NCERT Exemplar Solutions Class 11 Maths Statistics — Short and Long Answer Type Q1–23

NCERT Exemplar Problems, Class 11 Mathematics — Statistics, Short Answer and Long Answer Type questions 1 to 23. All 23 questions solved.

This page is the computation half of the unit: mean deviation about the mean or median (Q1 to Q4, Q16, Q17), variance and standard deviation from a table (Q9 to Q11, Q15, Q18, Q19), and the “combine two sets” and “correct a mistaken reading” questions that run on sums rather than raw data (Q6 to Q8, Q12 to Q14, Q22, Q23). Three formulae carry almost everything: $\text{M.D.}=\dfrac{\sum f_i|x_i-a|}{\sum f_i}$, $\sigma^2=\dfrac{\sum f_ix_i^2}{N}-\bar x^2$, and $\sum x_i^2=N(\sigma^2+\bar x^2)$.

Four printed answers (Q1, Q15, Q16, Q17) do not match the questions as printed, and the key’s verdict on consistency in Q21 is wrong. Each is explained where it occurs.

Key insight. Never carry raw data around when a question gives you $N$, $\bar x$ and $\sigma$. Convert to the two sums $\sum x=N\bar x$ and $\sum x^2=N(\sigma^2+\bar x^2)$, repair or combine those sums, and convert back.

Question 1

Find the mean deviation about the mean of the distribution:

Size	20	21	22	23	24
Frequency	6	4	5	1	4

Solution. The mean comes first, then the absolute distance of each size from it, weighted by frequency. Here $N=20$ and $\sum fx=120+84+110+23+96=433$, so $\bar x=\dfrac{433}{20}=21.65$.

$x$	$f$	$\lvert x-21.65\rvert$	$f\lvert x-21.65\rvert$
20	6	1.65	9.90
21	4	0.65	2.60
22	5	0.35	1.75
23	1	1.35	1.35
24	4	2.35	9.40

The total is $25.00$, so $\text{M.D.}=\dfrac{25}{20}=1.25$.

The printed key says $0.32$. The five deviations in the table add to $25$ over $20$ items, which can be checked line by line, so $0.32$ is a misprint (it is also far too small: the data run from $20$ to $24$ with a standard deviation of about $1.46$).

Question 2

Find the mean deviation about the median of the following distribution:

Marks obtained	10	11	12	14	15
No. of students	2	3	8	3	4

Solution. The cumulative frequencies are $2, 5, 13, 16, 20$. With $N=20$ the median is the mean of the 10th and 11th values, and both lie in the group with marks $12$, so the median is $12$. Then $$\sum f|x-12|=2(2)+3(1)+8(0)+3(2)+4(3)=4+3+0+6+12=25,$$ and $\text{M.D.}=\dfrac{25}{20}=1.25$.

Question 3

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

Solution. The mean of $1,2,\dots,n$ is $\dfrac{n+1}{2}$. When $n$ is odd this is itself one of the numbers, the middle one. The deviations on either side are $1,2,\dots,\dfrac{n-1}{2}$, each appearing twice, and the middle number contributes $0$. So $$\sum|x_i-\bar x|=2\cdot\frac{\frac{n-1}{2}\cdot\frac{n+1}{2}}{2}=\frac{(n-1)(n+1)}{4}=\frac{n^2-1}{4}.$$ Dividing by $n$ gives the mean deviation. As a check, $n=5$ gives deviations $2,1,0,1,2$, sum $6$, M.D. $=\dfrac65$, and the formula gives $\dfrac{24}{20}=\dfrac65$.

Question 4

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an even number.

Solution. The mean is still $\dfrac{n+1}{2}$, but now it falls halfway between two numbers, so no deviation is zero. The deviations are $\dfrac12,\dfrac32,\dots,\dfrac{n-1}{2}$, each appearing twice. The sum of the first $\dfrac n2$ odd numbers $1+3+\dots+(n-1)$ is $\left(\dfrac n2\right)^2$, so $$\sum|x_i-\bar x|=2\cdot\frac12\cdot\frac{n^2}{4}=\frac{n^2}{4},\qquad \text{M.D.}=\frac{n^2/4}{n}=\frac n4.$$ Check with $n=4$: the deviations $1.5,0.5,0.5,1.5$ sum to $4$, and $\dfrac44=1=\dfrac44$.

