Q.Find the variance and standard deviation of the data: 4,8,6,5,9,10,3,7.
Concept understanding — Variance and Standard Deviation of Data
Variance and Standard Deviation of Data
For a data set these measure spread about the mean. For ungrouped data
x1,…,xn with mean xˉ,
σ2=n1∑i=1n(xi−xˉ)2=n∑xi2−xˉ2,
and the standard deviation is σ=σ2. For a frequency distribution with values xi (class midpoints for grouped data) and frequencies
fi, N=∑fi,
σ2=N∑fixi2−(N∑fixi)2.
The coefficient of variation CV=xˉσ×100 compares
relative spread. Variance is unchanged by a shift of origin and scales as
a2 under x↦ax+b. For two groups combined, the pooled variance uses the
group means xˉ1,xˉ2 and the overall mean xˉ:
σ2=n1+n2n1(σ12+d12)+n2(σ22+d22) with
dj=xˉj−xˉ. These formulas cover ungrouped, grouped and combined data.
Variance and standard deviation of both ungrouped and grouped data are part of the NCERT/CBSE Class 11 Mathematics "Statistics" chapter, matching "variance and standard deviation formula class 11 maths" searches. These measures of dispersion are also frequently tested in JEE Main and CET quantitative-aptitude sections.
[!TLDR] Find xˉ, average the squared deviations, then square-root. [!ANSWER] Mean =6.5, Variance =5.25, Standard Deviation =5.25≈2.29.
The data 4,8,6,5,9,10,3,7 has n=8. Sum =4+8+6+5+9+10+3+7=52, so xˉ=52/8=6.5. The deviations are −2.5,1.5,−0.5,−1.5,2.5,3.5,−3.5,0.5, and their squares are 6.25,2.25,0.25,2.25,6.25,12.25,12.25,0.25, summing to 42. Therefore σ2=42/8=5.25, and σ=5.25≈2.29. (Cross-check: ∑xi2=16+64+36+25+81+100+9+49=380, so σ2=380/8−6.52=47.5−42.25=5.25, confirming the same result.) [!ANSWER] Mean =6.5; Variance =5.25; Standard Deviation ≈2.29.
Compute xˉ, form (xi−xˉ)2 for every observation, sum, divide by n for σ2, then take the square root for σ; cross-check with σ2=x2−(xˉ)2.
Squaring the deviations incorrectly for negative values (e.g. treating (−2.5)2 as −6.25 instead of the correct positive 6.25) is a common error.
- CBSE 2026Set ANNUAL1 markMCQQ.If the variance of the data is 144 then the standard deviation of the data is(a) 144(b) -12(c) 12(d) 14
›Reveal solutionSolution
Standard deviation is defined as the positive square root of variance.
By definition, standard deviation σ=variance. Given variance =144:
σ=144=12
Standard deviation is always taken as the non-negative root (a spread measure cannot be negative), which rules out option (b) −12.
✓Final answer(c) 12.
- CBSE 2026Set ANNUAL1 markMCQQ.The square of standard deviation is —(a) Mean-deviation(b) Range(c) Variance(d) Mean
›Reveal solutionSolution
The square of the standard deviation is called the variance, option (c).
Standard deviation (σ) measures the spread of data about the mean. By definition, variance (σ2) is exactly the square of the standard deviation: Variance=σ2=(Standard Deviation)2.
So among the given options, the square of standard deviation is the Variance, not mean-deviation, range, or mean.
✓Final answerThe correct option is (c) Variance.
- CBSE 2026Set ANNUAL1 markQ.Fill in the blank: If variance of given data is 225, then the standard deviation of the same data will be ______.
›Reveal solutionSolution
Standard deviation =Variance; here 225=15.
By definition, variance is the square of the standard deviation, so standard deviation =variance.
Given variance =225: standard deviation =225=15.
✓Final answerStandard deviation =15.
- CBSE 2024Set ANNUAL1 markMCQQ.If the standard deviation of a data is 0.012, find the variance.(a) 0.144(b) 0.00144(c) 0.000144(d) None of these
›Reveal solutionSolution
Variance is defined as the square of the standard deviation.
By definition, variance =(standard deviation)2.
Given standard deviation =0.012:
Variance=(0.012)2=0.012×0.012=0.000144
✓Final answer(c) 0.000144.
- CBSE 2024Set ANNUAL1 markQ.If the standard deviation of obtained marks of a class students is 1.6, then find the variance.
›Reveal solutionSolution
Given standard deviation =1.6, the variance is 2.56.
By definition, variance =(standard deviation)2. Here standard deviation =1.6, so variance =1.6×1.6=2.56.
✓Final answerVariance =2.56.
- CBSE 2024Set ANNUAL1 markQ.Fill in the blank: In a series, the difference between the maximum value and the minimum value is called ______.
›Reveal solutionSolution
This difference is called the range of the series, the simplest measure of dispersion.
Step 1. Range is defined as: Range = Maximum value − Minimum value.
Step 2. It measures how spread out the data is, from the smallest to the largest observation.
✓Final answerThe difference between the maximum value and the minimum value in a series is called its range.
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