\displaystyle \textbf{Question 1: }\text{A survey was conducted by a group of students as a part of their}
\displaystyle \text{environment awareness programme, in which they collected the following data regarding}
\displaystyle \text{the number of plants in }20\text{ houses in a locality. Find the mean number of plants}
\displaystyle \text{per house.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Number of plants}&0-2&2-4&4-6&6-8&8-10&10-12&12-14\\ \hline \text{Number of houses}&1&2&1&5&6&2&3\\ \hline\end{array}
\displaystyle \text{Which method did you use for finding the mean, and why?}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=2.\text{ Let the assumed mean be }A=7.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 0-2&1&1&-6&-3&-3\\ \hline 2-4&3&2&-4&-2&-4\\ \hline 4-6&5&1&-2&-1&-1\\ \hline 6-8&7&5&0&0&0\\ \hline 8-10&9&6&2&1&6\\ \hline 10-12&11&2&4&2&4\\ \hline 12-14&13&3&6&3&9\\ \hline \text{Total}&&\sum f_i=20&&&\sum f_iu_i=11\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =7+2\left(\frac{11}{20}\right)
\displaystyle =7+1.1=8.1
\displaystyle \therefore \text{The mean number of plants per house is }8.1.
\displaystyle \text{The step-deviation method is used because all the class intervals have the same width.}
\displaystyle \text{It reduces the deviations to small integers and makes the calculations simpler.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Thirty women were examined in a hospital by a doctor and the number of}
\displaystyle \text{heart beats per minute recorded and summarised as follows. Find the mean heart beats}
\displaystyle \text{per minute for these women, choosing a suitable method.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Number of heart beats per minute}&65-68&68-71&71-74&74-77&77-80&80-83&83-86\\ \hline \text{Number of women}&2&4&3&8&7&4&2\\ \hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=3.\text{ Let the assumed mean be }A=75.5.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 65-68&66.5&2&-9&-3&-6\\ \hline 68-71&69.5&4&-6&-2&-8\\ \hline 71-74&72.5&3&-3&-1&-3\\ \hline 74-77&75.5&8&0&0&0\\ \hline 77-80&78.5&7&3&1&7\\ \hline 80-83&81.5&4&6&2&8\\ \hline 83-86&84.5&2&9&3&6\\ \hline \text{Total}&&\sum f_i=30&&&\sum f_iu_i=4\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =75.5+3\left(\frac{4}{30}\right)
\displaystyle =75.5+0.4=75.9
\displaystyle \therefore \text{The mean heart beats per minute is }75.9.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the mean of the following frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&3-5&5-7&7-9&9-11&11-13\\ \hline \text{Frequency}&5&10&10&7&8\\ \hline\end{array}\hfill\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=2.\text{ Let the assumed mean be }A=8.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 3-5&4&5&-4&-2&-10\\ \hline 5-7&6&10&-2&-1&-10\\ \hline 7-9&8&10&0&0&0\\ \hline 9-11&10&7&2&1&7\\ \hline 11-13&12&8&4&2&16\\ \hline \text{Total}&&\sum f_i=40&&&\sum f_iu_i=3\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =8+2\left(\frac{3}{40}\right)
\displaystyle =8+\frac{6}{40}=8+0.15=8.15
\displaystyle \therefore \text{The mean of the given frequency distribution is }8.15.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The following distribution shows the daily pocket allowance given to the}
\displaystyle \text{children of a multistorey building. The average pocket allowance is Rs. }18.00.