Question 5

Find the standard deviation of the first $n$ natural numbers.

Solution. Use $\sigma^2=\dfrac{\sum x^2}{n}-\bar x^2$ with $\sum x=\dfrac{n(n+1)}{2}$ and $\sum x^2=\dfrac{n(n+1)(2n+1)}{6}$: $$\sigma^2=\frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}=\frac{(n+1)\left[2(2n+1)-3(n+1)\right]}{12}=\frac{(n+1)(n-1)}{12}=\frac{n^2-1}{12}.$$

Question 6

The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results: number of observations $=25$, mean $=18.2$ seconds, standard deviation $=3.25$ seconds. Further, another set of $15$ observations $x_1,x_2,\dots,x_{15}$, also in seconds, is now available and we have $\sum_{i=1}^{15}x_i=279$ and $\sum_{i=1}^{15}x_i^2=5524$. Calculate the standard deviation based on all $40$ observations.

Solution. The first set is only known through its summary, so rebuild its sums. $\sum x=25\times18.2=455$ and $\sum x^2=25(3.25^2+18.2^2)=25(10.5625+331.24)=8545.0625$. Add the second set: $$\sum x=455+279=734,\qquad \sum x^2=8545.0625+5524=14069.0625.$$ Over $N=40$: $\bar x=\dfrac{734}{40}=18.35$ and $$\sigma^2=\frac{14069.0625}{40}-18.35^2=351.7266-336.7225=15.0041,\qquad \sigma\approx3.87.$$

Question 7

The mean and standard deviation of a set of $n_1$ observations are $\bar x_1$ and $s_1$, while those of another set of $n_2$ observations are $\bar x_2$ and $s_2$. Show that the standard deviation of the combined set of $(n_1+n_2)$ observations is $$\text{S.D.}=\sqrt{\frac{n_1(s_1)^2+n_2(s_2)^2}{n_1+n_2}+\frac{n_1n_2(\bar x_1-\bar x_2)^2}{(n_1+n_2)^2}}.$$

Solution. Write $N=n_1+n_2$ and $d=\bar x_1-\bar x_2$. The combined mean is $\bar x=\dfrac{n_1\bar x_1+n_2\bar x_2}{N}$, so $$\bar x_1-\bar x=\frac{n_2d}{N},\qquad \bar x_2-\bar x=-\frac{n_1d}{N}.$$ For any set, $\sum(x-\bar x)^2=\sum(x-\bar x_1)^2+n_1(\bar x_1-\bar x)^2$ when the set has mean $\bar x_1$ (the cross term vanishes because $\sum(x-\bar x_1)=0$). Hence, for the first set, $\sum(x-\bar x)^2=n_1s_1^2+n_1\dfrac{n_2^2d^2}{N^2}$, and for the second, $n_2s_2^2+n_2\dfrac{n_1^2d^2}{N^2}$. Adding, $$N\sigma^2=n_1s_1^2+n_2s_2^2+\frac{n_1n_2d^2(n_2+n_1)}{N^2}=n_1s_1^2+n_2s_2^2+\frac{n_1n_2d^2}{N}.$$ Dividing by $N$ gives $$\sigma^2=\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2}+\frac{n_1n_2(\bar x_1-\bar x_2)^2}{(n_1+n_2)^2},$$ which is the stated result. Notice the second term: even if both sets have the same spread, combining sets with different means increases the spread.

Question 8

Two sets each of $20$ observations, have the same standard deviation $5$. The first set has a mean $17$ and the second a mean $22$. Determine the standard deviation of the set obtained by combining the given two sets.

Solution. Apply the result of Question 7 with $n_1=n_2=20$, $s_1=s_2=5$ and $\bar x_1-\bar x_2=-5$: $$\sigma^2=\frac{20\cdot25+20\cdot25}{40}+\frac{20\cdot20\cdot25}{40^2}=25+6.25=31.25,\qquad \sigma=\sqrt{31.25}\approx5.59.$$ The combined spread is larger than $5$ because the two sets sit around different centres.

Question 9

The frequency distribution

$x$	A	2A	3A	4A	5A	6A
$f$	2	1	1	1	1	1

where A is a positive integer, has a variance of $160$. Determine the value of A.

