Concept understanding — Mean Variance Natural Numbers
Mean and Variance of Natural Numbers
Let’s start with something you already know: the mean (average) and variance (spread) of a set of numbers. If I give you the first five natural numbers — 1, 2, 3, 4, 5 — you can compute their mean and variance easily. But what if I ask: What is the mean of all natural numbers? That’s infinite, so it doesn’t make sense directly. Instead, we ask: What is the mean of the first n natural numbers? And then we see how it behaves as n grows.
That’s the core idea: we study the mean and variance of the first n natural numbers as a function of n, and often look at what happens when n becomes very large.
Intuition First
Imagine you line up the numbers 1,2,3,…,n on a number line. Their average is somewhere in the middle — roughly n/2. More precisely, the mean of the first n natural numbers is 2n+1. For n=5, that’s 3, which matches your intuition.
Now, variance measures how spread out the numbers are around that mean. For small n, the spread is small; for large n, the spread grows. The variance of the first n natural numbers turns out to be 12n2−1. For n=5, that’s 1225−1=2, which is a moderate spread.
Note
These formulas assume we are using population variance (dividing by n, not n−1). In exam contexts, always check which variance definition is expected — but for natural numbers, population variance is standard.
Precise Statement
Let X be a random variable that takes values 1,2,3,…,n with equal probability 1/n. Then:
Mean: μn=2n+1
Variance: σn2=12n2−1
These are exact formulas for any positive integer n.
Derivation (Why These Formulas?)
Mean
The sum of the first n natural numbers is 1+2+⋯+n=2n(n+1).
Since there are n numbers, the mean is:
μn=n1⋅2n(n+1)=2n+1
Variance
Variance is the average of squared deviations from the mean:
σn2=n1∑k=1n(k−μn)2
A cleaner way uses the identity: σ2=E[X2]−(E[X])2.
First, E[X2]=n1∑k=1nk2. The sum of squares formula is ∑k=1nk2=6n(n+1)(2n+1). So:
E[X2]=n1⋅6n(n+1)(2n+1)=6(n+1)(2n+1)
Now, (E[X])2=(2n+1)2=4(n+1)2.
Therefore:
σn2=6(n+1)(2n+1)−4(n+1)2
Factor (n+1):
σn2=(n+1)[62n+1−4n+1]
Compute the bracket: common denominator 12:
122(2n+1)−3(n+1)=124n+2−3n−3=12n−1
Thus:
σn2=(n+1)⋅12n−1=12n2−1
What This Tells You
The mean grows linearly with n — roughly half of n.
The variance grows quadratically — roughly n2/12 for large n.
For large n, the standard deviation σn≈12n≈0.2887n, meaning the spread is about 29% of the range.
Tip
A quick memory aid: For the first n natural numbers, mean is 2n+1 and variance is 12n2−1. Notice the denominator 12 — it’s the same as the variance of a continuous uniform distribution over [0,1], which is 1/12.
Common Exam Pitfall
Watch out
Do not confuse the variance of the first n natural numbers with the variance of a sample from a larger population. Here, the set {1,2,…,n} is the entire population, so we divide by n, not n−1. If a problem says “variance of the first n natural numbers,” use 12n2−1.
Quick Check
For n=1: mean = 1, variance = 0 (only one number, no spread). Formula gives 1212−1=0 — correct.
For n=2: numbers 1,2, mean = 1.5, variance = 2(1−1.5)2+(2−1.5)2=20.25+0.25=0.25. Formula gives 124−1=0.25 — correct.
You now have the complete picture: from intuition to derivation to exam-ready formulas.
Mean and Variance of the First n Natural Numbers is a classic result taught in the NCERT Class 11 Mathematics chapter on Statistics, matching searches like "mean and variance of natural numbers formula" or "statistics important questions class 11 maths". Because it combines the sum-of-squares formula with statistics, it's a frequently asked derivation-and-apply question in both CBSE boards and JEE Main.
Concept: Mean & Variance of Natural Numbers
The first n natural numbers are 1,2,3,…,n. Their sum is 2n(n+1), and sum of squares is 6n(n+1)(2n+1).
The mean of the first n natural numbers is 2n+1, and the variance is 12n2−1. This follows from summing an arithmetic progression and using the formula for the sum of squares.
The first n natural numbers are 1,2,3,…,n. The mean is just the average — but the variance measures how spread out these numbers are around that average. Since the numbers are equally spaced, both quantities have neat closed forms.
Let’s derive them step by step.
Mean (average)
The sum of the first n natural numbers is the classic arithmetic series:
1+2+⋯+n=2n(n+1)
The mean xˉ is this sum divided by n:
xˉ=n1⋅2n(n+1)=2n+1
So the mean sits exactly halfway between 1 and n.
