Chapter 13 ยท Statistics

13.3 Variance and Standard Deviation

Square the deviations instead of dropping their signs: the mean of the squared deviations is the variance $\sigma^2$, and its square root, the standard deviation $\sigma$, is the measure of spread used across science, finance and industry.

The mean deviation fixed the problem of cancelling deviations by taking absolute values. It works, but the modulus is awkward to handle in algebra: $|x-a|$ has a corner at $x=a$, the sum $\sum|x_i-a|$ has to be split into cases, and there is no neat formula that turns $\sum|x_i-\bar x|$ into something simpler. There is a second way to make every deviation positive: square it. Squares are never negative, they are smooth, and $(x_i-\bar x)^2$ can be expanded by ordinary algebra.

Averaging the squared deviations gives the variance, and taking its square root brings us back to the units of the data: the standard deviation. It is the number a pharmaceutical company quotes for the variation in the amount of medicine in each tablet, that an engineer uses to set tolerances on machined parts, that a weather scientist uses to say how unusual a monsoon was, and that an investor uses to measure how risky a share is. In this section we compute it for lists, for discrete frequency tables and for class-interval tables, and we develop the shortcut that makes large tables manageable.

Variance and Standard Deviation of a List

๐Ÿ“˜ Definition 13.4 โ€” Variance and standard deviation

The variance of the observations $x_1,x_2,\dots,x_n$ with mean $\bar x$ is the mean of the squared deviations from the mean: $$\sigma^2=\frac1n\sum_{i=1}^{n}(x_i-\bar x)^2.$$ The standard deviation is its non-negative square root, $$\sigma=\sqrt{\frac1n\sum_{i=1}^{n}(x_i-\bar x)^2}.$$ $\sigma$ has the same unit as the data ($\sigma^2$ has the square of that unit); $\sigma=0$ exactly when all the observations are equal.

Squaring gives large deviations extra weight: a deviation of $4$ contributes $16$, four times as much as a deviation of $2$. So $\sigma$ reacts more strongly than the mean deviation to observations far from the mean, and for any data $\sigma\ge\text{M.D.}(\bar x)$. (We always divide by $n$ in this course. Some calculators and spreadsheets also offer a version that divides by $n-1$, used when the data are a sample from a larger population; you will meet it in later statistics courses.)

โœ๏ธ Example 1 โ€” Variance from the definition
Find the variance and standard deviation of $6, 9, 3, 12, 10$.
Mean $\bar x=\dfrac{40}{5}=8$.
Table
$x_i$6931210
$x_i-\bar x$โˆ’21โˆ’542
$(x_i-\bar x)^2$4125164
Answer $\sum(x_i-\bar x)^2=50$, so $\sigma^2=\dfrac{50}{5}=10$ and $\sigma=\sqrt{10}=3.16$ (to $2$ d.p.).
Compare The mean deviation is $\tfrac{2+1+5+4+2}{5}=2.8$, a little less than $\sigma$, as expected.

When the mean is not a whole number, the squared deviations become messy decimals. Expanding the square gives a formula that needs only $\sum x_i$ and $\sum x_i^2$.

๐Ÿ“ Theorem 13.5 โ€” "Mean of the squares minus the square of the mean"

For observations $x_1,\dots,x_n$ with mean $\bar x$, $$\sigma^2=\frac1n\sum_{i=1}^{n}x_i^2-\bar x^{\,2},\qquad\text{equivalently}\qquad \sigma=\frac1n\sqrt{n\sum x_i^2-\left(\sum x_i\right)^2}.$$

Proof. Expand each square: $(x_i-\bar x)^2=x_i^2-2\bar x\,x_i+\bar x^{\,2}$. Adding over $i$, and using $\sum x_i=n\bar x$, $$\sum(x_i-\bar x)^2=\sum x_i^2-2\bar x\sum x_i+n\bar x^{\,2}=\sum x_i^2-2n\bar x^{\,2}+n\bar x^{\,2}=\sum x_i^2-n\bar x^{\,2}.$$ Divide by $n$ for the first form. For the second, multiply $\sigma^2=\tfrac1n\sum x_i^2-\tfrac{1}{n^2}\left(\sum x_i\right)^2$ by $n^2$ and take square roots. $\blacksquare$

