NCERT Solutions: Class 11 Maths Chapter 13 - Statistics

The mean is $\bar{x}=\dfrac{80}{8}=10$. The absolute deviations from 10 are $6,3,2,1,0,2,3,7$, whose sum is $24$. Therefore M.D. about mean $=\dfrac{24}{8}=3$.

The mean is $\bar{x}=\dfrac{500}{10}=50$. The absolute deviations from 50 sum to $84$. Hence M.D. about mean $=\dfrac{84}{10}=8.4$.

Arranged data are $10,11,11,12,13,13,14,16,16,17,17,18$. The median is $\dfrac{13+14}{2}=13.5$. The sum of absolute deviations from 13.5 is $28$. Thus M.D. about median $=\dfrac{28}{12}=\dfrac73$.

Arranged data are $36,42,45,46,46,49,51,53,60,72$. The median is $\dfrac{46+49}{2}=47.5$. The sum of absolute deviations from 47.5 is $70$. Hence M.D. about median $=\dfrac{70}{10}=7$.

Here $N=25$ and $\sum f_ix_i=350$, so $\bar{x}=14$. Also $\sum f_i|x_i-14|=158$. Therefore M.D. about mean $=\dfrac{158}{25}=6.32$.

Here $N=80$ and $\sum f_ix_i=4000$, so $\bar{x}=50$. Then $\sum f_i|x_i-50|=1280$. Thus M.D. about mean $=\dfrac{1280}{80}=16$.

Here $N=26$. The median item lies at the $13.5$th position, so the median is $7$. Now $\sum f_i|x_i-7|=84$. Therefore M.D. about median $=\dfrac{84}{26}=\dfrac{42}{13}$.

Here $N=29$. The median is the $15$th item, which is $30$. The sum $\sum f_i|x_i-30|=148$. Hence M.D. about median $=\dfrac{148}{29}\approx5.10$.

Using class marks $50,150,250,350,450,550,650,750$, $N=50$ and $\bar{x}=358$. The weighted sum of absolute deviations is $7896$. Hence M.D. about mean $=\dfrac{7896}{50}=157.92$.

Using class marks $100,110,120,130,140,150$, $N=100$ and $\bar{x}=125.3$. The weighted sum of absolute deviations is $1128.8$. Therefore M.D. about mean $=11.288$.

Here $N=50$. The median class is $20-30$, so median $=20+\dfrac{25-14}{14}\times10=27.857...$. Using class marks $5,15,25,35,45,55$, $\sum f_i|x_i-M|=517.142...$. Thus M.D. about median $=\dfrac{517.142...}{50}\approx10.34$.

Convert to continuous classes $15.5-20.5,20.5-25.5,\ldots,50.5-55.5$. Here $N=100$, and the median class is $35.5-40.5$. Median $=35.5+\dfrac{50-37}{26}\times5=38$. Using class marks $18,23,28,33,38,43,48,53$, $\sum f_i|x_i-38|=735$. Hence M.D. about median $=\dfrac{735}{100}=7.35$.

Here $n=8$, $\sum x_i=72$, so $\bar{x}=9$. Also $\sum (x_i-9)^2=74$. Therefore variance $=\dfrac{74}{8}=9.25$.

For $1,2,\ldots,n$, $\bar{x}=\dfrac{1+2+\cdots+n}{n}=\dfrac{n+1}{2}$. Also $\dfrac{\sum x_i^2}{n}=\dfrac{(n+1)(2n+1)}{6}$. Hence variance $=\dfrac{(n+1)(2n+1)}6-\left(\dfrac{n+1}{2}\right)^2=\dfrac{n^2-1}{12}$.

The data are $3,6,9,\ldots,30$, i.e. 3 times the first 10 natural numbers. Mean $=3\cdot\dfrac{11}{2}=16.5$. Variance $=3^2\cdot\dfrac{10^2-1}{12}=9\cdot\dfrac{99}{12}=74.25$.

Here $N=40$ and $\sum f_ix_i=760$, so $\bar{x}=19$. Also $\sum f_i(x_i-19)^2=1736$. Therefore variance $=\dfrac{1736}{40}=43.4$.

