs twice while the others occur once, so the unique mode is 4.
Replacing 9 by 29 increases the sum by 20, so the mean increases by 20/5=4 to 9.
The sorted middle observation and the frequency maximum remain unchanged. The new median and mode are still 4, illustrating sensitivity of the mean to the size of an extreme observation.
Answer Original: mean 5, median 4, mode 4. After replacement: mean 9, median 4, mode 4.
03 / UNDERSTAND & APPLY
3. Spread needs both a center and a convention
The range is maximum minus minimum. It is simple but depends only on two observations. Mean absolute deviation about a chosen center c is (1/n)Σ|xᵢ−c|; the chosen center must be stated. Mean deviation about the mean and mean deviation about the median need not have the same value.
The population-style variance of a finite data list is σ²=(1/n)Σ(xᵢ−x̄)². It is nonnegative and equals zero exactly when all observations are identical. Standard deviation σ is its nonnegative square root. If the observations are measured in centimeters, variance is in square centimeters and standard deviation is in centimeters.
Expanding the squares gives Σ(xᵢ−x̄)²=Σxᵢ²−n x̄², since Σxᵢ=n x̄. Therefore σ²=(Σxᵢ²)/n−x̄². This raw-moment formula is algebraically convenient, though direct subtraction of nearly equal large rounded numbers can be numerically unstable.
When estimating a population variance from a random sample, the usual sample variance is s²=Σ(xᵢ−x̄)²/(n−1), for n≥2. It differs from the descriptive divisor-n variance by n/(n−1). The unbiasedness motivation uses a sampling model; it is not a reason to change the denominator silently in a deterministic data-summary question. This lesson states which convention each problem uses.
σ²=Σfⱼ(xⱼ−x̄)²/n=Σfⱼxⱼ²/n−x̄², n=Σfⱼ
The divisor-n descriptive variance; frequencies represent repeated observations.
ORIGINAL WORKED EXAMPLE
Compute spread in two independent ways
For 2,4,4,6,9, find divisor-n variance, standard deviation, sample variance and mean absolute deviation about the mean.
The mean is 5. Squared deviations are 9,1,1,1,16, totaling 28. Thus σ²=28/5 and σ=√(28/5).
The raw sum of squares is 4+16+16+36+81=153. The alternate calculation gives 153/5−25=28/5, agreeing exactly.
The sample variance uses n−1=4, so s²=28/4=7. This is a different requested quantity, not a competing arithmetic answer.
Absolute deviations about 5 are 3,1,1,1,4, totaling 10. Their mean is 2. Squared and absolute deviations measure spread differently.
Answer σ²=28/5; σ=√(28/5); s²=7; mean absolute deviation about the mean=2.
04 / UNDERSTAND & APPLY
4. Centering removes the cross term
For any real center c, write xᵢ−c=(xᵢ−x̄)+(x̄−c). Squaring and summing produces a cross term containing Σ(xᵢ−x̄), which is zero. The result is Σ(xᵢ−c)²=Σ(xᵢ−x̄)²+n(x̄−c)².
The final term is nonnegative and vanishes only when c=x̄. Thus the arithmetic mean uniquely minimizes the sum of squared deviations for a nonempty finite data set. This identity explains why variance is centered at the mean and why using an assumed mean requires a correction term.
For an assumed center a, let dᵢ=xᵢ−a. Then x̄=a+(Σdᵢ)/n and σ²=(Σdᵢ²)/n−((Σdᵢ)/n)². Merely averaging dᵢ² gives deviations about a, not the variance unless a equals the actual mean.
Absolute deviations behave differently: a median minimizes Σ|xᵢ−c|. Pairing the smallest and largest observations shows why: their total distance to c is minimal whenever c lies between them. Intersecting the central intervals gives the middle observation for odd n or the interval between the central two for even n. The conventional even-sample median is one minimizer among that whole interval.
ORIGINAL WORKED EXAMPLE
Recover the mean and variance from shifted totals
Five observations satisfy Σ(xᵢ−10)=15 and Σ(xᵢ−10)²=65. Find their mean and divisor-n variance.