Two consequences are used constantly: $\sum x_i^2=n\left(\sigma^2+\bar x^{\,2}\right)$, which recovers the sum of squares from a given mean and variance; and $\sigma^2\ge0$ gives $\tfrac1n\sum x_i^2\ge\bar x^{\,2}$ โ€” the mean of the squares is never less than the square of the mean.

โœ๏ธ Example 2 โ€” Using $\sum x_i$ and $\sum x_i^2$
Find the variance and standard deviation of $1, 2, 4, 6, 8$.
Sums $\sum x_i=21$, so $\bar x=4.2$; $\sum x_i^2=1+4+16+36+64=121$.
Theorem 13.5 $\sigma^2=\dfrac{121}{5}-4.2^2=24.2-17.64=6.56$.
Answer $\sigma=\sqrt{6.56}=2.56$ (to $2$ d.p.).
Check Directly: deviations $-3.2,-2.2,-0.2,1.8,3.8$ have squares $10.24+4.84+0.04+3.24+14.44=32.8$, and $\tfrac{32.8}{5}=6.56$ โœ“.
โœ๏ธ Example 3 โ€” Working backwards to missing observations
Five observations have mean $7$ and variance $5.2$. Three of them are $5, 7, 10$. Find the other two.
Sum Call them $a,b$. $\sum x_i=5\times7=35$, so $a+b=35-22=13$.
Sum of squares $\sum x_i^2=n(\sigma^2+\bar x^{\,2})=5(5.2+49)=271$, so $a^2+b^2=271-(25+49+100)=97$.
Solve $2ab=(a+b)^2-(a^2+b^2)=169-97=72$, $ab=36$. So $a,b$ are the roots of $t^2-13t+36=0$, i.e. $(t-4)(t-9)=0$: the observations are $4$ and $9$.
Check $5,7,10,4,9$: mean $7$; squared deviations $4,0,9,9,4$ add to $26$, and $\tfrac{26}{5}=5.2$ โœ“.
โœ๏ธ Example 4 โ€” Correcting a wrongly copied observation
The mean and standard deviation of $25$ observations were calculated as $40$ and $3$. It was later found that one observation, $34$, had been copied as $43$. Find the correct mean and standard deviation.
Wrong sums $\sum x_i=25\times40=1000$; $\sum x_i^2=25(3^2+40^2)=25\times1609=40\,225$.
Correct sums Remove $43$ and put in $34$: $\sum x_i=1000-43+34=991$; $\sum x_i^2=40\,225-43^2+34^2=40\,225-1849+1156=39\,532$.
Correct mean $\bar x=\dfrac{991}{25}=39.64$.
Correct ฯƒ $\sigma^2=\dfrac{39\,532}{25}-39.64^2=1581.28-1571.3296=9.9504$, so $\sigma=3.15$ (to $2$ d.p.).
Interactive 1 โ€” Variance as an Average of Squares

Each observation's deviation from the mean is drawn as a square with that deviation as its side, so its area is $(x_i-\bar x)^2$. The variance is the average area; the standard deviation is the side of a square with that average area (the purple square on the right). The shaded band is $\bar x\pm\sigma$. Drag any dot and watch one far-away point blow up the variance, because its square grows with the square of its distance. Press Play to stretch and squeeze the data.

dataโ€”
$\bar x$โ€”
$\sigma^2=\tfrac1n\sum(x_i-\bar x)^2$โ€”
$\sigma$โ€”
M.D.$(\bar x)$โ€”

Frequency Distributions and the Shortcut Method

For a frequency table we weight each squared deviation by its frequency, exactly as we weighted the distances in Section 13.2: with $N=\sum f_i$, $$\sigma^2=\frac1N\sum f_i(x_i-\bar x)^2=\frac1N\sum f_ix_i^2-\bar x^{\,2},$$ where for class intervals $x_i$ is the class mark. (The second form follows by the same expansion as Theorem 13.5, with every sum weighted by $f_i$.) For a table with large class marks the squares $x_i^2$ are huge; coding the class marks as small integers makes the work light.