\displaystyle \text{Find out the missing frequency.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Class interval}&11-13&13-15&15-17&17-19&19-21&21-23&23-25\\ \hline \text{Frequency}&7&6&9&13&p&5&4\\ \hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=2.\text{ Let the assumed mean be }A=18.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 11-13&12&7&-6&-3&-21\\ \hline 13-15&14&6&-4&-2&-12\\ \hline 15-17&16&9&-2&-1&-9\\ \hline 17-19&18&13&0&0&0\\ \hline 19-21&20&p&2&1&p\\ \hline 21-23&22&5&4&2&10\\ \hline 23-25&24&4&6&3&12\\ \hline \text{Total}&&\sum f_i=44+p&&&\sum f_iu_i=p-20\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle 18=18+2\left(\frac{p-20}{44+p}\right)
\displaystyle 0=2\left(\frac{p-20}{44+p}\right)
\displaystyle p-20=0
\displaystyle p=20
\displaystyle \therefore \text{The missing frequency is }20.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If the mean of the following distribution is }27,\text{ find the value of }p.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class}&0-10&10-20&20-30&30-40&40-50\\ \hline \text{Frequency}&8&p&12&13&10\\ \hline\end{array}\hfill\text{[CBSE 2006C]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=10.\text{ Let the assumed mean be }A=25.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 0-10&5&8&-20&-2&-16\\ \hline 10-20&15&p&-10&-1&-p\\ \hline 20-30&25&12&0&0&0\\ \hline 30-40&35&13&10&1&13\\ \hline 40-50&45&10&20&2&20\\ \hline \text{Total}&&\sum f_i=43+p&&&\sum f_iu_i=17-p\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle 27=25+10\left(\frac{17-p}{43+p}\right)
\displaystyle 2=10\left(\frac{17-p}{43+p}\right)
\displaystyle \frac{1}{5}=\frac{17-p}{43+p}
\displaystyle 43+p=85-5p
\displaystyle 6p=42
\displaystyle p=7
\displaystyle \therefore \text{The value of }p\text{ is }7.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The table below shows the daily expenditure on food of }25\text{ households in a locality.}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Daily expenditure (in Rs.)}&100-150&150-200&200-250&250-300&300-350\\ \hline \text{Number of households}&4&5&12&2&2\\ \hline\end{array}
\displaystyle \text{Find the mean daily expenditure on food by a suitable method.}\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=50.\text{ Let the assumed mean be }A=225.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 100-150&125&4&-100&-2&-8\\ \hline 150-200&175&5&-50&-1&-5\\ \hline 200-250&225&12&0&0&0\\ \hline 250-300&275&2&50&1&2\\ \hline 300-350&325&2&100&2&4\\ \hline \text{Total}&&\sum f_i=25&&&\sum f_iu_i=-7\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =225+50\left(\frac{-7}{25}\right)
\displaystyle =225-14=211
\displaystyle \therefore \text{The mean daily expenditure on food is Rs. }211.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A class teacher has the following absentee record of }40\text{ students of a class }
\displaystyle \text{for the whole term. Find the mean number of days a student was absent.}
\displaystyle \text{(i)}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Number of days}&0-6&6-12&12-18&18-24&24-30&30-36&36-42\\ \hline \text{Number of students}&10&11&7&4&4&3&1\\ \hline\end{array}\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=6.\text{ Let the assumed mean be }A=15.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 0-6&3&10&-12&-2&-20\\ \hline 6-12&9&11&-6&-1&-11\\ \hline 12-18&15&7&0&0&0\\ \hline 18-24&21&4&6&1&4\\ \hline 24-30&27&4&12&2&8\\ \hline 30-36&33&3&18&3&9\\ \hline 36-42&39&1&24&4&4\\ \hline \text{Total}&&\sum f_i=40&&&\sum f_iu_i=-6\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =15+6\left(\frac{-6}{40}\right)
\displaystyle =15-0.9=14.1
\displaystyle \therefore \text{The mean number of days a student was absent is }14.1\text{ days.}
\displaystyle \\

\displaystyle \text{(ii)}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Number of days}&0-6&6-10&10-14&14-20&20-28&28-38&38-40\\ \hline \text{Number of students}&11&10&7&4&4&3&1\\ \hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Since the class intervals have unequal widths, we use the assumed mean method.}
\displaystyle \text{Let the assumed mean be }A=12.
