\displaystyle \textbf{Question 1. }\text{The following table shows the number of patients of different age group who }
\displaystyle \text{were discharged from the hospital in a particular month:} \hspace{0.2cm}\text{[CBSE 2025]}

\displaystyle \begin{array}{|c|c|}\hline \text{Age\ (in\ years)} & \text{Number\ of\ Patients\ Discharged}\\ \hline 5\text{-}15 & 6\\ \hline 15\text{-}25 & 11\\ \hline 25\text{-}35 & 21\\ \hline 35\text{-}45 & 23\\ \hline 45\text{-}55 & 14\\ \hline 55\text{-}65 & 5\\ \hline \text{Total} & 80\\ \hline \end{array}

\displaystyle \text{Find the mean and the mode of the above data.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Age (In years)} & x_i & \text{No. of patients discharged }(f_i) & d_i=x_i-a & f_id_i\\ \hline 5-15 & 10 & 6 & -30 & -180\\ \hline 15-25 & 20 & 11 & -20 & -220\\ \hline 25-35 & 30 & 21=f_{0} & -10 & -210\\ \hline 35-45 & 40=a & 23=f_{1} & 0 & 0\\ \hline 45-55 & 50 & 14=f_{2} & 10 & 140\\ \hline 55-65 & 60 & 5 & 20 & 100\\ \hline \text{Total} & & \sum f_i=80 & & \sum f_id_i=-370\\ \hline \end{array}

\displaystyle \text{Now,}
\displaystyle \bar{x}=a+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =40+\frac{-370}{80}
\displaystyle =40-4.625
\displaystyle \text{Mean}=35.375
\displaystyle \text{Now,}
\displaystyle \text{Modal class}=35-45
\displaystyle l=35,\ f_{0}=21,\ f_{1}=23,\ f_{2}=14,\ h=10
\displaystyle \text{Mode}=l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h
\displaystyle =35+10\left(\frac{23-21}{2\times23-21-14}\right)
\displaystyle =35+\frac{20}{11}
\displaystyle =36.81
\\

\displaystyle \textbf{Question 2. }\text{The cumulative frequency for calculating median is obtained by} \\ \text{adding the frequencies of all the:} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) classes up to the median class}
\displaystyle \text{(b) classes following the median class}
\displaystyle \text{(c) classes preceding the median class}
\displaystyle \text{(d) all classes}
\displaystyle \text{Answer:}
\displaystyle \text{In the median formula for grouped data, }cf\text{ denotes the cumulative frequency of the class}
\displaystyle \text{preceding the median class. It is obtained by adding the frequencies of all the classes}
\displaystyle \text{preceding the median class.}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 3. }\text{If mean and median of given set of observations are }10\text{ and } \\ 11\text{ respectively, then the value of mode is:}
\displaystyle \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) }10.5
\displaystyle \text{(b) }8
\displaystyle \text{(c) }13
\displaystyle \text{(d) }21
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle \text{Mean}=10,\text{ median}=11,\text{ mode}=?
\displaystyle 3\text{ median}=\text{mode}+2\text{ mean}
\displaystyle 3\times11=\text{mode}+20
\displaystyle \Rightarrow \text{mode}=33-20=13
\\

\displaystyle \textbf{Question 4. }\text{BINGO is game of chance. The host has }75\text{ balls numbered }1\text{ through }75.
\displaystyle \text{Each player has a BINGO card with some numbers written on it. The participant cancels}
\displaystyle \text{the number on the card when called out a number written on the ball selected at random.}
\displaystyle \text{Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.}
\displaystyle \text{The table given below, shows the data of one such game where }48\text{ balls were used}
\displaystyle \text{before Tara said BINGO.}

\displaystyle \begin{array}{|c|c|}\hline \text{Numbers\ announced} & \text{Number\ of\ times}\\ \hline 0\text{-}15 & 8\\ \hline 15\text{-}30 & 9\\ \hline 30\text{-}45 & 10\\ \hline 45\text{-}60 & 12\\ \hline 60\text{-}75 & 9\\ \hline \end{array}

\displaystyle \text{Based on the above information, answer the following:} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(i) Write the median class.}
\displaystyle \text{(ii) When first ball was picked up, what was the probability of calling out an even number?}
\displaystyle \text{(iii) (a) Find median of the given data.}
\displaystyle \text{OR}
\displaystyle \text{(b) Find mode of the given data.}
\displaystyle \text{Answer:}

\displaystyle \textbf{(i)}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval} & \text{Frequency} & \text{Cumulative frequency }(cf)\\ \hline 0-15 & 8 & 8\\ \hline 15-30 & 9 & 17\\ \hline 30-45 & 10 & 27\\ \hline 45-60 & 12 & 39\\ \hline 60-75 & 9 & 48\\ \hline & N=48 & \\ \hline \end{array}

\displaystyle \text{Now, }\frac{N}{2}=\frac{48}{2}=24
\displaystyle \text{Now, cumulative frequency just greater than }\frac{N}{2}\text{ (i.e. }24\text{) is }27\text{ and corresponding class is }30-45.
\displaystyle \text{So, the median class is }30-45.