Solution. The total frequency is $7$. $\sum fx=2A+2A+3A+4A+5A+6A=22A$ and $\sum fx^2=2A^2+4A^2+9A^2+16A^2+25A^2+36A^2=92A^2$. So $$\sigma^2=\frac{92A^2}{7}-\left(\frac{22A}{7}\right)^2=\frac{644A^2-484A^2}{49}=\frac{160A^2}{49}.$$ Setting this equal to $160$ gives $A^2=49$, and since A is positive, $A=7$.

Question 10

For the frequency distribution

$x$	2	3	4	5	6	7
$f$	4	9	16	14	11	6

Find the standard deviation. (The book prints “standard distribution”; it means standard deviation, as the key’s answer confirms.)

Solution. $N=60$. $\sum fx=8+27+64+70+66+42=277$ and $\sum fx^2=16+81+256+350+396+294=1393$. Then $$\sigma^2=\frac{1393}{60}-\left(\frac{277}{60}\right)^2=\frac{83580-76729}{3600}=\frac{6851}{3600},\qquad \sigma=\frac{\sqrt{6851}}{60}\approx1.38.$$

Question 11

There are $60$ students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

Marks	0	1	2	3	4	5
Frequency	$x-2$	$x$	$x^2$	$(x+1)^2$	$2x$	$x+1$

where $x$ is a positive integer. Determine the mean and standard deviation of the marks.

Solution. The frequencies must add to $60$: $(x-2)+x+x^2+(x+1)^2+2x+(x+1)=2x^2+7x=60$. So $2x^2+7x-60=0$, which factorises as $(x-4)(2x+15)=0$. Only $x=4$ is a positive integer. The frequencies are $2,4,16,25,8,5$, which indeed add to $60$.

Then $\sum fx=0+4+32+75+32+25=168$, so $\bar x=\dfrac{168}{60}=2.8$. And $\sum fx^2=0+4+64+225+128+125=546$, so $\sigma^2=\dfrac{546}{60}-2.8^2=9.1-7.84=1.26$ and $\sigma=\sqrt{1.26}\approx1.12$.

Question 12

The mean life of a sample of $60$ bulbs was $650$ hours and the standard deviation was $8$ hours. A second sample of $80$ bulbs has a mean life of $660$ hours and standard deviation $7$ hours. Find the overall standard deviation.

Solution. Use the combined-set formula from Question 7 with $n_1=60$, $n_2=80$, $s_1=8$, $s_2=7$, $\bar x_1-\bar x_2=-10$: $$\sigma^2=\frac{60\cdot64+80\cdot49}{140}+\frac{60\cdot80\cdot100}{140^2}=\frac{7760}{140}+\frac{480000}{19600}=55.4286+24.4898=79.918.$$ So $\sigma=\sqrt{79.918}\approx8.94$ hours (the key rounds this to $8.9$).

Question 13

Mean and standard deviation of $100$ items are $50$ and $4$, respectively. Find the sum of all the items and the sum of the squares of the items.

Solution. $\sum x=N\bar x=100\times50=5000$. From $\sigma^2=\dfrac{\sum x^2}{N}-\bar x^2$, $\sum x^2=N(\sigma^2+\bar x^2)=100(16+2500)=251600$.

Question 14

If for a distribution $\sum(x-5)=3$, $\sum(x-5)^2=43$ and the total number of items is $18$, find the mean and standard deviation.

Solution. Put $d=x-5$. The mean of $d$ is $\dfrac{3}{18}=\dfrac16$, so $\bar x=5+\dfrac16=\dfrac{31}{6}\approx5.17$. Shifting the data does not change the spread, so $\sigma_x=\sigma_d$: $$\sigma^2=\frac{43}{18}-\left(\frac16\right)^2=\frac{86-1}{36}=\frac{85}{36},\qquad \sigma=\frac{\sqrt{85}}{6}\approx1.54.$$ The key prints $1.53$, which truncates $1.5366$ instead of rounding it.

Question 15

Find the mean and variance of the frequency distribution given below:

$x$	$1\le x<3$	$3\le x<5$	$5\le x<7$	$7\le x<10$
$f$	6	4	5	1

Solution. For a grouped distribution, use the class marks $2,4,6,8.5$ as the values. $N=16$, $\sum fx=12+16+30+8.5=66.5$, so $\bar x=\dfrac{66.5}{16}=4.156$. And $\sum fx^2=24+64+180+72.25=340.25$, so $$\sigma^2=\frac{340.25}{16}-4.15625^2=21.2656-17.2744=3.991.$$ The printed key gives mean $5.5$ and variance $4.26$. No reading of this table produces those values (even the mean of the four class marks is only $5.125$), so the key seems to belong to a different version of the question. The answer for the table as printed is mean $\approx4.16$ and variance $\approx3.99$.