\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&f_id_i\\ \hline 0-6&3&11&-9&-99\\ \hline 6-10&8&10&-4&-40\\ \hline 10-14&12&7&0&0\\ \hline 14-20&17&4&5&20\\ \hline 20-28&24&4&12&48\\ \hline 28-38&33&3&21&63\\ \hline 38-40&39&1&27&27\\ \hline \text{Total}&&\sum f_i=40&&\sum f_id_i=19\\ \hline\end{array}
\displaystyle \bar{x}=A+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =12+\frac{19}{40}
\displaystyle =12+0.475=12.475
\displaystyle \therefore \text{The mean number of days a student was absent is }12.475\text{ days.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The following table gives the literacy rate (in percentage) of }35\text{ cities. }
\displaystyle \text{Find the mean literacy rate.}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Literacy rate (in \%)}&45-55&55-65&65-75&75-85&85-95\\ \hline \text{Number of cities}&3&10&11&8&3\\ \hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=10.\text{ Let the assumed mean be }A=70.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 45-55&50&3&-20&-2&-6\\ \hline 55-65&60&10&-10&-1&-10\\ \hline 65-75&70&11&0&0&0\\ \hline 75-85&80&8&10&1&8\\ \hline 85-95&90&3&20&2&6\\ \hline \text{Total}&&\sum f_i=35&&&\sum f_iu_i=-2\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =70+10\left(\frac{-2}{35}\right)
\displaystyle =70-\frac{20}{35}=70-0.5714=69.4286
\displaystyle \therefore \text{The mean literacy rate is }69.43\%\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The following is the cumulative frequency distribution (of less than type) of }
\displaystyle 1000\text{ persons} \ \text{each of age }20\text{ years and above. Determine the mean age.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Age below (in years)}&30&40&50&60&70&80\\ \hline \text{Number of persons}&100&220&350&750&950&1000\\ \hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{First, we convert the cumulative frequency distribution into an ordinary frequency distribution.}
\displaystyle \begin{array}{|c|c|}\hline \text{Age (in years)}&\text{Frequency}\\ \hline 20-30&100\\ \hline 30-40&220-100=120\\ \hline 40-50&350-220=130\\ \hline 50-60&750-350=400\\ \hline 60-70&950-750=200\\ \hline 70-80&1000-950=50\\ \hline\end{array}
\displaystyle \text{The class size is }h=10.\text{ Let the assumed mean be }A=55.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 20-30&25&100&-30&-3&-300\\ \hline 30-40&35&120&-20&-2&-240\\ \hline 40-50&45&130&-10&-1&-130\\ \hline 50-60&55&400&0&0&0\\ \hline 60-70&65&200&10&1&200\\ \hline 70-80&75&50&20&2&100\\ \hline \text{Total}&&\sum f_i=1000&&&\sum f_iu_i=-370\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =55+10\left(\frac{-370}{1000}\right)
\displaystyle =55-3.7=51.3
\displaystyle \therefore \text{The mean age is }51.3\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The marks obtained by }110\text{ students in an examination are given below:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Marks}&30-35&35-40&40-45&45-50&50-55&55-60&60-65\\ \hline \text{Frequency}&14&16&28&23&18&8&3\\ \hline\end{array}
\displaystyle \text{Find the mean marks of the students.}\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=5.\text{ Let the assumed mean be }A=47.5.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 30-35&32.5&14&-15&-3&-42\\ \hline 35-40&37.5&16&-10&-2&-32\\ \hline 40-45&42.5&28&-5&-1&-28\\ \hline 45-50&47.5&23&0&0&0\\ \hline 50-55&52.5&18&5&1&18\\ \hline 55-60&57.5&8&10&2&16\\ \hline 60-65&62.5&3&15&3&9\\ \hline \text{Total}&&\sum f_i=110&&&\sum f_iu_i=-59\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =47.5+5\left(\frac{-59}{110}\right)
\displaystyle =47.5-\frac{295}{110}=47.5-2.6818=44.8182
\displaystyle \therefore \text{The mean marks of the students are }44.82\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the mean of the following data using assumed mean method:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Classes}&0-5&5-10&10-15&15-20&20-25\\ \hline \text{Frequency}&8&7&10&13&12\\ \hline\end{array}\hfill\text{[CBSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the assumed mean be }A=12.5.
