\displaystyle \textbf{Question 1: }\text{Find the mode of the following data:}
\displaystyle \text{(i) }3,\ 5,\ 7,\ 4,\ 5,\ 3,\ 5,\ 6,\ 8,\ 9,\ 5,\ 3,\ 5,\ 3,\ 6,\ 9,\ 7,\ 4
\displaystyle \text{(ii) }3,\ 3,\ 7,\ 4,\ 5,\ 3,\ 5,\ 6,\ 8,\ 9,\ 5,\ 3,\ 5,\ 3,\ 6,\ 9,\ 7,\ 4
\displaystyle \text{(iii) }15,\ 8,\ 26,\ 25,\ 24,\ 15,\ 18,\ 20,\ 24,\ 15,\ 19,\ 15
\displaystyle \text{Answer:}
\displaystyle \text{(i) The frequency table for the given data is:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline x_i & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline f_i & 4 & 2 & 5 & 2 & 2 & 1 & 2 \\ \hline \end{array}
\displaystyle \text{Here, }5\text{ has the maximum frequency, i.e., }5.
\displaystyle \therefore\ \text{Mode}=5.
\displaystyle \text{(ii) The frequency table for the given data is:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline x_i & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline f_i & 5 & 2 & 4 & 2 & 2 & 1 & 2 \\ \hline \end{array}
\displaystyle \text{Here, }3\text{ has the maximum frequency, i.e., }5.
\displaystyle \therefore\ \text{Mode}=3.
\displaystyle \text{(iii) The frequency table for the given data is:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x_i & 8 & 15 & 18 & 19 & 20 & 24 & 25 & 26 \\ \hline f_i & 1 & 4 & 1 & 1 & 1 & 2 & 1 & 1 \\ \hline \end{array}
\displaystyle \text{Here, }15\text{ has the maximum frequency, i.e., }4.
\displaystyle \therefore\ \text{Mode}=15.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The shirt sizes worn by a group of }200\text{ persons, who bought}
\displaystyle \text{the shirt from a store, are as follows. Find the modal shirt size worn by the group.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Shirt size} & 37 & 38 & 39 & 40 & 41 & 42 & 43 & 44 \\ \hline \text{Number of persons} & 15 & 25 & 39 & 41 & 36 & 17 & 15 & 12 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{From the given frequency distribution, the maximum frequency is }41.
\displaystyle \text{The shirt size corresponding to the maximum frequency }41\text{ is }40.
\displaystyle \therefore\ \text{Modal shirt size}=40.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the mode of the following distribution.}
\displaystyle \text{(i)}\quad\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Class interval} & 25-30 & 30-35 & 35-40 & 40-45 & 45-50 & 50-55 \\ \hline \text{Frequency} & 25 & 34 & 50 & 42 & 38 & 14 \\ \hline \end{array}
\displaystyle \text{(ii)}\quad\begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Class} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 \\ \hline \text{Frequency} & 8 & 10 & 10 & 16 & 12 & 6 & 7 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) The maximum frequency is }50,\text{ so }35-40\text{ is the modal class.}
\displaystyle l=35,\quad h=5,\quad f=50,\quad f_1=34,\quad f_2=42
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =35+\frac{50-34}{2(50)-34-42}\times5
\displaystyle =35+\frac{16}{24}\times5
\displaystyle =35+\frac{10}{3}=38.33
\displaystyle \therefore\ \text{Mode}=38.33\text{ (approximately).}
\displaystyle \text{(ii) The maximum frequency is }16,\text{ so }30-40\text{ is the modal class.}
\displaystyle l=30,\quad h=10,\quad f=16,\quad f_1=10,\quad f_2=12
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =30+\frac{16-10}{2(16)-10-12}\times10
\displaystyle =30+\frac{6}{10}\times10
\displaystyle =30+6=36
\displaystyle \therefore\ \text{Mode}=36.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Compare the modal ages of two groups of students appearing for}
