(33)

$a,b,c\in N$ and $a<b<c$

Since, mean = 18 $\Rightarrow \frac{9+25+a+b+c}{5}=18$

$\Rightarrow 34+a+b+c=90\Rightarrow a+b+c=56$

Mean deviation about the mean is 4

$\Rightarrow \frac{|9-18|+|25-18|+|a-18|+|b-18|+|c-18|}{5}=4$

$\Rightarrow 9+7+|a-18|+|b-18|+|c-18|=20$

$\Rightarrow |a-18|+|b-18|+|c-18|=4$

Variance $=\frac{136}{5}$

$\Rightarrow \frac{\sum |x_i-\overline{x}{|}^2}{n}=\frac{136}{5}$

$\Rightarrow \frac{81+49+|a-18{|}^2+|b-18{|}^2+|c-18{|}^2}{5}=\frac{136}{5}$

$\Rightarrow |a-18{|}^2+|b-18{|}^2+|c-18{|}^2=136-130=6$

Possible values are $|a-18{|}^2=1, |b-18{|}^2=1$ and $|c-18{|}^2=4$

$\because$   $a<b<c$

$\therefore$  $18-a=1, b-18=1$ and $c-18=2$

$\Rightarrow a=17, b=19$ and $c=20$

$\text{Hence, }2a+b-c=34+19-20=33$

(2521)

We have, mean = 12, standard deviation = 3

Let $x_1,x_2, \dots , x_{15}$ be 15 observations

$\therefore$    $n=15$

$\frac{x_1+x_2+\dots +x_{14}+10}{15}=12$$\Rightarrow x_1+x_2+\dots +x_{14}=170$

Now, correct mean = $\mu$

$\frac{x_1+x_2+\dots +x_{14}+12}{15}=\mu \Rightarrow \frac{170+12}{15}=\mu \Rightarrow \mu =\frac{182}{15}$

Standard deviation $=\frac{1}{15}\sqrt{15\sum _{i=1}^{15}{x}_i^2-{\left(\sum _{i=1}^{15}{x}_i\right)}^2}$

$\Rightarrow 3=\frac{1}{15}\sqrt{15\times \sum _{i=1}^{15}{x}_i^2-{\left(180\right)}^2}$$\Rightarrow 45=\sqrt{15\times \sum _{i=1}^{15}{x}_i^2-32400}$

$\Rightarrow 2025=15\sum _{i=1}^{15}{x}_i^2-32400$

$\Rightarrow 2025+32400=15(x_1^2+x_2^2+\dots +x_{14}^2+100)$

$\Rightarrow 2295=x_1^2+x_2^2+\dots +x_{14}^2+100$

$\Rightarrow x_1^2+x_2^2+\dots +x_{14}^2=2195$

Correct Variance, ${\sigma }^2=\frac{1}{n^2}(15\times \sum _{i=1}^{15}{x}_i^2-{\left(\sum _{i=1}^n{x}_i\right)}^2)$

$=\frac{1}{{\left(15\right)}^2}(15\times (x_1^2+x_2^2+\dots +x_{14}^2+144)-{\left(182\right)}^2)$

$=\frac{1}{225}(15\times (2339)-33124)$$=\frac{1}{225}(35085-33124)$

${\sigma }^2=\frac{1961}{225}$

$\therefore$   $15(\mu +{\mu }^2+{\sigma }^2)=15(\frac{182}{15}+{\left(\frac{182}{15}\right)}^2+\frac{1961}{225})$

$=15\left(\frac{{\displaystyle 2730+33124+1961}}{{\displaystyle 225}}\right)$$=\frac{37815}{15}=2521$

(29)

Mean $\left(\overline{x}\right)$$=\frac{\sum f_i{x}_i}{\sum f_i}$

$=\frac{0\times 3+1\times 2+5\times 3+6\times 2+10\times 6+12\times 3+17\times 3}{3+2+3+2+6+3+3}$

$=\frac{176}{22}=8$

$\therefore$     Variance ${\sigma }^2$$=\frac{\sum f_i{(x_i-\overline{x})}^2}{\sum f_i}$

       $=\frac{3{(0-8)}^2+2{(1-8)}^2+3{(5-8)}^2+2{(6-8)}^2+6{(10-8)}^2+3{(12-8)}^2+3{(17-8)}^2}{22}$

       $=\frac{3\times 64+2\times 49+3\times 9+2\times 4+6\times 4+3\times 16+3\times 81}{22}$

      $=\frac{640}{22}=29.09\approx 29$