\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&f_id_i\\ \hline 0-5&2.5&8&-10&-80\\ \hline 5-10&7.5&7&-5&-35\\ \hline 10-15&12.5&10&0&0\\ \hline 15-20&17.5&13&5&65\\ \hline 20-25&22.5&12&10&120\\ \hline \text{Total}&&\sum f_i=50&&\sum f_id_i=70\\ \hline\end{array}
\displaystyle \bar{x}=A+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =12.5+\frac{70}{50}
\displaystyle =12.5+1.4=13.9
\displaystyle \therefore \text{The mean of the given data is }13.9.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the mean of the following frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Classes}&25-29&30-34&35-39&40-44&45-49&50-54&55-59\\ \hline \text{Frequency}&14&22&16&6&5&3&4\\ \hline\end{array}\hfill\text{[CBSE 2006C]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=5.\text{ Let the assumed mean be }A=42.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 25-29&27&14&-15&-3&-42\\ \hline 30-34&32&22&-10&-2&-44\\ \hline 35-39&37&16&-5&-1&-16\\ \hline 40-44&42&6&0&0&0\\ \hline 45-49&47&5&5&1&5\\ \hline 50-54&52&3&10&2&6\\ \hline 55-59&57&4&15&3&12\\ \hline \text{Total}&&\sum f_i=70&&&\sum f_iu_i=-79\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =42+5\left(\frac{-79}{70}\right)
\displaystyle =42-\frac{395}{70}
\displaystyle =42-5.6429=36.3571
\displaystyle \therefore \text{The mean of the given frequency distribution is }36.36\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The mean of the following frequency distribution is }62.8\text{ and the sum }
\displaystyle \text{of all the frequencies is }50.\text{ Compute the missing frequencies }f_1\text{ and }f_2.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class}&0-20&20-40&40-60&60-80&80-100&100-120\\ \hline \text{Frequency}&5&f_1&10&f_2&7&8\\ \hline\end{array}\hfill\text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=20.\text{ Let the assumed mean be }A=50.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 0-20&10&5&-40&-2&-10\\ \hline 20-40&30&f_1&-20&-1&-f_1\\ \hline 40-60&50&10&0&0&0\\ \hline 60-80&70&f_2&20&1&f_2\\ \hline 80-100&90&7&40&2&14\\ \hline 100-120&110&8&60&3&24\\ \hline \text{Total}&&50&&&28-f_1+f_2\\ \hline\end{array}
\displaystyle 5+f_1+10+f_2+7+8=50
\displaystyle f_1+f_2=20\qquad\ldots\text{(i)}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle 62.8=50+20\left(\frac{28-f_1+f_2}{50}\right)
\displaystyle 12.8=\frac{2}{5}(28-f_1+f_2)
\displaystyle 32=28-f_1+f_2
\displaystyle f_2-f_1=4\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2f_2=24
\displaystyle f_2=12
\displaystyle \text{From (i),}\quad f_1+12=20
\displaystyle f_1=8
\displaystyle \therefore \text{The missing frequencies are }f_1=8\text{ and }f_2=12.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In a retail market, fruit vendors were selling mangoes kept in}
\displaystyle \text{packing boxes. These boxes contained varying number of mangoes. The following was the}
\displaystyle \text{distribution of mangoes according to the number of boxes.}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Number of mangoes}&50-52&53-55&56-58&59-61&62-64\\ \hline \text{Number of boxes}&15&110&135&115&25\\ \hline\end{array}
\displaystyle \text{Find the mean number of mangoes kept in a packing box. Which method of finding the mean}
\displaystyle \text{did you choose?}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks differ by }3.\text{ Hence, take }h=3\text{ and assumed mean }A=57.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 50-52&51&15&-6&-2&-30\\ \hline 53-55&54&110&-3&-1&-110\\ \hline 56-58&57&135&0&0&0\\ \hline 59-61&60&115&3&1&115\\ \hline 62-64&63&25&6&2&50\\ \hline \text{Total}&&\sum f_i=400&&&\sum f_iu_i=25\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =57+3\left(\frac{25}{400}\right)
\displaystyle =57+\frac{75}{400}
\displaystyle =57+0.1875=57.1875
\displaystyle \therefore \text{The mean number of mangoes per packing box is }57.1875\approx57.19.
\displaystyle \text{The step-deviation method is used because the class marks differ by a common factor of }3.