\displaystyle \text{an entrance test:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Age (in years)} & 16-18 & 18-20 & 20-22 & 22-24 & 24-26 \\ \hline \text{Group A} & 50 & 78 & 46 & 28 & 23 \\ \hline \text{Group B} & 54 & 89 & 40 & 25 & 17 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{For Group A, the maximum frequency is }78,\text{ so }18-20\text{ is the modal class.}
\displaystyle l=18,\quad h=2,\quad f=78,\quad f_1=50,\quad f_2=46
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =18+\frac{78-50}{2(78)-50-46}\times2
\displaystyle =18+\frac{28}{60}\times2=18+\frac{14}{15}=18.93
\displaystyle \therefore\ \text{Modal age of Group A}=18.93\text{ years (approximately).}
\displaystyle \text{For Group B, the maximum frequency is }89,\text{ so }18-20\text{ is the modal class.}
\displaystyle l=18,\quad h=2,\quad f=89,\quad f_1=54,\quad f_2=40
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =18+\frac{89-54}{2(89)-54-40}\times2
\displaystyle =18+\frac{35}{84}\times2=18+\frac{5}{6}=18.83
\displaystyle \therefore\ \text{Modal age of Group B}=18.83\text{ years (approximately).}
\displaystyle \therefore\ \text{Group A has a slightly higher modal age than Group B.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The following data gives the information on the observed lifetimes}
\displaystyle \text{(in hours) of }225\text{ electrical components. Determine the modal lifetimes of the}
\displaystyle \text{components.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Lifetimes (in hours)} & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 & 100-120 \\ \hline \text{No. of components} & 10 & 35 & 52 & 61 & 38 & 29 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The maximum frequency is }61,\text{ so }60-80\text{ is the modal class.}
\displaystyle l=60,\quad h=20,\quad f=61,\quad f_1=52,\quad f_2=38
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =60+\frac{61-52}{2(61)-52-38}\times20
\displaystyle =60+\frac{9}{32}\times20
\displaystyle =60+\frac{45}{8}=65.625
\displaystyle \therefore\ \text{Modal lifetime of the components}=65.625\text{ hours.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The following table gives the daily income of }50\text{ workers of a}
\displaystyle \text{factory. Find the mean, mode and median of the above data.}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Daily income (in Rs.)} & 100-120 & 120-140 & 140-160 & 160-180 & 180-200 \\ \hline \text{Number of workers} & 12 & 14 & 8 & 6 & 10 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{First, we calculate the mean using the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Daily income} & f_i & x_i & f_i x_i \\ \hline 100-120 & 12 & 110 & 1320 \\ \hline 120-140 & 14 & 130 & 1820 \\ \hline 140-160 & 8 & 150 & 1200 \\ \hline 160-180 & 6 & 170 & 1020 \\ \hline 180-200 & 10 & 190 & 1900 \\ \hline \text{Total} & 50 & & 7260 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{7260}{50}=145.2
\displaystyle \therefore\ \text{Mean daily income}=Rs.\ 145.20.
\displaystyle \text{The maximum frequency is }14,\text{ so }120-140\text{ is the modal class.}
\displaystyle l=120,\quad h=20,\quad f=14,\quad f_1=12,\quad f_2=8
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =120+\frac{14-12}{2(14)-12-8}\times20
\displaystyle =120+\frac{2}{8}\times20=125
\displaystyle \therefore\ \text{Mode}=Rs.\ 125.
\displaystyle \text{Now, preparing the cumulative frequency table for finding the median, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Daily income} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 100-120 & 12 & 12 \\ \hline 120-140 & 14 & 26 \\ \hline 140-160 & 8 & 34 \\ \hline 160-180 & 6 & 40 \\ \hline 180-200 & 10 & 50 \\ \hline \end{array}
\displaystyle N=50\quad\Rightarrow\quad\frac{N}{2}=\frac{50}{2}=25
\displaystyle \text{The cumulative frequency just greater than }25\text{ is }26.
