Class 10: Statistics – Exercise 15.1

$\displaystyle \textbf{Question 1: }\text{Calculate the mean for the following distribution:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline x&5&6&7&8&9\\ \hline f&4&8&14&11&3\\ \hline\end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 5&4&20\\ \hline 6&8&48\\ \hline 7&14&98\\ \hline 8&11&88\\ \hline 9&3&27\\ \hline \text{Total}&\sum f_i=40&\sum f_ix_i=281\\ \hline\end{array}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{281}{40}=7.025$
$\displaystyle \therefore \text{The mean of the given distribution is }7.025.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the missing value of }p\text{ for the following distribution whose mean is }12.58.$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline x&5&8&10&12&p&20&25\\ \hline f&2&5&8&22&7&4&2\\ \hline\end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 5&2&10\\ \hline 8&5&40\\ \hline 10&8&80\\ \hline 12&22&264\\ \hline p&7&7p\\ \hline 20&4&80\\ \hline 25&2&50\\ \hline \text{Total}&\sum f_i=50&\sum f_ix_i=524+7p\\ \hline\end{array}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}$
$\displaystyle 12.58=\frac{524+7p}{50}$
$\displaystyle 12.58\times50=524+7p$
$\displaystyle 629=524+7p$
$\displaystyle 7p=629-524=105$
$\displaystyle p=15$
$\displaystyle \therefore \text{The missing value of }p\text{ is }15.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{The following table gives the number of boys of a particular age in a class of }$
$\displaystyle 40\text{ students.} \ \text{Calculate the mean age of the students.}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Age (in years)}&15&16&17&18&19&20\\ \hline \text{No. of students}&3&8&10&10&5&4\\ \hline\end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 15&3&45\\ \hline 16&8&128\\ \hline 17&10&170\\ \hline 18&10&180\\ \hline 19&5&95\\ \hline 20&4&80\\ \hline \text{Total}&\sum f_i=40&\sum f_ix_i=698\\ \hline\end{array}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}$
$\displaystyle =\frac{698}{40}=17.45\text{ years}$
$\displaystyle \therefore \text{The mean age of the students is }17.45\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Five coins were simultaneously tossed }1000\text{ times and at each toss the}$
$\displaystyle \text{number of heads was observed. The number of tosses during which }0,1,2,3,4\text{ and }5\text{ heads}$
$\displaystyle \text{were obtained are shown in the table below. Find the mean number of heads per toss.}$
$\displaystyle \begin{array}{|c|c|}\hline \text{No. of heads per toss}&\text{No. of tosses}\\ \hline 0&38\\ \hline 1&144\\ \hline 2&342\\ \hline 3&287\\ \hline 4&164\\ \hline 5&25\\ \hline \text{Total}&1000\\ \hline\end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 0&38&0\\ \hline 1&144&144\\ \hline 2&342&684\\ \hline 3&287&861\\ \hline 4&164&656\\ \hline 5&25&125\\ \hline \text{Total}&\sum f_i=1000&\sum f_ix_i=2470\\ \hline\end{array}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}$
$\displaystyle =\frac{2470}{1000}=2.47$
$\displaystyle \therefore \text{The mean number of heads per toss is }2.47.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{The arithmetic mean of the following data is }14.\text{ Find the value of }k.$
$\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline x_i&5&10&15&20&25\\ \hline f_i&7&k&8&4&5\\ \hline\end{array}\hfill\text{[CBSE 2002C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 5&7&35\\ \hline 10&k&10k\\ \hline 15&8&120\\ \hline 20&4&80\\ \hline 25&5&125\\ \hline \text{Total}&\sum f_i=24+k&\sum f_ix_i=360+10k\\ \hline\end{array}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}$
$\displaystyle 14=\frac{360+10k}{24+k}$
$\displaystyle 14(24+k)=360+10k$