๐Ÿ“ Theorem 13.6 โ€” Shortcut (step-deviation) method for the variance

Let the class marks (or values) $x_i$ be equally spaced with common difference $h$, let $A$ be any one of them, and put $u_i=\dfrac{x_i-A}{h}$. Then $$\sigma_x^2=h^2\,\sigma_u^2=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right],$$ that is, $$\sigma_x=\frac hN\sqrt{N\sum f_iu_i^2-\left(\sum f_iu_i\right)^2}.$$

Proof. $x_i=A+hu_i$, and by Theorem 13.3 $\bar x=A+h\bar u$, where $\bar u=\tfrac1N\sum f_iu_i$. Subtracting, $x_i-\bar x=h(u_i-\bar u)$, so $(x_i-\bar x)^2=h^2(u_i-\bar u)^2$. Multiply by $f_i$, add and divide by $N$: $\sigma_x^2=h^2\sigma_u^2$. Finally apply Theorem 13.5 (weighted by $f_i$) to the $u_i$: $\sigma_u^2=\tfrac1N\sum f_iu_i^2-\bar u^{\,2}$. $\blacksquare$

The proof shows why the method works: subtracting $A$ slides every value and the mean by the same amount, which changes no deviation; dividing by $h$ shrinks every deviation by the factor $h$, so the variance shrinks by $h^2$ and we multiply back at the end. Whatever $A$ you choose, the answer is the same (try it in Interactive 2).

โœ๏ธ Example 5 โ€” A discrete distribution
Find the variance and standard deviation of the goals data of Section 13.2: $x_i: 0,1,2,3,4,5$ with $f_i: 5,10,12,8,3,2$ (mean $2$, mean deviation $1$).
Squares $(x_i-2)^2=4,1,0,1,4,9$; $f_i(x_i-2)^2=20,10,0,8,12,18$; sum $68$.
Answer $\sigma^2=\dfrac{68}{40}=1.7$ and $\sigma=\sqrt{1.7}=1.30$ goals (to $2$ d.p.).
Check $\sum f_ix_i^2=0+10+48+72+48+50=228$, and $\tfrac{228}{40}-2^2=5.7-4=1.7$ โœ“. As expected, $\sigma=1.30\gt\text{M.D.}=1$.
โœ๏ธ Example 6 โ€” A continuous distribution by the shortcut method
Find the mean and standard deviation of the marks of $50$ students (Section 13.2, Example 3):
Marks0โ€“1010โ€“2020โ€“3030โ€“4040โ€“5050โ€“6060โ€“70
Students361114853
Code Class marks $5,15,\dots,65$; $A=35$, $h=10$, $u_i=\dfrac{x_i-35}{10}$.
Table
Class$f_i$$x_i$$u_i$$f_iu_i$$f_iu_i^2$
0โ€“1035โˆ’3โˆ’927
10โ€“20615โˆ’2โˆ’1224
20โ€“301125โˆ’1โˆ’1111
30โ€“401435000
40โ€“50845188
50โ€“6055521020
60โ€“703653927
Total50โˆ’5117
Mean $\bar x=35+10\times\dfrac{-5}{50}=34$ marks.
ฯƒ $\sigma=\dfrac{10}{50}\sqrt{50\times117-(-5)^2}=\dfrac15\sqrt{5825}=15.26$ marks (to $2$ d.p.). Equivalently $\sigma_u^2=\tfrac{117}{50}-(-0.1)^2=2.33$ and $\sigma=10\sqrt{2.33}$.
Compare The mean deviation of these marks was $12$; again $\sigma$ is larger.
Interactive 2 โ€” Shortcut Calculator for a Frequency Table