\displaystyle \therefore\ 120-140\text{ is the median class.}
\displaystyle l=120,\quad f=14,\quad F=12,\quad h=20
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =120+\frac{25-12}{14}\times20
\displaystyle =120+\frac{260}{14}=138.57\text{ (approximately)}
\displaystyle \therefore\ \text{Median daily income}=Rs.\ 138.57.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The following distribution gives the state-wise teacher-student ratio}
\displaystyle \text{in higher secondary schools of India. Find the mode and mean of this data. Interpret}
\displaystyle \text{the two measures.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Number of students per teacher} & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 & 40-45 & 45-50 & 50-55 \\ \hline \text{Number of states/UT} & 3 & 8 & 9 & 10 & 3 & 0 & 0 & 2 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The maximum frequency is }10,\text{ so }30-35\text{ is the modal class.}
\displaystyle l=30,\quad h=5,\quad f=10,\quad f_1=9,\quad f_2=3
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =30+\frac{10-9}{2(10)-9-3}\times5
\displaystyle =30+\frac{1}{8}\times5=30.625
\displaystyle \therefore\ \text{Mode}=30.63\text{ students per teacher (approximately).}
\displaystyle \text{For calculating the mean, we use the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Students per teacher} & f_i & x_i & f_i x_i \\ \hline 15-20 & 3 & 17.5 & 52.5 \\ \hline 20-25 & 8 & 22.5 & 180 \\ \hline 25-30 & 9 & 27.5 & 247.5 \\ \hline 30-35 & 10 & 32.5 & 325 \\ \hline 35-40 & 3 & 37.5 & 112.5 \\ \hline 40-45 & 0 & 42.5 & 0 \\ \hline 45-50 & 0 & 47.5 & 0 \\ \hline 50-55 & 2 & 52.5 & 105 \\ \hline \text{Total} & 35 & & 1022.5 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{1022.5}{35}=29.21\text{ (approximately)}
\displaystyle \therefore\ \text{Mean}=29.21\text{ students per teacher (approximately).}
\displaystyle \text{Thus, on average, there are about }29.21\text{ students per teacher.}
\displaystyle \text{The most common teacher-student ratio is about }30.63\text{ students per teacher.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the mean, median and mode of the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Classes} & 0-50 & 50-100 & 100-150 & 150-200 & 200-250 & 250-300 & 300-350 \\ \hline \text{Frequency} & 2 & 3 & 5 & 6 & 5 & 3 & 1 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{First, we calculate the mean using the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Class interval} & f_i & x_i & f_i x_i \\ \hline 0-50 & 2 & 25 & 50 \\ \hline 50-100 & 3 & 75 & 225 \\ \hline 100-150 & 5 & 125 & 625 \\ \hline 150-200 & 6 & 175 & 1050 \\ \hline 200-250 & 5 & 225 & 1125 \\ \hline 250-300 & 3 & 275 & 825 \\ \hline 300-350 & 1 & 325 & 325 \\ \hline \text{Total} & 25 & & 4225 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{4225}{25}=169
\displaystyle \therefore\ \text{Mean}=169.
\displaystyle \text{Now, preparing the cumulative frequency table for finding the median, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Class interval} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 0-50 & 2 & 2 \\ \hline 50-100 & 3 & 5 \\ \hline 100-150 & 5 & 10 \\ \hline 150-200 & 6 & 16 \\ \hline 200-250 & 5 & 21 \\ \hline 250-300 & 3 & 24 \\ \hline 300-350 & 1 & 25 \\ \hline \end{array}
\displaystyle N=25\quad\Rightarrow\quad\frac{N}{2}=\frac{25}{2}=12.5
\displaystyle \text{The cumulative frequency just greater than }12.5\text{ is }16.