Use a=10 and dᵢ=xᵢ−10. The mean deviation is 15/5=3, so x̄=10+3=13.
The mean squared deviation about 10 is 65/5=13, but this is not yet the variance.
Subtract the squared mean deviation: σ²=13−3²=4.
The centering identity checks the total: squared deviations about the true mean total 5·4=20, and shifting the center by 3 adds 5·9=45, producing 65.
Answer Mean 13; divisor-n variance 4.
05 / UNDERSTAND & APPLY
5. A shift changes location; a scale changes spread
If each observation is transformed by yᵢ=axᵢ+b, then ȳ=a x̄+b by linearity of the sum. Also yᵢ−ȳ=a(xᵢ−x̄), so Var(y)=a²Var(x) and SD(y)=|a|SD(x). The absolute value is essential because standard deviation cannot be negative.
Adding a constant changes the mean and median by that constant but leaves variance, standard deviation and range unchanged. Multiplying by a negative constant reverses the order of observations while multiplying the range and standard deviation by its absolute value. The conventional median still transforms as a·median+b.
The same variance scaling holds for sample variance because the common divisor n−1 is unchanged. Mean absolute deviation about the correspondingly transformed center scales by |a|. These rules can replace long recalculations, but only when the same affine transformation is applied to every observation.
The coefficient of variation is commonly CV=100·σ/x̄ percent when x̄>0 on a meaningful ratio scale. It compares relative spread. It is unchanged by multiplication by a positive constant, but not by translation. Near-zero means or arbitrary-origin scales make this ratio unstable or inappropriate; do not use it as a universal ranking device.
ORIGINAL WORKED EXAMPLE
A negative scale still produces positive spread
A data set has mean 7 and standard deviation 3. Each value becomes y=5−2x. Find the transformed mean and standard deviation.
The mean transforms linearly: ȳ=5−2·7=−9.
The original variance is 3²=9. Multiplication by −2 multiplies variance by (−2)²=4, giving 36.
The transformed standard deviation is √36=6, equivalently |−2|·3.
The constant 5 shifts every value and its mean together, so it contributes nothing to the deviations or spread.
Answer Transformed mean −9; standard deviation 6; variance 36.
06 / UNDERSTAND & APPLY
6. Overall spread includes differences between group means
For groups of sizes n₁,n₂ with means μ₁,μ₂, the combined mean is μ=(n₁μ₁+n₂μ₂)/(n₁+n₂). Averaging the two means without weights is valid only when the group sizes are equal or when the values happen to make the weighting irrelevant.
If σ₁²,σ₂² are divisor-n variances within the groups, the combined variance is [n₁(σ₁²+(μ₁−μ)²)+n₂(σ₂²+(μ₂−μ)²)]/(n₁+n₂). Apply the centering identity within each group to derive it. The first part measures within-group scatter; the second measures separation of the group centers.
A weighted average of within-group variances alone omits the between-group contribution. Even if every observation within each group is identical, the combined variance can be positive when the group means differ. This is why pooling spread requires more than pooling the variance numbers.
If sample variances are supplied instead, first recover each within-group squared-deviation total as (nⱼ−1)sⱼ². Add the between-group terms and then divide by N or N−1 according to the requested combined convention. Mixing n and n−1 formulas without conversion is a common source of error.
ORIGINAL WORKED EXAMPLE
Pool two groups with different means
Group A has 3 observations, mean 4 and divisor-n variance 2. Group B has 2 observations, mean 9 and divisor-n variance 1. Find combined mean and divisor-n variance.
The total size is 5 and total sum is 3·4+2·9=30, giving μ=6.
Group A contributes 3[2+(4−6)²]=3(2+4)=18 to the squared deviations about 6.
Group B contributes 2[1+(9−6)²]=2(1+9)=20. Total squared deviation is 38.
Divide by 5 to obtain 38/5. A raw-moment check gives total squares 3(2+16)+2(1+81)=54+164=218, and 218/5−36=38/5.