\displaystyle \textbf{(ii)}
\displaystyle \text{The possible outcomes are:}
\displaystyle 1,2,3,4,\ldots,74,75\text{ i.e. }75
\displaystyle \text{Now, even numbered balls are favourable, then favourable outcomes are:}
\displaystyle 2,4,6,8,10,12,\ldots,72,74
\displaystyle \text{Number of favourable outcomes}=37
\displaystyle \text{Required probability}=\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}=\frac{37}{75}

\displaystyle \textbf{28. (iii)(a)}
\displaystyle \text{Here }l=30,\ cf=17,\ f=10,\ h=15
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)\times h
\displaystyle =30+\left(\frac{24-17}{10}\right)\times15
\displaystyle =30+0.7\times15=40.5

\displaystyle \textbf{(iii)(b)}
\displaystyle \text{As maximum frequency is }12.\text{ So modal class is }45-60.
\displaystyle \text{Lower limit of modal class, }l=45
\displaystyle \text{Frequency of modal class, }f_{1}=12
\displaystyle \text{Frequency of class preceding modal class, }f_{0}=10
\displaystyle \text{Frequency of class succeeding modal class, }f_{2}=9
\displaystyle \text{Class size, }h=15
\displaystyle \text{Mode}=l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h
\displaystyle =45+\frac{12-10}{2\times12-10-9}\times15=45+\frac{2}{5}\times15=51
\\

\displaystyle \textbf{Question 5. }\text{For some data }x_{1},x_{2},\ldots,x_{n}\text{ with respective frequencies } 
\displaystyle f_{1}, f_{2}, \ldots, f_{n}, \text{ the value of }  \sum_{i=1}^{n}f_{i}(x_{i}-\bar{x})\text{ is equal to:} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(a) }n\bar{x} \qquad   \text{(b) }1 \qquad  \text{(c) }\sum f_{i} \qquad  \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \sum_{i=1}^{n}f_{i}(x_{i}-\bar{x})=f_{1}(x_{1}-\bar{x})+f_{2}(x_{2}-\bar{x})+f_{3}(x_{3}-\bar{x})+\cdots+f_{n}(x_{n}-\bar{x})
\displaystyle =[f_{1}x_{1}+f_{2}x_{2}+f_{3}x_{3}+\cdots+f_{n}x_{n}]-\bar{x}[f_{1}+f_{2}+f_{3}+\cdots+f_{n}]
\displaystyle =\sum f_{i}x_{i}-\bar{x}\sum f_{i}
\displaystyle =\bar{x}\sum f_{i}-\bar{x}\sum f_{i}\qquad \left[\because \bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}\right]
\displaystyle =0
\\

\displaystyle \textbf{Question 6. }\text{The middle most observation of every data arranged in order} \\ \text{is called:} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(a) mode}
\displaystyle \text{(b) median}
\displaystyle \text{(c) mean}
\displaystyle \text{(d) deviation}
\displaystyle \text{Answer:}
\displaystyle \text{The middle most observation of a data arranged in ascending or descending} \\ \text{order is called the median.}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 7. }\text{The empirical relation between the mode, median and mean} \\ \text{of a distribution is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) Mode }=3\text{ Median }-2\text{ Mean}
\displaystyle \text{(b) Mode }=3\text{ Mean }-2\text{ Median}
\displaystyle \text{(c) Mode }=2\text{ Median }-3\text{ Mean}
\displaystyle \text{(d) Mode }=2\text{ Mean }-3\text{ Median}
\displaystyle \text{Answer:}
\displaystyle \text{The empirical relation among mean, median and mode is}
\displaystyle \text{Mode}=3\text{ Median}-2\text{ Mean}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 8. }\text{Using empirical relationship, the mode of a distribution whose mean is } \\ 7.2\text{ and the median }7.1\text{ is }   \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{(a) }6.2  \qquad \text{(b) }6.3 \qquad  \text{(c) }6.5 \qquad  \text{(d) }6.9
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \text{Mode}=3\text{ median}-2\text{ mean}
\displaystyle =3\times7.1-2\times7.2
\displaystyle =21.3-14.4
\displaystyle =6.9
\\