Type the lower limit of the first class, the class width and the frequencies; the table fills in $x_i$, $u_i=\tfrac{x_i-A}{h}$, $f_iu_i$ and $f_iu_i^2$, and the histogram marks $\bar x$ and the band $\bar x\pm\sigma$. Change the assumed mean $A$ โ€” or press Play to step it through every class mark. Every $u_i$ and both column totals change, yet $\bar x$ and $\sigma$ never do; the last readout repeats $\sigma$ from $\tfrac1N\sum f_i(x_i-\bar x)^2$ as a check.

$\sum f_iu_i$ ยท $\sum f_iu_i^2$โ€”
$\bar x=A+h\tfrac{\sum f_iu_i}{N}$โ€”
$\sigma$ (shortcut)โ€”
$\sigma$ (direct check)โ€”
โš ๏ธ Common mistakes with the variance and standard deviation

1. $\left(\sum x_i\right)^2$ for $\sum x_i^2$. $\sum x_i^2$ squares first and then adds ($1^2+2^2=5$); $\left(\sum x_i\right)^2$ adds first ($3^2=9$).

2. Forgetting to subtract $\bar x^{\,2}$ in Theorem 13.5, or subtracting $\bar x$ instead of $\bar x^{\,2}$.

3. Not multiplying back by $h^2$ (or $h$ for $\sigma$) in the shortcut method. $\sigma_u$ is the spread of the coded values, not of the data.

4. Reporting the variance as the spread. If marks are out of $70$, $\sigma^2=233$ "marksยฒ" is not a typical distance; $\sigma=15.26$ marks is.

5. A negative variance. $\sigma^2\ge0$ always. A negative answer from $\tfrac1n\sum x_i^2-\bar x^{\,2}$ means an arithmetic slip โ€” often a rounded $\bar x$ squared too early. Keep full accuracy until the end.

6. Correcting only $\sum x_i$. When an observation is corrected, update both $\sum x_i$ and $\sum x_i^2$ (Example 4).

๐Ÿš€ Beyond the board syllabus (JEE Advanced)

Not in the CBSE syllabus, and not in JEE Main (which lists only mean deviation, variance and standard deviation); JEE Advanced lists the analysis of frequency distributions with the same mean but different variances, and the coefficient of variation is the tool for it. A standard deviation of $10$ runs means a lot for a batsman who averages $15$ and little for one who averages $80$. To compare the variability of data with different means (or different units), divide the standard deviation by the mean: $$\text{C.V.}=\frac{\sigma}{\bar x}\times100\%\qquad(\bar x\gt0).$$ The series with the smaller C.V. is the more consistent. (The proof of Theorem 13.6 also shows that if $y_i=ax_i+b$ then $\bar y=a\bar x+b$ and $\sigma_y=|a|\,\sigma_x$: adding a constant changes neither $\sigma$ nor the variance.)

Worked example. Over a season batsman P scores a mean of $50$ runs per innings with $\sigma=15$; batsman Q a mean of $40$ with $\sigma=10$. Who is more consistent?
Solution. C.V.(P) $=\tfrac{15}{50}\times100=30\%$; C.V.(Q) $=\tfrac{10}{40}\times100=25\%$. Q is more consistent, although P scores more on average. Check: P's typical swing is $\tfrac{3}{10}$ of his average, Q's only $\tfrac14$ of his โœ“.

