Class 10: Statistics – Exercise 15.2

$\displaystyle \textbf{Question 1: }\text{The number of telephone calls received at an exchange per interval for }250$
$\displaystyle \text{ successive one-} \ \text{minute intervals are given in the following frequency table:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{No. of calls }(x)&0&1&2&3&4&5&6\\ \hline \text{No. of intervals }(f)&15&24&29&46&54&43&39\\ \hline\end{array}$
$\displaystyle \text{Compute the mean number of calls per interval.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the assumed mean be }A=3.$
$\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&d_i=x_i-A&f_id_i\\ \hline 0&15&-3&-45\\ \hline 1&24&-2&-48\\ \hline 2&29&-1&-29\\ \hline 3&46&0&0\\ \hline 4&54&1&54\\ \hline 5&43&2&86\\ \hline 6&39&3&117\\ \hline \text{Total}&\sum f_i=250&&\sum f_id_i=135\\ \hline\end{array}$
$\displaystyle \bar{x}=A+\frac{\sum f_id_i}{\sum f_i}$
$\displaystyle =3+\frac{135}{250}$
$\displaystyle =3+0.54=3.54$
$\displaystyle \therefore \text{The mean number of calls per interval is }3.54.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Five coins were simultaneously tossed }1000\text{ times, and at each toss the}$
$\displaystyle \text{number of heads was observed. The number of tosses during which }0,1,2,3,4\text{ and } 5$
$\displaystyle \text{ heads were obtained are shown} \ \text{in the table below. Find the mean number of heads per toss.}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{No. of heads per toss }(x)&0&1&2&3&4&5\\ \hline \text{No. of tosses }(f)&38&144&342&287&164&25\\ \hline\end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the assumed mean be }A=3.$
$\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&d_i=x_i-A&f_id_i\\ \hline 0&38&-3&-114\\ \hline 1&144&-2&-288\\ \hline 2&342&-1&-342\\ \hline 3&287&0&0\\ \hline 4&164&1&164\\ \hline 5&25&2&50\\ \hline \text{Total}&\sum f_i=1000&&\sum f_id_i=-530\\ \hline\end{array}$
$\displaystyle \bar{x}=A+\frac{\sum f_id_i}{\sum f_i}$
$\displaystyle =3+\frac{-530}{1000}$
$\displaystyle =3-0.53=2.47$
$\displaystyle \therefore \text{The mean number of heads per toss is }2.47.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{The marks obtained out of }50,\text{ by }102\text{ students in a Physics }$
$\displaystyle \text{test are given in the frequency table below:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline \text{Marks }(x)&15&20&22&24&25&30&33&38&45\\ \hline \text{Frequency }(f)&5&8&11&20&23&18&13&3&1\\ \hline\end{array}$
$\displaystyle \text{Find the average number of marks.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the assumed mean be }A=25.$
$\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&d_i=x_i-A&f_id_i\\ \hline 15&5&-10&-50\\ \hline 20&8&-5&-40\\ \hline 22&11&-3&-33\\ \hline 24&20&-1&-20\\ \hline 25&23&0&0\\ \hline 30&18&5&90\\ \hline 33&13&8&104\\ \hline 38&3&13&39\\ \hline 45&1&20&20\\ \hline \text{Total}&\sum f_i=102&&\sum f_id_i=110\\ \hline\end{array}$
$\displaystyle \bar{x}=A+\frac{\sum f_id_i}{\sum f_i}$
$\displaystyle =25+\frac{110}{102}$
$\displaystyle =25+1.0784=26.0784\approx26.08$
$\displaystyle \therefore \text{The average number of marks is }26.08\text{ (approx.).}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{The number of students absent in a class were recorded every day for }$
$\displaystyle 120\text{ days and the} \ \text{information is given in the following frequency table:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{No. of students absent }(x)&0&1&2&3&4&5&6&7\\ \hline \text{No. of days }(f)&1&4&10&50&34&15&4&2\\ \hline\end{array}$
$\displaystyle \text{Find the mean number of students absent per day.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the assumed mean be }A=3.$
$\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&d_i=x_i-A&f_id_i\\ \hline 0&1&-3&-3\\ \hline 1&4&-2&-8\\ \hline 2&10&-1&-10\\ \hline 3&50&0&0\\ \hline 4&34&1&34\\ \hline 5&15&2&30\\ \hline 6&4&3&12\\ \hline 7&2&4&8\\ \hline \text{Total}&\sum f_i=120&&\sum f_id_i=63\\ \hline\end{array}$
$\displaystyle \bar{x}=A+\frac{\sum f_id_i}{\sum f_i}$
$\displaystyle =3+\frac{63}{120}$
$\displaystyle =3+0.525=3.525$
$\displaystyle \therefore \text{The mean number of students absent per day is }3.525.$
$\displaystyle \\$