\displaystyle \text{This gives small integral values of }u_i\text{ and simplifies the calculation.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{To find out the concentration of }SO_2\text{ in the air (in parts per million, i.e., ppm),}
\displaystyle \text{ the data was collected for }30\text{ localities in a certain city and is presented below:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Concentration of }SO_2\text{ (in ppm)}&0.00-0.04&0.04-0.08&0.08-0.12&0.12-0.16&0.16-0.20&0.20-0.24\\ \hline \text{Frequency}&4&9&9&2&4&2\\ \hline\end{array}
\displaystyle \text{Find the mean concentration of }SO_2\text{ in the air.}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=0.04.\text{ Let the assumed mean be }A=0.10.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 0.00-0.04&0.02&4&-0.08&-2&-8\\ \hline 0.04-0.08&0.06&9&-0.04&-1&-9\\ \hline 0.08-0.12&0.10&9&0&0&0\\ \hline 0.12-0.16&0.14&2&0.04&1&2\\ \hline 0.16-0.20&0.18&4&0.08&2&8\\ \hline 0.20-0.24&0.22&2&0.12&3&6\\ \hline \text{Total}&&\sum f_i=30&&&\sum f_iu_i=-1\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =0.10+0.04\left(\frac{-1}{30}\right)
\displaystyle =0.10-0.00133=0.09867\text{ ppm}
\displaystyle \therefore \text{The mean concentration of }SO_2\text{ in the air is }0.099\text{ ppm (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If the mean of the following frequency distribution is }18,\text{ find the missing frequency.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Class interval}&11-13&13-15&15-17&17-19&19-21&21-23&23-25\\ \hline \text{Frequency}&3&6&9&13&p&5&4\\ \hline\end{array}\hfill\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=2.\text{ Let the assumed mean be }A=18.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 11-13&12&3&-6&-3&-9\\ \hline 13-15&14&6&-4&-2&-12\\ \hline 15-17&16&9&-2&-1&-9\\ \hline 17-19&18&13&0&0&0\\ \hline 19-21&20&p&2&1&p\\ \hline 21-23&22&5&4&2&10\\ \hline 23-25&24&4&6&3&12\\ \hline \text{Total}&&\sum f_i=40+p&&&\sum f_iu_i=p-8\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle 18=18+2\left(\frac{p-8}{40+p}\right)
\displaystyle 0=2\left(\frac{p-8}{40+p}\right)
\displaystyle p-8=0
\displaystyle p=8
\displaystyle \therefore \text{The missing frequency is }8.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The daily income of a sample of }50\text{ employees are tabulated as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Income (in Rs.)}&1-200&201-400&401-600&601-800\\ \hline \text{No. of employees}&14&15&14&7\\ \hline\end{array}
\displaystyle \text{Find the mean daily income of employees.}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks differ by }200.\text{ Hence, take }h=200\text{ and assumed mean }A=300.5.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 1-200&100.5&14&-200&-1&-14\\ \hline 201-400&300.5&15&0&0&0\\ \hline 401-600&500.5&14&200&1&14\\ \hline 601-800&700.5&7&400&2&14\\ \hline \text{Total}&&\sum f_i=50&&&\sum f_iu_i=14\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =300.5+200\left(\frac{14}{50}\right)
\displaystyle =300.5+56=356.5
\displaystyle \therefore \text{The mean daily income of the employees is Rs. }356.50.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The mean of the following frequency distribution is }25.\text{ Find the value of }f.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class}&0-10&10-20&20-30&30-40&40-50\\ \hline \text{Frequency}&5&18&15&f&6\\ \hline\end{array}\hfill\text{[CBSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=10.\text{ Let the assumed mean be }A=25.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 0-10&5&5&-20&-2&-10\\ \hline 10-20&15&18&-10&-1&-18\\ \hline 20-30&25&15&0&0&0\\ \hline 30-40&35&f&10&1&f\\ \hline 40-50&45&6&20&2&12\\ \hline \text{Total}&&\sum f_i=44+f&&&\sum f_iu_i=f-16\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle 25=25+10\left(\frac{f-16}{44+f}\right)
\displaystyle 0=10\left(\frac{f-16}{44+f}\right)
\displaystyle f-16=0
\displaystyle f=16
\displaystyle \therefore \text{The value of }f\text{ is }16.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The following table shows the age of patients admitted in a hospital}