Exercises

I
Variance and standard deviation of a list. (Give $\sigma$ to $2$ decimal places where needed.)
  • 1 $2, 4, 6, 8, 10$
  • 2 $5, 12, 3, 18, 6, 8, 2, 10$
  • 3 the first ten multiples of $3$
  • 4 $1.2, 2.4, 1.8, 3.0, 2.1$
  • 5 $3, 7, 8, 12, 15$, using $\sum x_i$ and $\sum x_i^2$
  • 6 $10, 20, 30, 40, 50$
  • 7 Six observations have mean $7$ and variance $7$. Four of them are $3, 6, 8, 11$. Find the other two.
  • 8 The mean and variance of $10$ observations were found to be $15$ and $9$. Later it was noticed that the observation $21$ had been read as $12$. Find the correct mean and standard deviation.
Check: mean, variance and standard deviation of your own list
$n$ ยท $\sum x_i$ ยท $\sum x_i^2$โ€”
$\bar x$โ€”
$\sigma^2$โ€”
$\sigma$โ€”
1 $\bar x=6$; squared deviations $16,4,0,4,16$; $\sigma^2=\tfrac{40}{5}=8$, $\sigma=2.83$.
2 $\bar x=\tfrac{64}{8}=8$; deviations $-3,4,-5,10,-2,0,-6,2$; squares add to $194$; $\sigma^2=24.25$, $\sigma=4.92$.
3 $3,6,\dots,30$: $\sum x_i=165$, $\bar x=16.5$; $\sum x_i^2=9(1^2+2^2+\cdots+10^2)=9\times385=3465$; $\sigma^2=346.5-272.25=74.25$, $\sigma=8.62$.
4 $\bar x=\tfrac{10.5}{5}=2.1$; deviations $-0.9,0.3,-0.3,0.9,0$; squares add to $1.8$; $\sigma^2=0.36$, $\sigma=0.6$.
5 $\sum x_i=45$, $\bar x=9$; $\sum x_i^2=9+49+64+144+225=491$; $\sigma^2=98.2-81=17.2$, $\sigma=4.15$.
6 $\bar x=30$; squared deviations $400,100,0,100,400$; $\sigma^2=200$, $\sigma=10\sqrt2=14.14$.
7 $a+b=42-28=14$; $\sum x_i^2=6(7+49)=336$, so $a^2+b^2=336-230=106$; $ab=\tfrac{196-106}{2}=45$; $t^2-14t+45=0$ gives $5$ and $9$.
8 Wrong: $\sum x_i=150$, $\sum x_i^2=10(9+225)=2340$. Correct: $\sum x_i=150-12+21=159$, $\sum x_i^2=2340-144+441=2637$. $\bar x=15.9$; $\sigma^2=263.7-252.81=10.89$; $\sigma=3.3$.
II
Frequency distributions. Find the mean and the standard deviation.
  • 1
    $x_i$48121620
    $f_i$361074
  • 2
    $x_i$12345
    $f_i$4812106
  • 3
    Class0โ€“1010โ€“2020โ€“3030โ€“4040โ€“50
    $f_i$5815166
  • 4 Daily wages of $50$ workers:
    Wage (โ‚น)100โ€“200200โ€“300300โ€“400400โ€“500500โ€“600
    Workers8122064
  • 5 Heights of $50$ students:
    Height (cm)140โ€“145145โ€“150150โ€“155155โ€“160160โ€“165165โ€“170
    Students49151273
  • 6
    Class0โ€“2020โ€“4040โ€“6060โ€“8080โ€“100
    $f_i$5122094
  • 7 For $20$ observations, $\sum(x_i-10)=12$ and $\sum(x_i-10)^2=180$.
  • 8 For $40$ observations, $\sum x_i=200$ and $\sum x_i^2=1160$.
1 $A=12$, $h=4$, $u_i=-2,\dots,2$: $\sum f_iu_i=3$, $\sum f_iu_i^2=41$, $N=30$. $\bar x=12+4\cdot\tfrac{3}{30}=12.4$; $\sigma=\tfrac{4}{30}\sqrt{30\cdot41-9}=\tfrac{4}{30}\sqrt{1221}=4.66$.
2 $N=40$, $\sum f_ix_i=126$, $\bar x=3.15$; $\sum f_ix_i^2=454$; $\sigma^2=11.35-9.9225=1.4275$, $\sigma=1.19$.
3 $A=25$, $h=10$: $\sum f_iu_i=10$, $\sum f_iu_i^2=68$, $N=50$. $\bar x=25+10\cdot\tfrac{10}{50}=27$; $\sigma=\tfrac{10}{50}\sqrt{50\cdot68-100}=\tfrac15\sqrt{3300}=11.49$.
4 $A=350$, $h=100$: $\sum f_iu_i=-14$, $\sum f_iu_i^2=66$. $\bar x=350-28=\text{โ‚น}\,322$; $\sigma=\tfrac{100}{50}\sqrt{50\cdot66-196}=2\sqrt{3104}=\text{โ‚น}\,111.43$.