\displaystyle \therefore\ 150-200\text{ is the median class.}
\displaystyle l=150,\quad f=6,\quad F=10,\quad h=50
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =150+\frac{12.5-10}{6}\times50
\displaystyle =150+\frac{125}{6}=170.83\text{ (approximately)}
\displaystyle \therefore\ \text{Median}=170.83\text{ (approximately).}
\displaystyle \text{The maximum frequency is }6,\text{ so }150-200\text{ is the modal class.}
\displaystyle l=150,\quad h=50,\quad f=6,\quad f_1=5,\quad f_2=5
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =150+\frac{6-5}{2(6)-5-5}\times50
\displaystyle =150+\frac{1}{2}\times50=175
\displaystyle \therefore\ \text{Mode}=175.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A student noted the number of cars passing through a spot on a}
\displaystyle \text{road for }100\text{ periods each of }3\text{ minutes and summarised it in the table}
\displaystyle \text{given below. Find the mode of the data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Number of cars} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \text{Frequency} & 7 & 14 & 13 & 12 & 20 & 11 & 15 & 8 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The maximum frequency is }20,\text{ so }40-50\text{ is the modal class.}
\displaystyle l=40,\quad h=10,\quad f=20,\quad f_1=12,\quad f_2=11
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =40+\frac{20-12}{2(20)-12-11}\times10
\displaystyle =40+\frac{8}{17}\times10
\displaystyle =40+\frac{80}{17}=44.71\text{ (approximately)}
\displaystyle \therefore\ \text{Mode}=44.71\text{ (approximately).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the mean, median and mode of the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Classes} & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 & 100-120 & 120-140 \\ \hline \text{Frequency} & 6 & 8 & 10 & 12 & 6 & 5 & 3 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2008, 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{First, we calculate the mean using the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Class interval} & f_i & x_i & f_i x_i \\ \hline 0-20 & 6 & 10 & 60 \\ \hline 20-40 & 8 & 30 & 240 \\ \hline 40-60 & 10 & 50 & 500 \\ \hline 60-80 & 12 & 70 & 840 \\ \hline 80-100 & 6 & 90 & 540 \\ \hline 100-120 & 5 & 110 & 550 \\ \hline 120-140 & 3 & 130 & 390 \\ \hline \text{Total} & 50 & & 3120 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{3120}{50}=62.4
\displaystyle \therefore\ \text{Mean}=62.4.
\displaystyle \text{Now, preparing the cumulative frequency table for finding the median, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Class interval} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 0-20 & 6 & 6 \\ \hline 20-40 & 8 & 14 \\ \hline 40-60 & 10 & 24 \\ \hline 60-80 & 12 & 36 \\ \hline 80-100 & 6 & 42 \\ \hline 100-120 & 5 & 47 \\ \hline 120-140 & 3 & 50 \\ \hline \end{array}
\displaystyle N=50\quad\Rightarrow\quad\frac{N}{2}=\frac{50}{2}=25
\displaystyle \text{The cumulative frequency just greater than }25\text{ is }36.
\displaystyle \therefore\ 60-80\text{ is the median class.}
\displaystyle l=60,\quad f=12,\quad F=24,\quad h=20
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =60+\frac{25-24}{12}\times20
\displaystyle =60+\frac{20}{12}=61.67\text{ (approximately)}
\displaystyle \therefore\ \text{Median}=61.67\text{ (approximately).}
\displaystyle \text{The maximum frequency is }12,\text{ so }60-80\text{ is the modal class.}
\displaystyle l=60,\quad h=20,\quad f=12,\quad f_1=10,\quad f_2=6
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =60+\frac{12-10}{2(12)-10-6}\times20
\displaystyle =60+\frac{2}{8}\times20=65
\displaystyle \therefore\ \text{Mode}=65.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The frequency distribution table of agriculture holdings in a}
\displaystyle \text{village is given below. Find the modal agriculture holdings of the village.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Area of land (in hectares)} & 1-3 & 3-5 & 5-7 & 7-9 & 9-11 & 11-13 \\ \hline \text{Number of families} & 20 & 45 & 80 & 55 & 40 & 12 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The maximum frequency is }80,\text{ so }5-7\text{ is the modal class.}
\displaystyle l=5,\quad h=2,\quad f=80,\quad f_1=45,\quad f_2=55