$\displaystyle 336+14k=360+10k$
$\displaystyle 4k=24$
$\displaystyle k=6$
$\displaystyle \therefore \text{The value of }k\text{ is }6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{The arithmetic mean of the following data is }25.\text{ Find the value of }k.$
$\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline x_i&5&15&25&35&45\\ \hline f_i&3&k&3&6&2\\ \hline\end{array}\hfill\text{[CBSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 5&3&15\\ \hline 15&k&15k\\ \hline 25&3&75\\ \hline 35&6&210\\ \hline 45&2&90\\ \hline \text{Total}&\sum f_i=14+k&\sum f_ix_i=390+15k\\ \hline\end{array}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}$
$\displaystyle 25=\frac{390+15k}{14+k}$
$\displaystyle 25(14+k)=390+15k$
$\displaystyle 350+25k=390+15k$
$\displaystyle 10k=40$
$\displaystyle k=4$
$\displaystyle \therefore \text{The value of }k\text{ is }4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If the mean of the following data is }18.75.\text{ Find the value of }p.$
$\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline x_i&10&15&p&25&30\\ \hline f_i&5&10&7&8&2\\ \hline\end{array}\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 10&5&50\\ \hline 15&10&150\\ \hline p&7&7p\\ \hline 25&8&200\\ \hline 30&2&60\\ \hline \text{Total}&\sum f_i=32&\sum f_ix_i=460+7p\\ \hline\end{array}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}$
$\displaystyle 18.75=\frac{460+7p}{32}$
$\displaystyle 18.75\times32=460+7p$
$\displaystyle 600=460+7p$
$\displaystyle 7p=140$
$\displaystyle p=20$
$\displaystyle \therefore \text{The value of }p\text{ is }20.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Find the value of }p,\text{ if the mean of the following distribution is }20.$
$\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline x&15&17&19&20+p&23\\ \hline f&2&3&4&5p&6\\ \hline\end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 15&2&30\\ \hline 17&3&51\\ \hline 19&4&76\\ \hline 20+p&5p&5p(20+p)\\ \hline 23&6&138\\ \hline \text{Total}&\sum f_i=15+5p&\sum f_ix_i=295+100p+5p^2\\ \hline\end{array}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}$
$\displaystyle 20=\frac{295+100p+5p^2}{15+5p}$
$\displaystyle 20(15+5p)=295+100p+5p^2$
$\displaystyle 300+100p=295+100p+5p^2$
$\displaystyle 5p^2=5$
$\displaystyle p^2=1$
$\displaystyle p=\pm1$
$\displaystyle \text{Since }5p\text{ is a frequency, it cannot be negative.}$
$\displaystyle \therefore p=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find the missing frequencies in the following frequency distribution if it is}$
$\displaystyle \text{ known that the mean of the distribution is }50.$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline x&10&30&50&70&90&\text{Total}\\ \hline f&17&f_1&32&f_2&19&120\\ \hline\end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\ \hline 10&17&170\\ \hline 30&f_1&30f_1\\ \hline 50&32&1600\\ \hline 70&f_2&70f_2\\ \hline 90&19&1710\\ \hline \text{Total}&\sum f_i=120&\sum f_ix_i=3480+30f_1+70f_2\\ \hline\end{array}$
$\displaystyle 17+f_1+32+f_2+19=120$
$\displaystyle f_1+f_2=52\qquad\ldots\text{(i)}$
$\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}$
$\displaystyle 50=\frac{3480+30f_1+70f_2}{120}$
$\displaystyle 6000=3480+30f_1+70f_2$
$\displaystyle 30f_1+70f_2=2520$
$\displaystyle 3f_1+7f_2=252\qquad\ldots\text{(ii)}$
$\displaystyle \text{Multiplying (i) by }3,\quad 3f_1+3f_2=156\qquad\ldots\text{(iii)}$
$\displaystyle \text{Subtracting (iii) from (ii),}$
$\displaystyle 4f_2=96$
$\displaystyle f_2=24$
$\displaystyle \text{From (i),}\quad f_1+24=52$
$\displaystyle f_1=28$
$\displaystyle \therefore \text{The missing frequencies are }f_1=28\text{ and }f_2=24.$
$\displaystyle \\$