\displaystyle \text{during a particular week:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Age (in years)}&5-15&15-25&25-35&35-45&45-55&55-65\\ \hline \text{Number of patients}&5&12&20&24&15&4\\ \hline\end{array}\hfill\text{[CBSE 2022]}
\displaystyle \text{Find the mean age of patients.}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=10.\text{ Let the assumed mean be }A=40.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 5-15&10&5&-30&-3&-15\\ \hline 15-25&20&12&-20&-2&-24\\ \hline 25-35&30&20&-10&-1&-20\\ \hline 35-45&40&24&0&0&0\\ \hline 45-55&50&15&10&1&15\\ \hline 55-65&60&4&20&2&8\\ \hline \text{Total}&&\sum f_i=80&&&\sum f_iu_i=-36\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =40+10\left(\frac{-36}{80}\right)
\displaystyle =40-4.5=35.5
\displaystyle \therefore \text{The mean age of the patients is }35.5\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Find the mean of the following frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Classes}&25-30&30-35&35-40&40-45&45-50&50-55&55-60\\ \hline \text{Frequency}&14&22&16&6&5&3&4\\ \hline\end{array}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=5.\text{ Let the assumed mean be }A=42.5.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 25-30&27.5&14&-15&-3&-42\\ \hline 30-35&32.5&22&-10&-2&-44\\ \hline 35-40&37.5&16&-5&-1&-16\\ \hline 40-45&42.5&6&0&0&0\\ \hline 45-50&47.5&5&5&1&5\\ \hline 50-55&52.5&3&10&2&6\\ \hline 55-60&57.5&4&15&3&12\\ \hline \text{Total}&&\sum f_i=70&&&\sum f_iu_i=-79\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =42.5+5\left(\frac{-79}{70}\right)
\displaystyle =42.5-\frac{395}{70}
\displaystyle =42.5-5.6429=36.8571
\displaystyle \therefore \text{The mean of the given frequency distribution is }36.86\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In a test, the marks obtained by }100\text{ students (out of }50\text{ marks) are given below:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Marks obtained}&0-10&10-20&20-30&30-40&40-50\\ \hline \text{Number of students}&12&23&34&25&6\\ \hline\end{array}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=10.\text{ Let the assumed mean be }A=25.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 0-10&5&12&-20&-2&-24\\ \hline 10-20&15&23&-10&-1&-23\\ \hline 20-30&25&34&0&0&0\\ \hline 30-40&35&25&10&1&25\\ \hline 40-50&45&6&20&2&12\\ \hline \text{Total}&&\sum f_i=100&&&\sum f_iu_i=-10\\ \hline\end{array}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle =25+10\left(\frac{-10}{100}\right)
\displaystyle =25-1=24
\displaystyle \therefore \text{The mean marks obtained by the students are }24.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The following distribution shows the weekly pocket allowance of }64\text{ children}
\displaystyle \text{ of a locality. If the mean pocket allowance is Rs. }180,\text{ find the values of }x\text{ and }y.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Pocket allowance (in Rs.)}&110-130&130-150&150-170&170-190&190-210&210-230&230-250\\ \hline \text{Number of children}&7&6&9&13&x&5&y\\ \hline\end{array} \\ \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{The class size is }h=20.\text{ Let the assumed mean be }A=180.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval}&x_i&f_i&d_i=x_i-A&u_i=\frac{d_i}{h}&f_iu_i\\ \hline 110-130&120&7&-60&-3&-21\\ \hline 130-150&140&6&-40&-2&-12\\ \hline 150-170&160&9&-20&-1&-9\\ \hline 170-190&180&13&0&0&0\\ \hline 190-210&200&x&20&1&x\\ \hline 210-230&220&5&40&2&10\\ \hline 230-250&240&y&60&3&3y\\ \hline \text{Total}&&64&&&x+3y-32\\ \hline\end{array}
\displaystyle 7+6+9+13+x+5+y=64
\displaystyle x+y=24\qquad\ldots\text{(i)}
\displaystyle \bar{x}=A+h\left(\frac{\sum f_iu_i}{\sum f_i}\right)
\displaystyle 180=180+20\left(\frac{x+3y-32}{64}\right)
\displaystyle x+3y-32=0
\displaystyle x+3y=32\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle 2y=8
\displaystyle y=4
\displaystyle \text{From (i),}\quad x+4=24
\displaystyle x=20
\displaystyle \therefore x=20\text{ and }y=4.
\displaystyle \\

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