5 $A=152.5$, $h=5$, $u_i=-2,\dots,3$: $\sum f_iu_i=18$, $\sum f_iu_i^2=92$. $\bar x=152.5+5\cdot\tfrac{18}{50}=154.3$ cm; $\sigma=\tfrac{5}{50}\sqrt{50\cdot92-324}=\tfrac1{10}\sqrt{4276}=6.54$ cm.
6 $A=50$, $h=20$: $\sum f_iu_i=-5$, $\sum f_iu_i^2=57$. $\bar x=50-20\cdot\tfrac{5}{50}=48$; $\sigma=\tfrac{20}{50}\sqrt{50\cdot57-25}=0.4\sqrt{2825}=21.26$.
7 With $u_i=x_i-10$ ($h=1$): $\bar x=10+\tfrac{12}{20}=10.6$; $\sigma^2=\tfrac{180}{20}-\left(\tfrac{12}{20}\right)^2=9-0.36=8.64$, $\sigma=2.94$.
8 $\bar x=5$; $\sigma^2=\tfrac{1160}{40}-25=29-25=4$, $\sigma=2$.
III
Applications and reasoning.
  • 1 The daily screen times (in hours) of eight students were $2, 3, 3, 4, 5, 6, 6, 7$. Find the mean and the standard deviation.
  • 2 For $1, 2, 3, 4, 5$ find the standard deviation and the mean deviation about the mean, and check that $\sigma\ge\text{M.D.}(\bar x)$.
  • 3 Prove that $\sigma=0$ if and only if all the observations are equal.
  • 4 In a test the mean mark is $60$ and the standard deviation is $8$. How many standard deviations above the mean is a mark of $76$? Which mark is $1.5$ standard deviations below the mean?
  • 5 For $15$ observations the sum of the squares of the deviations from their mean is $240$. Find the standard deviation.
  • 6 Find the standard deviation of the volumes (ml) filled by machine P in Section 13.1: $498, 503, 500, 497, 502, 501, 499, 500$.
  • 7 For the screen-time data of part 1, how many of the eight observations lie within one standard deviation of the mean, i.e. in $[\bar x-\sigma,\ \bar x+\sigma]$?
  • 8 Show that $\sum x_i^2\ge n\bar x^{\,2}$ for any observations, with equality only when they are all equal.
  • 9 Data on rainfall are in millimetres. In what unit is the variance? Why do weather reports quote the standard deviation instead?
1 $\bar x=\tfrac{36}{8}=4.5$ h; squared deviations $6.25,2.25,2.25,0.25,0.25,2.25,2.25,6.25$ add to $22$; $\sigma^2=2.75$, $\sigma=1.66$ h.
2 $\bar x=3$; $\sigma^2=\tfrac{4+1+0+1+4}{5}=2$, $\sigma=1.41$; M.D. $=\tfrac{2+1+0+1+2}{5}=1.2$; $1.41\ge1.2$ โœ“.
3 $\sigma^2=\tfrac1n\sum(x_i-\bar x)^2$ is a sum of non-negative terms, so it is $0$ exactly when every term is $0$, i.e. every $x_i=\bar x$. Conversely, if all $x_i=c$ then $\bar x=c$ and every deviation is $0$.
4 $\tfrac{76-60}{8}=2$ standard deviations above. $60-1.5\times8=48$.
5 $\sigma^2=\tfrac{240}{15}=16$, $\sigma=4$.
6 $\bar x=500$; squared deviations $4,9,0,9,4,1,1,0$ add to $28$; $\sigma^2=3.5$, $\sigma=1.87$ ml.
7 $\sigma=1.658\ldots$, so the interval is about $[2.84,\ 6.16]$; it contains $3,3,4,5,6,6$ โ€” six of the eight observations.
8 By Theorem 13.5, $\tfrac1n\sum x_i^2-\bar x^{\,2}=\sigma^2\ge0$, so $\sum x_i^2\ge n\bar x^{\,2}$; equality means $\sigma=0$, i.e. all observations equal (part 3).
9 mmยฒ. A "typical squared distance" in mmยฒ has no direct meaning for rainfall; its square root $\sigma$ is back in mm and can be compared with the mean rainfall itself.
๐ŸŽฏ Exam corner Board ยท 1 markBoard ยท 2 marksBoard ยท 3 marksBoard ยท 5 marksJEE MainJEE Advanced