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =5+\frac{80-45}{2(80)-45-55}\times2
\displaystyle =5+\frac{35}{60}\times2
\displaystyle =5+\frac{7}{6}=6.17\text{ (approximately)}
\displaystyle \therefore\ \text{Modal agriculture holding}=6.17\text{ hectares (approximately).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The given distribution shows the number of runs scored by}
\displaystyle \text{some top batsmen of the world in one-day international cricket matches. Find the}
\displaystyle \text{mode of the data.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Runs scored} & 3000 & 4000 & 5000 & 6000 & 7000 & 8000 & 9000 & 10000 \\ & -4000 & -5000 & -6000 & -7000 & -8000 & -9000 & -10000 & -11000 \\ \hline \text{Number of batsmen} & 4 & 18 & 9 & 7 & 6 & 3 & 1 & 1 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The maximum frequency is }18,\text{ so }4000-5000\text{ is the modal class.}
\displaystyle l=4000,\quad h=1000,\quad f=18,\quad f_1=4,\quad f_2=9
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =4000+\frac{18-4}{2(18)-4-9}\times1000
\displaystyle =4000+\frac{14}{23}\times1000
\displaystyle =4000+608.70=4608.70\text{ (approximately)}
\displaystyle \therefore\ \text{Mode}=4608.70\text{ runs (approximately).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The monthly income of }100\text{ families are given as below.}
\displaystyle \text{Calculate the modal income.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Income (in Rs.)} & 0 & 5000 & 10000 & 15000 & 20000 & 25000 & 30000 & 35000 \\ & -5000 & -10000 & -15000 & -20000 & -25000 & -30000 & -35000 & -40000 \\ \hline \text{Number of families} & 8 & 26 & 41 & 16 & 3 & 3 & 2 & 1 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The maximum frequency is }41,\text{ so }10000-15000\text{ is the modal class.}
\displaystyle l=10000,\quad h=5000,\quad f=41,\quad f_1=26,\quad f_2=16
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =10000+\frac{41-26}{2(41)-26-16}\times5000
\displaystyle =10000+\frac{15}{40}\times5000
\displaystyle =10000+1875=11875
\displaystyle \therefore\ \text{Modal monthly income}=Rs.\ 11875.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{250 apples of a box were weighed and the distribution of masses}
\displaystyle \text{of the apples is given in the following table:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Mass (in grams)} & 80-100 & 100-120 & 120-140 & 140-160 & 160-180 \\ \hline \text{Number of apples} & 20 & 60 & 70 & x & 60 \\ \hline \end{array}
\displaystyle \text{(i) Find the value of }x\text{ and mean mass of the apples.}
\displaystyle \text{(ii) Find the modal mass of the apples.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since the total number of apples is }250,
\displaystyle 20+60+70+x+60=250
\displaystyle 210+x=250\quad\Rightarrow\quad x=40
\displaystyle \therefore\ \text{The missing frequency is }40.
\displaystyle \text{For finding the mean, we calculate the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Mass (in grams)} & f_i & x_i & f_i x_i \\ \hline 80-100 & 20 & 90 & 1800 \\ \hline 100-120 & 60 & 110 & 6600 \\ \hline 120-140 & 70 & 130 & 9100 \\ \hline 140-160 & 40 & 150 & 6000 \\ \hline 160-180 & 60 & 170 & 10200 \\ \hline \text{Total} & 250 & & 33700 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{33700}{250}=134.8
\displaystyle \therefore\ \text{Mean mass of the apples}=134.8\text{ grams.}
\displaystyle \text{(ii) The maximum frequency is }70,\text{ so }120-140\text{ is the modal class.}
\displaystyle l=120,\quad h=20,\quad f=70,\quad f_1=60,\quad f_2=40
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =120+\frac{70-60}{2(70)-60-40}\times20
\displaystyle =120+\frac{10}{40}\times20=125
\displaystyle \therefore\ \text{Modal mass of the apples}=125\text{ grams.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The following table shows the ages of the patients admitted in a}
\displaystyle \text{hospital during a year. Find the mode and mean of the given data.}
\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} & 6 & 11 & 21 & 23 & 14 & 5 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{The maximum frequency is }23,\text{ so }35-45\text{ is the modal class.}
\displaystyle l=35,\quad h=10,\quad f=23,\quad f_1=21,\quad f_2=14