Question 16

Calculate the mean deviation about the mean for the following frequency distribution:

Class interval	0–4	4–8	8–12	12–16	16–20
Frequency	4	6	8	5	2

Solution. Class marks are $2,6,10,14,18$. $N=25$ and $\sum fx=8+36+80+70+36=230$, so $\bar x=9.2$.

mark $x$	$f$	$\lvert x-9.2\rvert$	$f\lvert x-9.2\rvert$
2	4	7.2	28.8
6	6	3.2	19.2
10	8	0.8	6.4
14	5	4.8	24.0
18	2	8.8	17.6

The total is $96$, so $\text{M.D.}=\dfrac{96}{25}=3.84$. The key’s $0.99$ is not possible: the data run from $0$ to $20$, so the average distance from the mean is several units.

Question 17

Calculate the mean deviation from the median of the following data:

Class interval	0–6	6–12	12–18	18–24	24–30
Frequency	4	5	3	6	2

Solution. For grouped data the median comes from the median-class formula. $N=20$, $\dfrac N2=10$, the cumulative frequencies are $4,9,12,18,20$, so the median class is $12$–$18$ with $l=12$, $c.f.=9$, $f=3$, $h=6$: $$\text{Median}=12+\frac{10-9}{3}\times6=14.$$ Now take the class marks $3,9,15,21,27$ and their distances from $14$: $11,5,1,7,13$. $$\sum f|x-14|=4(11)+5(5)+3(1)+6(7)+2(13)=44+25+3+42+26=140,\qquad \text{M.D.}=\frac{140}{20}=7.$$ The key prints $7.08$. Working with the median exactly at $14$ the answer is exactly $7$, and I could not reproduce $7.08$ by any consistent method, so the key looks like a rounding or arithmetic slip.

Question 18

Determine the mean and standard deviation for the following distribution:

Marks	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
Frequency	1	6	6	8	8	2	2	3	0	2	1	0	0	0	1

Solution. $N=40$. $\sum fx=2+18+24+40+48+14+16+27+0+22+12+0+0+0+16=239$, so $\bar x=\dfrac{239}{40}=5.975$. $\sum fx^2=4+54+96+200+288+98+128+243+0+242+144+0+0+0+256=1753$, so $$\sigma^2=\frac{1753}{40}-\left(\frac{239}{40}\right)^2=\frac{70120-57121}{1600}=\frac{12999}{1600}=8.124,\qquad \sigma\approx2.85.$$

Question 19

The weights of coffee in $70$ jars is shown in the following table:

Weight (in grams)	200–201	201–202	202–203	203–204	204–205	205–206
Frequency	13	27	18	10	1	1

Determine variance and standard deviation of the above distribution.

Solution. Use the step-deviation method with class marks $200.5,\dots,205.5$ and the assumed mean $202.5$, so $u=x-202.5$ takes the values $-2,-1,0,1,2,3$. $\sum fu=-26-27+0+10+2+3=-38$ and $\sum fu^2=52+27+0+10+4+9=102$. Variance is unchanged by the shift: $$\sigma^2=\frac{102}{70}-\left(\frac{-38}{70}\right)^2=1.4571-0.2947=1.1624\approx1.16,\qquad \sigma\approx1.08\text{ g}.$$ (The mean is $202.5-0.543=201.96$ g.)

Question 20

Determine mean and standard deviation of first $n$ terms of an A.P. whose first term is $a$ and common difference is $d$.

Solution. The terms are $a,\,a+d,\dots,a+(n-1)d$. Their mean is the average of the first and last, $a+\dfrac{(n-1)d}{2}$. For the spread, adding $a$ shifts all terms and does not change the variance, and multiplying by $d$ multiplies the variance by $d^2$. The variance of $0,1,\dots,n-1$ is the same as that of $1,\dots,n$, namely $\dfrac{n^2-1}{12}$ by Question 5. So $$\sigma^2=d^2\cdot\frac{n^2-1}{12},\qquad \sigma=|d|\sqrt{\frac{n^2-1}{12}}.$$

Question 21

Following are the marks obtained, out of $100$, by two students Ravi and Hashina in $10$ tests.