How this is examined: the Board's long question is usually the mean and standard deviation of a class-interval table by the shortcut method (5 marks); shorter questions correct a wrongly recorded observation or work back from a given mean and variance. JEE Main asks for missing observations from $\bar x$ and $\sigma^2$; JEE Advanced, whose syllabus includes comparing distributions with the same mean but different variances, can also ask for coefficients of variation (see the ๐Ÿš€ box).

E1. Board ยท 1 mark The variance of $1, 3, 5, 7$ is (A) $\sqrt5$   (B) $5$   (C) $20$   (D) $4$

Idea $\bar x=4$; squared deviations $9,1,1,9$ add to $20$; $\sigma^2=\tfrac{20}{4}=5$.
Trap (A) is the standard deviation, (C) forgets to divide by $n$, (D) is the mean.
Answer (B).

E2. Board ยท 2 marks The mean and variance of $10$ observations are $12$ and $4$. Find $\sum x_i$ and $\sum x_i^2$.

Sum $\sum x_i=n\bar x=120$. (1 mark)
Sum of squares $\sigma^2=\tfrac1n\sum x_i^2-\bar x^{\,2}$ gives $\sum x_i^2=n(\sigma^2+\bar x^{\,2})=10(4+144)=1480$. (1 mark)

E3. Board ยท 3 marks The mean and standard deviation of $10$ observations were found to be $50$ and $5$. On checking, it was found that the observation $45$ had been copied as $54$. Find the correct mean and standard deviation.

Wrong sums $\sum x_i=500$; $\sum x_i^2=10(25+2500)=25\,250$. (1 mark)
Correct sums $\sum x_i=500-54+45=491$; $\sum x_i^2=25\,250-2916+2025=24\,359$. (1 mark)
Answer $\bar x=49.1$; $\sigma^2=2435.9-49.1^2=2435.9-2410.81=25.09$; $\sigma=5.01$ (to $2$ d.p.). (1 mark)

E4. Board ยท 5 marks Find the mean and standard deviation of the following distribution of the weights (kg) of $50$ adults.

Weight30โ€“4040โ€“5050โ€“6060โ€“7070โ€“8080โ€“90
Adults37121585
Code Class marks $35,45,\dots,85$; $A=65$, $h=10$; $u_i=-3,-2,-1,0,1,2$. (1 mark)
Columns $f_iu_i=-9,-14,-12,0,8,10$, sum $-17$; $f_iu_i^2=27,28,12,0,8,20$, sum $95$. (1 mark)
Mean $\bar x=65+10\times\dfrac{-17}{50}=65-3.4=61.6$ kg. (1 mark)
Variance $\sigma_u^2=\dfrac{95}{50}-\left(\dfrac{-17}{50}\right)^2=1.9-0.1156=1.7844$, so $\sigma^2=100\times1.7844=178.44$. (1 mark)
Answer $\sigma=\sqrt{178.44}=13.36$ kg (to $2$ d.p.). (1 mark)