Here $N=22$ and $\sum f_ix_i=2200$, so $\bar{x}=100$. Also $\sum f_i(x_i-100)^2=640$. Thus variance $=\dfrac{640}{22}=\dfrac{320}{11}\approx29.09$.

Here $N=100$ and $\sum f_ix_i=6400$, so mean $=64$. Also $\sum f_i(x_i-64)^2=286$. Hence variance $=\dfrac{286}{100}=2.86$ and standard deviation $=\sqrt{2.86}\approx1.69$.

Using class marks $15,45,75,105,135,165,195$, $N=30$ and $\sum f_ix_i=3210$, so mean $=107$. Also $\sum f_i(x_i-107)^2=68280$. Therefore variance $=\dfrac{68280}{30}=2276$.

Using class marks $5,15,25,35,45$, $N=50$ and $\sum f_ix_i=1350$, so mean $=27$. Also $\sum f_i(x_i-27)^2=6600$. Hence variance $=\dfrac{6600}{50}=132$.

Using class marks $72.5,77.5,82.5,87.5,92.5,97.5,102.5,107.5,112.5$, $N=60$ and $\sum f_ix_i=5580$, so mean $=93$. The variance is $\dfrac{\sum f_i(x_i-93)^2}{60}=\dfrac{1267}{12}\approx105.58$. Therefore standard deviation $\approx10.28$.

Using the hint, convert the classes to $32.5-36.5,36.5-40.5,40.5-44.5,44.5-48.5,48.5-52.5$, whose class marks are $34.5,38.5,42.5,46.5,50.5$. With $N=100$, mean $=43.5$. The variance is $30.84$, so standard deviation $=\sqrt{30.84}\approx5.55$ mm.

Total sum $=8\times9=72$. The six known observations sum to $60$, so the remaining two have sum $12$. Also $\sum x_i^2=8(9.25+9^2)=722$. The six known squares sum to $642$, so the remaining squares sum to $80$. If the two observations are $p,q$, then $p+q=12$ and $p^2+q^2=80$. Thus $2pq=(p+q)^2-(p^2+q^2)=144-80=64$, so $pq=32$. Hence $p,q$ are roots of $t^2-12t+32=0$, giving $4$ and $8$.

Total sum $=7\times8=56$. The five known observations sum to $42$, so the remaining two have sum $14$. Also $\sum x_i^2=7(16+8^2)=560$. The five known squares sum to $460$, so the remaining squares sum to $100$. Let the two observations be $p,q$. Then $p+q=14$ and $p^2+q^2=100$, so $2pq=196-100=96$ and $pq=48$. Thus $p,q$ are roots of $t^2-14t+48=0$, giving $4$ and $10$.

Multiplying every observation by 3 multiplies the mean by 3 and the standard deviation by $|3|$. Therefore the new mean is $3\times8=24$ and the new standard deviation is $3\times4=12$.

Let the original mean be $\bar{x}$. The mean of $ax_1,ax_2,\ldots,ax_n$ is $\dfrac{a\sum x_i}{n}=a\bar{x}$. The new variance is $\dfrac1n\sum (ax_i-a\bar{x})^2=\dfrac1n\sum a^2(x_i-ar{x})^2=a^2\sigma^2$.

Original $\sum x_i=20\times10=200$ and $\sum x_i^2=20(2^2+10^2)=2080$. (i) Omitting 8 gives $n=19$, sum $192$, and sum of squares $2016$. Mean $=192/19\approx10.105$ and variance $=2016/19-(192/19)^2\approx3.9889$, so S.D. $\approx1.997$. (ii) Replacing 8 by 12 gives sum $204$ and sum of squares $2160$. Mean $=204/20=10.2$ and variance $=2160/20-10.2^2=3.96$, so S.D. $\approx1.990$.

Original $\sum x_i=100\times20=2000$ and $\sum x_i^2=100(3^2+20^2)=40900$. Omitting $21,21,18$ gives $n=97$, sum $=2000-60=1940$, and sum of squares $=40900-(441+441+324)=39694$. The corrected mean is $1940/97=20$. Variance $=39694/97-20^2\approx9.2165$, so standard deviation $\approx3.036$.