Ravi	25	50	45	30	70	42	36	48	35	60
Hashina	10	70	50	20	95	55	42	60	48	80

Who is more intelligent and who is more consistent?

Solution. “More intelligent” is read as the higher average, and “more consistent” as the lower coefficient of variation, $\text{C.V.}=\dfrac{\sigma}{\bar x}\times100$. (Coefficient of variation is no longer in the current NCERT Class 11 syllabus, but it is the method the question expects.)

Ravi: $\sum x=441$, $\bar x=44.1$; $\sum(x-\bar x)^2=1690.9$, so $\sigma^2=169.09$, $\sigma\approx13.00$, and $\text{C.V.}=\dfrac{13.00}{44.1}\times100\approx29.5$.

Hashina: $\sum x=530$, $\bar x=53$; $\sum(x-\bar x)^2=5928$, so $\sigma^2=592.8$, $\sigma\approx24.35$, and $\text{C.V.}=\dfrac{24.35}{53}\times100\approx45.9$.

Hashina’s mean is higher, so she is the better performer on average. But her C.V. is much larger, so her marks swing more; Ravi’s smaller C.V. makes him more consistent. The key says Hashina is both, which is wrong on consistency: a larger standard deviation relative to the mean means less consistent, not more.

Question 22

Mean and standard deviation of $100$ observations were found to be $40$ and $10$, respectively. If at the time of calculation two observations were wrongly taken as $30$ and $70$ in place of $3$ and $27$ respectively, find the correct standard deviation.

Solution. Recover the wrong sums, then swap the two readings. $\sum x=100\times40=4000$ and $\sum x^2=100(10^2+40^2)=170000$. Correcting: $$\sum x=4000-30-70+3+27=3930,\qquad \sum x^2=170000-900-4900+9+729=164938.$$ Then $\bar x=39.3$ and $\sigma^2=1649.38-39.3^2=1649.38-1544.49=104.89$, so $\sigma\approx10.24$.

Question 23

While calculating the mean and variance of $10$ readings, a student wrongly used the reading $52$ for the correct reading $25$. He obtained the mean and variance as $45$ and $16$ respectively. Find the correct mean and the variance.

Solution. The wrong data gave $\sum x=450$ and $\sum x^2=10(16+45^2)=20410$. Replace $52$ by $25$: $$\sum x=450-52+25=423,\qquad \sum x^2=20410-2704+625=18331.$$ So the correct mean is $42.3$ and the correct variance is $1833.1-42.3^2=1833.1-1789.29=43.81$.

Common mistakes

• Question 1 and 16: dividing by the number of classes. The mean deviation is $\sum f|x-\bar x|/\sum f$, not divided by the count of rows; students forget the frequency weight on each deviation.
• Question 3 and 4: using one formula for both parities. When $n$ is odd the mean is a data value (one deviation is zero); when $n$ is even it is not. That is why the answers are $\dfrac{n^2-1}{4n}$ and $\dfrac n4$ rather than a single expression.
• Question 6, 8 and 12: averaging the standard deviations. The combined standard deviation is not the mean of the two; you must add the variances and also the term for the difference between the two means, which is missing if you treat the sets as having the same centre.
• Question 11: not solving for $x$ first. The frequencies must add to $60$; the negative root $-7.5$ of $2x^2+7x-60=0$ has to be rejected because $x$ is a positive integer.
• Question 15 and 19: using class limits instead of class marks. For a grouped table every observation is represented by the middle of its class; using $1,3,5,7$ or the upper limits gives a wrong mean.
• Question 22 and 23: correcting only the mean. The variance depends on $\sum x^2$, so the squares of the wrong and the right readings have to be swapped too, and the corrected mean has to be squared at the end.
• Question 21: reading “larger standard deviation” as more consistent. A larger spread means less consistency; compare $\sigma/\bar x$, not $\sigma$ alone, when the means differ.

Practise next

• The objective questions, fill-in-the-blanks and coefficient-of-variation questions are on Objective and Fill in the Blanks, Q24–46.
• NCERT Class 11 Chapter 13: Exercise 13.1 for mean deviation, Exercise 13.2 for variance and standard deviation, and the Miscellaneous Exercise for combined-set and corrected-reading problems.