E5. JEE Main Six observations have mean $5$ and variance $6$. Four of them are $1, 4, 6, 9$. The other two are (A) $4$ and $6$   (B) $3$ and $7$   (C) $2$ and $8$   (D) $5$ and $5$

Sum $a+b=30-20=10$ โ€” all four options pass this test.
Squares $\sum x_i^2=6(6+25)=186$; $1+16+36+81=134$, so $a^2+b^2=52$, $ab=\tfrac{100-52}{2}=24$: $a,b=4,6$.
Trap (D) keeps the mean but gives $a^2+b^2=50$, i.e. variance $\tfrac{184}{6}-25=5.67$; the variance condition is what decides.
Answer (A).

E6. JEE Advanced Only one option correct. The mean of a distribution is $20$ and its coefficient of variation is $25\%$. Its variance is (A) $25$   (B) $5$   (C) $4$   (D) $16$

Idea C.V. $=\tfrac{\sigma}{\bar x}\times100$ (๐Ÿš€ box): $25=\tfrac{\sigma}{20}\times100$, so $\sigma=5$.
Trap (B) is $\sigma$, not $\sigma^2$.
Answer (A) $\sigma^2=25$.
๐Ÿ”ฌ Researchโ€“Teaching NexusMini project ยท about 1 hour

Research question. Has the monsoon rainfall in your region become more variable over the last few decades?

Why it matters. Farmers, reservoir engineers and disaster-management agencies plan for the spread of rainfall, not just its average: a district that receives its average every year can plan crops easily, while one that swings between drought and flood cannot. Climate scientists in India and worldwide track the standard deviation of seasonal rainfall to detect such changes.

What to do.

  1. Obtain the Juneโ€“September (monsoon) rainfall totals for your state or district for at least $30$ consecutive years โ€” public meteorological data portals publish them, or your teacher can provide a table.
  2. Split the years into three decades. For each decade find the mean and the standard deviation (Theorem 13.5 in a spreadsheet: columns $x$ and $x^2$).
  3. Mark on a time-series graph the band $\bar x\pm\sigma$ for the whole period, and list the years outside it. Are they spread evenly, or bunched in one decade?
  4. Group all the years into classes of width $100\ \text{mm}$ and find $\sigma$ by the shortcut method (Theorem 13.6). How close is it to the value from the raw data?

Tools: A spreadsheet, a public rainfall data table, graph paper or a plotting app

Hand in: The data table, the decade-by-decade means and standard deviations, the time-series graph with the $\bar x\pm\sigma$ band, the grouped calculation, and a paragraph answering the research question with the evidence and its limits.

Going further. A decade with a smaller mean and a smaller $\sigma$ may still be relatively more variable. If you are preparing for JEE Advanced, read the ๐Ÿš€ box and compare the decades using the relative measure introduced there.

Instructor note: all-India monsoon totals are close to $900\ \text{mm}$ with a standard deviation of roughly $10\%$ of the mean; individual districts are far more variable. Ten-year standard deviations fluctuate a lot just by chance, so students should be cautious before claiming a trend โ€” a good discussion of how much data a conclusion needs. The grouped $\sigma$ normally agrees with the raw one to within a few millimetres.

๐Ÿงช Self-check quiz

0/5 correct

Q1. The standard deviation of $7, 7, 7, 7, 7$ is

Q2. The variance of $1, 3, 5$ is

Q3. For $5$ observations, $\sum x_i^2=50$ and $\bar x=3$. The variance is

Q4. In the shortcut method with class width $h=5$, the coded values $u_i$ have standard deviation $1.2$. The standard deviation of the data is

Q5. Heights are measured in cm. The variance of the heights is measured in

โ† 13.2 Mean Deviation for Grouped DataChapter 14 โ€” Probability โ†’