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =35+\frac{23-21}{2(23)-21-14}\times10
\displaystyle =35+\frac{2}{11}\times10
\displaystyle =35+\frac{20}{11}=36.82\text{ (approximately)}
\displaystyle \therefore\ \text{Mode}=36.82\text{ years (approximately).}
\displaystyle \text{For finding the mean, we calculate the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Age (in years)} & f_i & x_i & f_i x_i \\ \hline 5-15 & 6 & 10 & 60 \\ \hline 15-25 & 11 & 20 & 220 \\ \hline 25-35 & 21 & 30 & 630 \\ \hline 35-45 & 23 & 40 & 920 \\ \hline 45-55 & 14 & 50 & 700 \\ \hline 55-65 & 5 & 60 & 300 \\ \hline \text{Total} & 80 & & 2830 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2830}{80}=35.375
\displaystyle \therefore\ \text{Mean age}=35.375\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the median and mode of the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Class} & 1-3 & 3-5 & 5-7 & 7-9 & 9-11 \\ \hline \text{Frequency} & 7 & 8 & 2 & 2 & 1 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{First, we prepare the cumulative frequency table to find the median.}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Class interval} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 1-3 & 7 & 7 \\ \hline 3-5 & 8 & 15 \\ \hline 5-7 & 2 & 17 \\ \hline 7-9 & 2 & 19 \\ \hline 9-11 & 1 & 20 \\ \hline \end{array}
\displaystyle N=20\quad\Rightarrow\quad\frac{N}{2}=\frac{20}{2}=10
\displaystyle \text{The cumulative frequency just greater than }10\text{ is }15.
\displaystyle \therefore\ 3-5\text{ is the median class.}
\displaystyle l=3,\quad f=8,\quad F=7,\quad h=2
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =3+\frac{10-7}{8}\times2
\displaystyle =3+\frac{6}{8}=3.75
\displaystyle \therefore\ \text{Median}=3.75.
\displaystyle \text{The maximum frequency is }8,\text{ so }3-5\text{ is the modal class.}
\displaystyle l=3,\quad h=2,\quad f=8,\quad f_1=7,\quad f_2=2
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =3+\frac{8-7}{2(8)-7-2}\times2
\displaystyle =3+\frac{1}{7}\times2=3+\frac{2}{7}=3.29\text{ (approximately)}
\displaystyle \therefore\ \text{Mode}=3.29\text{ (approximately).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the mean and mode of the following frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Class} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 \\ \hline \text{Frequency} & 8 & 7 & 15 & 20 & 12 & 8 & 10 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{First, we calculate the mean using the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Class interval} & f_i & x_i & f_i x_i \\ \hline 0-10 & 8 & 5 & 40 \\ \hline 10-20 & 7 & 15 & 105 \\ \hline 20-30 & 15 & 25 & 375 \\ \hline 30-40 & 20 & 35 & 700 \\ \hline 40-50 & 12 & 45 & 540 \\ \hline 50-60 & 8 & 55 & 440 \\ \hline 60-70 & 10 & 65 & 650 \\ \hline \text{Total} & 80 & & 2850 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2850}{80}=35.625
\displaystyle \therefore\ \text{Mean}=35.625.
\displaystyle \text{The maximum frequency is }20,\text{ so }30-40\text{ is the modal class.}
\displaystyle l=30,\quad h=10,\quad f=20,\quad f_1=15,\quad f_2=12
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =30+\frac{20-15}{2(20)-15-12}\times10
\displaystyle =30+\frac{5}{13}\times10
\displaystyle =30+\frac{50}{13}=33.85\text{ (approximately)}
\displaystyle \therefore\ \text{Mode}=33.85\text{ (approximately).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find the mean and mode of the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Class} & 4-8 & 8-12 & 12-16 & 16-20 & 20-24 & 24-28 & 28-32 & 32-36 \\ \hline \text{Frequency} & 2 & 12 & 15 & 25 & 18 & 12 & 13 & 3 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{First, we calculate the mean using the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Class interval} & f_i & x_i & f_i x_i \\ \hline 4-8 & 2 & 6 & 12 \\ \hline 8-12 & 12 & 10 & 120 \\ \hline 12-16 & 15 & 14 & 210 \\ \hline 16-20 & 25 & 18 & 450 \\ \hline 20-24 & 18 & 22 & 396 \\ \hline 24-28 & 12 & 26 & 312 \\ \hline 28-32 & 13 & 30 & 390 \\ \hline 32-36 & 3 & 34 & 102 \\ \hline \text{Total} & 100 & & 1992 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{1992}{100}=19.92
\displaystyle \therefore\ \text{Mean}=19.92.