Variance — the definition
Variance is the average of the squared deviations from the mean. For a population (which these n numbers are), we use:
σ2=n1∑i=1n(xi−xˉ)2
A more convenient computational form is:
σ2=n1∑i=1nxi2−xˉ2
This avoids subtracting the mean from each term individually.
Sum of squares
The sum of squares of the first n natural numbers is a standard result:
∑i=1ni2=6n(n+1)(2n+1)
So the average of the squares is:
n1∑i=1ni2=6(n+1)(2n+1)
Plug into the variance formula
Using xˉ=2n+1, we have xˉ2=4(n+1)2. Therefore:
σ2=6(n+1)(2n+1)−4(n+1)2
Simplify
Factor out (n+1):
σ2=(n+1)[62n+1−4n+1]
Find a common denominator (12):
62n+1=124n+2,4n+1=123n+3
Subtract:
124n+2−(3n+3)=12n−1
So:
σ2=(n+1)⋅12n−1=12n2−1
Watch out
A common mistake is to use the sample variance formula (dividing by n−1) instead of the population variance. Here, since we are considering the entire set of the first n natural numbers, we divide by n, not n−1. Using n−1 would give 12n(n+1), which is incorrect for this problem.
Tip
Notice that the variance grows roughly as n2/12 — so the spread increases quadratically with n, while the mean grows linearly. This makes sense: as n gets larger, the numbers are more spread out relative to their average.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KCET 2020Set A-11 markMCQ
Q.The standard deviation of the data 6,7,8,9,10 is
(A) 2
(B) 10
(C) 2
(D) 10
›Reveal solutionSolution
The standard deviation of the data 6,7,8,9,10 is 2, which corresponds to option (A).
The concept here is standard deviation — a measure of how spread out the numbers are from their mean. For a small, evenly spaced set like this, you could almost guess the answer: the numbers are symmetric around 8, each step is 1 unit away, and the average squared deviation works out neatly. But let’s do it properly.
The standard deviation σ for a population (or ungrouped data) is defined as:
σ=n∑(xi−xˉ)2
where xˉ is the mean and n is the number of observations. We’ll compute step by step.
Find the meanxˉ:
xˉ=56+7+8+9+10=540=8
Compute each deviation from the mean and square it:
6−8=−2, square = 4
7−8=−1, square = 1
8−8=0, square = 0
9−8=1, square = 1
10−8=2, square = 4
Sum the squared deviations:
4+1+0+1+4=10
Divide by the number of observations (n=5) to get the variance:
Variance=510=2
Take the square root to get the standard deviation:
σ=2
Watch out
A common mistake is to forget to take the square root at the end — that would give variance 2, not standard deviation. Also, some students divide by n−1 (for sample standard deviation), but here the data is treated as the whole set, so divide by n.
Tip
For an arithmetic progression like this, the standard deviation is 12(d)2(n2−1) where d is the common difference. Here d=1, n=5, so σ=121⋅(25−1)=1224=2. A quick check!
✓Final answer
The standard deviation is 2, which is option (A).
KCET 2019Set A-11 markMCQ
Q.Mean and standard deviation of 100 items are 50 and 4 respectively. The sum of all squares of the items is
(A) 251600
(B) 256100
(C) 266000
(D) 261600
›Reveal solutionSolution
Given the mean and standard deviation of 100 items, the sum of squares is found using the formula σ2=n∑xi2−(xˉ)2. The correct sum is 251600.
The key idea here is that the standard deviation is a measure of spread, and it connects directly to the sum of squares through the variance formula. You don't need the individual items — just the mean, the standard deviation, and the number of items.
The variance σ2 is defined as the average of the squared deviations from the mean. But there's an equivalent computational form that's much more practical when you have summary statistics:
σ2=n∑xi2−(xˉ)2
This formula comes from expanding ∑(xi−xˉ)2=∑xi2−nxˉ2 and then dividing by n. It lets you jump straight to the sum of squares without ever seeing the raw data.
Let's work through it step by step.
Write down what you know.
Number of items: n=100
Mean: xˉ=50
Standard deviation: σ=4
So variance: σ2=42=16
Plug into the variance formula.
16=100∑xi2−(50)2
Simplify the known term.(50)2=2500, so:
16=100∑xi2−2500
Solve for the sum of squares.
Add 2500 to both sides:
2516=100∑xi2
Multiply both sides by 100:
∑xi2=251600
Watch out
A common mistake is to forget that the standard deviation is the square root of the variance. If you plug in σ=4 directly into the formula without squaring it first, you'll get a wildly wrong answer. Always square the standard deviation to get the variance before using it.
Tip
Notice that the sum of squares is a large number — that's expected because you're adding 100 squared numbers, each around 50, so each contributes roughly 2500. The total should be around 100×2500=250000, and the variance correction adds a bit more. This quick sanity check would immediately rule out options (B) and (C) as too large.
✓Final answer
The sum of all squares of the items is 251600, which corresponds to option (A).