Answer Combined mean 6; divisor-n variance 38/5.
07 / UNDERSTAND & APPLY
7. Correct totals before recomputing statistics
If one recorded value w should have been c, the corrected sum is S−w+c and the corrected squared sum is Q−w²+c². Both totals change. Correcting only the mean and leaving the old squared total produces an inconsistent variance.
For a data set of known size n, mean μ and divisor-n variance σ², recover S=nμ and Q=n(σ²+μ²). These totals are sufficient to recompute mean and variance after replacing, adding or removing known observations. The new denominator changes when an observation is added or removed.
Missing observations can often be found from a sum and a squared sum. If two unknown values have sum s and squared sum q, their product is (s²−q)/2. They are roots of t²−st+(s²−q)/2=0. A negative discriminant means no real values satisfy the reported summaries.
The final result should be checked for nonnegative variance and any constraints on the original data, such as nonnegativity or integrality. Rounded reported means and variances may not determine exact missing observations; distinguish exact summaries from rounded measurements.
ORIGINAL WORKED EXAMPLE
One wrong observation changes two totals
Five recorded observations have mean 10 and divisor-n variance 8. A recorded value 12 should have been 2. Find the corrected mean and variance.
The recorded sum is S=5·10=50. The recorded squared sum is Q=5(8+10²)=540.
Replace 12 by 2: S_new=50−12+2=40 and Q_new=540−144+4=400.
The corrected mean is 40/5=8. The corrected variance is 400/5−8²=80−64=16.
The number of observations remains 5 because this is a replacement, not a removal. The variance is nonnegative and the corrected squared-deviation total is 5·16=80.
When only class intervals and frequencies are known, the individual observations are not. Replacing each observation by its class midpoint gives an estimated mean and variance. These are exact for the midpoint-representative data set, but generally approximate for the unknown original data.
For a continuous grouped distribution, a common median estimate is L+h(N/2−C)/f, where L is the lower boundary of the median class, h its width, C the cumulative frequency before it and f its frequency. This linearly interpolates the cumulative count within that class, effectively assuming a uniform distribution there. The class boundaries must be consistent and non-overlapping.
For equal-width classes with a unique modal class and neighboring frequencies f₀,f₁,f₂, a conventional grouped mode estimate is L+h(f₁−f₀)/(2f₁−f₀−f₂). It is a histogram-based interpolation, not an exact recovery of the original most frequent value. Unequal widths require density-aware treatment; simply choosing the largest raw class frequency can be misleading.
Quartiles and percentiles split ordered data by position, but several interpolation conventions exist for finite samples. State the convention before computing. The interquartile range Q₃−Q₁ measures the middle spread. A 1.5-IQR fence is a screening rule for unusual observations, not proof that a point is erroneous or must be deleted.
ORIGINAL WORKED EXAMPLE
Keep midpoint estimates distinct from exact raw-data values
Classes [0,10), [10,20), [20,30) have frequencies 2,5,3. Estimate mean, divisor-N variance and grouped median.
The midpoints are 5,15,25 and total frequency N=10. Weighted sum is 2·5+5·15+3·25=160, so estimated mean is 16.
The weighted squared sum is 2·25+5·225+3·625=3050. Estimated divisor-N variance is 3050/10−16²=49.
Cumulative frequencies are 2,7,10, so the median class is [10,20). Use L=10,h=10,C=2,f=5.
The median estimate is 10+10(5−2)/5=16. Its agreement with the estimated mean is specific to these frequencies, not a general identity.
Answer Estimated mean 16; estimated variance 49; interpolated grouped median 16.
09 / UNDERSTAND & APPLY
9. Original practice with full solutions
Unless a question explicitly requests sample variance, use the divisor-n descriptive convention. Retain exact fractions until the final step.
ORIGINAL PRACTICE · 1
Find mean, median and mode of 1,3,3,5,8.
Reveal answer & reasoning
Mean 4; median 3; mode 3.
The sum is 20 over five observations. The third sorted value is 3, and 3 is the unique repeated value.