\displaystyle \text{The maximum frequency is }25,\text{ so }16-20\text{ is the modal class.}
\displaystyle l=16,\quad h=4,\quad f=25,\quad f_1=15,\quad f_2=18
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =16+\frac{25-15}{2(25)-15-18}\times4
\displaystyle =16+\frac{10}{17}\times4
\displaystyle =16+\frac{40}{17}=18.35\text{ (approximately)}
\displaystyle \therefore\ \text{Mode}=18.35\text{ (approximately).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The following table shows the number of patients of different}
\displaystyle \text{age groups who were discharged from the hospital in a particular month. Find the}
\displaystyle \text{mean and modal age.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Age (in years)} & 5-15 & 15-25 & 25-35 & 35-45 & 45-55 & 55-65 & \text{Total} \\ \hline \text{Number of patients discharged} & 6 & 11 & 21 & 23 & 14 & 5 & 80 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{First, we calculate the mean using the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Age (in years)} & f_i & x_i & f_i x_i \\ \hline 5-15 & 6 & 10 & 60 \\ \hline 15-25 & 11 & 20 & 220 \\ \hline 25-35 & 21 & 30 & 630 \\ \hline 35-45 & 23 & 40 & 920 \\ \hline 45-55 & 14 & 50 & 700 \\ \hline 55-65 & 5 & 60 & 300 \\ \hline \text{Total} & 80 & & 2830 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2830}{80}=35.375
\displaystyle \therefore\ \text{Mean age}=35.375\text{ years.}
\displaystyle \text{The maximum frequency is }23,\text{ so }35-45\text{ is the modal class.}
\displaystyle l=35,\quad h=10,\quad f=23,\quad f_1=21,\quad f_2=14
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =35+\frac{23-21}{2(23)-21-14}\times10
\displaystyle =35+\frac{2}{11}\times10
\displaystyle =35+\frac{20}{11}=36.82\text{ (approximately)}
\displaystyle \therefore\ \text{Modal age}=36.82\text{ years (approximately).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The following table gives the daily income of }50\text{ cab}
\displaystyle \text{drivers of a particular city. Find the mean income and the modal income.}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Income (in Rs.)} & 500-600 & 600-700 & 700-800 & 800-900 & 900-1000 \\ \hline \text{No. of drivers} & 12 & 14 & 8 & 6 & 10 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{First, we calculate the mean using the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Income (in Rs.)} & f_i & x_i & f_i x_i \\ \hline 500-600 & 12 & 550 & 6600 \\ \hline 600-700 & 14 & 650 & 9100 \\ \hline 700-800 & 8 & 750 & 6000 \\ \hline 800-900 & 6 & 850 & 5100 \\ \hline 900-1000 & 10 & 950 & 9500 \\ \hline \text{Total} & 50 & & 36300 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{36300}{50}=726
\displaystyle \therefore\ \text{Mean income}=Rs.\ 726.
\displaystyle \text{The maximum frequency is }14,\text{ so }600-700\text{ is the modal class.}
\displaystyle l=600,\quad h=100,\quad f=14,\quad f_1=12,\quad f_2=8
\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h
\displaystyle =600+\frac{14-12}{2(14)-12-8}\times100
\displaystyle =600+\frac{2}{8}\times100=625
\displaystyle \therefore\ \text{Modal income}=Rs.\ 625.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.