ORIGINAL PRACTICE · 2
Find divisor-n variance of 1,3,5.
Reveal answer & reasoning
8/3.
The mean is 3 and squared deviations are 4,0,4. Their average is 8/3.
ORIGINAL PRACTICE · 3
Find the sample variance of the same three observations.
Reveal answer & reasoning
4.
The squared-deviation total is 8, now divided by n−1=2.
ORIGINAL PRACTICE · 4
A data set has mean 12 and variance 9. Add 7 to every observation. What are the new mean and variance?
Reveal answer & reasoning
Mean 19; variance 9.
A common translation shifts the center but leaves each deviation unchanged.
ORIGINAL PRACTICE · 5
A data set has standard deviation 5. Multiply every value by −3. Find the new standard deviation.
Reveal answer & reasoning
15.
Standard deviation scales by the absolute multiplier, |−3|, not by its signed value.
ORIGINAL PRACTICE · 6
Two groups have sizes 4 and 6 and means 10 and 15. Find the combined mean.
Reveal answer & reasoning
13.
The total sum is 4·10+6·15=130 over 10 observations, giving 13.
ORIGINAL PRACTICE · 7
Four observations have mean 6 and squared sum 160. Find divisor-n variance.
Reveal answer & reasoning
4.
The mean square is 160/4=40; subtract the squared mean 36 to obtain 4.
ORIGINAL PRACTICE · 8
The five values 2,4,x,8,10 have mean 6. Find x.
Reveal answer & reasoning
6.
The total must be 5·6=30. The four known values total 24, leaving x=6.
ORIGINAL PRACTICE · 9
Two missing real values have sum 10 and squared sum 58. Find them.
Reveal answer & reasoning
3 and 7, in either order.
Their product is (100−58)/2=21. They solve t²−10t+21=(t−3)(t−7)=0.
ORIGINAL PRACTICE · 10
For sorted data 1,2,3,4,5,6,7,8, use medians of the lower and upper halves for Q₁,Q₃. Find IQR.
Reveal answer & reasoning
Q₁=2.5, Q₃=6.5, IQR=4.
The lower half is 1,2,3,4 and upper half is 5,6,7,8. Average each half’s central pair and subtract.
ORIGINAL PRACTICE · 11
A positive ratio-scale data set has mean 20 and standard deviation 4. Find CV.
Reveal answer & reasoning
20%.
CV=100·4/20=20%. This interpretation assumes a meaningful positive mean and ratio scale.
ORIGINAL PRACTICE · 12
Three identical observations all equal 7. Find their variance and standard deviation.
Reveal answer & reasoning
Both are zero.
Every deviation from their mean 7 is zero. This characterizes zero variance for a nonempty finite list.
10 / UNDERSTAND & APPLY
10. Audit the denominator, units and information loss
A complete statistical calculation states the data count, the weighting and the variance convention. Sorting is necessary for positional summaries, while frequencies must be respected for weighted sums. A variance cannot be negative, and a mean with nonnegative weights cannot lie outside the observed range.
Use transformations and raw totals to avoid unnecessary repetition. A common shift leaves spread unchanged; a common scale changes variance by its square and standard deviation by its absolute value. Combined-group variance includes both internal variation and separation of the means.
For grouped data, call midpoint calculations estimates and explain interpolation assumptions. For quartiles, identify the chosen convention. Do not turn an outlier flag into a claim of error, or an empirical mean-median-mode approximation into an exact theorem.
Finally, interpret the number in the original units and context. A center without spread may conceal a widely scattered distribution, and spread without the center may conceal a scale difference. When two summaries seem contradictory, return to the observations, the denominator and the definition before changing arithmetic.
KEEP THE BIG IDEAS
Quick revision
Means use every magnitude; medians use ordered positions; modes use frequencies.
Variance conventions must be stated, and standard deviation restores the original units.
Centering at the mean minimizes squared deviations.
Pooling groups requires both within-group and between-group variation.
Grouped summaries and finite-sample quantiles depend on stated approximation or convention.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.