\displaystyle \textbf{Question 9. }\text{Heights of }50\text{ students of class X of a school are recorded and} \\ \text{following data is obtained:} \hspace{0.2cm}\text{[CBSE 2023]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Height\ (in\ cm)} & 130\text{-}135 & 135\text{-}140 & 140\text{-}145 & 145\text{-}150 & 150\text{-}155 & 155\text{-}160\\ \hline \text{Number\ of\ Students} & 4 & 11 & 12 & 7 & 10 & 6\\ \hline \end{array}

\displaystyle \text{Find the median height of the students.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|}\hline \text{Height (in cm)} & \text{No. of students }(f) & cf\\ \hline 130-135 & 4 & 4\\ \hline 135-140 & 11 & 15\\ \hline 140-145 & 12 & 27\\ \hline 145-150 & 7 & 34\\ \hline 150-155 & 10 & 44\\ \hline 155-160 & 6 & 50\\ \hline \end{array}

\displaystyle N=\sum f=50;\quad \frac{N}{2}=25
\displaystyle cf\ 27\text{ just greater than }25\text{ and its corresponding interval is }140-150.
\displaystyle \text{So median class is }140-145.
\displaystyle \text{Here }l=140,\ cf=15,\ f=12,\ h=5
\displaystyle \text{Median}=l+\frac{\left(\frac{N}{2}-cf\right)}{f}\times h
\displaystyle \text{Median}=140+\frac{(25-15)}{12}\times5
\displaystyle =140+\frac{10\times5}{12}
\displaystyle =140+4.17=144.17\text{ cm}
\\

\displaystyle \textbf{Question 10. }\text{The table below shows the salaries of }280\text{ persons.} \hspace{0.2cm}\text{[CBSE 2023]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline \text{Salary\ (in\ thousand\ Rs)} & 5\text{-}10 & 10\text{-}15 & 15\text{-}20 & 20\text{-}25 & 25\text{-}30 & 30\text{-}35 & 35\text{-}40 & 40\text{-}45 & 45\text{-}50\\ \hline \text{No.\ of\ Persons} & 49 & 133 & 63 & 15 & 6 & 7 & 4 & 2 & 1\\ \hline \end{array}

\displaystyle \text{Calculate the median salary of the data.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|}\hline \text{Salary (In thousand Rs. )} & \text{No. of Persons} & cf\\ \hline 5-10 & 49 & 49\\ \hline 10-15 & 133 & 182\\ \hline 15-20 & 63 & 245\\ \hline 20-25 & 15 & 260\\ \hline 25-30 & 6 & 266\\ \hline 30-35 & 7 & 273\\ \hline 35-40 & 4 & 277\\ \hline 40-45 & 2 & 279\\ \hline 45-50 & 1 & 280\\ \hline \text{Total} & N=280 & \\ \hline \end{array}

\displaystyle \text{Here,}
\displaystyle N=280
\displaystyle \frac{N}{2}=\frac{280}{2}=140
\displaystyle cf\ 182\text{ is just greater than }140\text{ and its corresponding class is }10-15.
\displaystyle \text{Median class}=10-15
\displaystyle \therefore l=10,\ cf=49,\ h=5,\ f=133
\displaystyle \therefore \text{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)\times h
\displaystyle =10+\left(\frac{140-49}{133}\right)\times5
\displaystyle =10+3.42=13.42
\displaystyle \text{Therefore, median salary}=13.42\text{ (in thousand Rs. ).}
\\

\displaystyle \textbf{Question 11. }\text{The monthly expenditure on milk in }200\text{ families of a Housing} \\ \text{Society is given below:} \hspace{0.2cm}\text{[CBSE 2023]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{Monthly\ Expenditure\ (in\ Rs)} & 1000\text{-}1500 & 1500\text{-}2000 & 2000\text{-}2500 & 2500\text{-}3000 & 3000\text{-}3500 & 3500\text{-}4000 & 4000\text{-}4500 & 4500\text{-}5000\\ \hline \text{Number\ of\ families} & 24 & 40 & 33 & x & 30 & 22 & 16 & 7\\ \hline \end{array}

\displaystyle \text{Find the value of }x\text{ and also, find the median and mean expenditure on milk.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|}\hline \text{Monthly Expenditure (in \text{Rs. })} & \text{Number of families}\\ \hline 1000-1500 & 24\\ \hline 1500-2000 & 40\\ \hline 2000-2500 & 33\\ \hline 2500-3000 & x\\ \hline 3000-3500 & 30\\ \hline 3500-4000 & 22\\ \hline 4000-4500 & 16\\ \hline 4500-5000 & 7\\ \hline \end{array}

\displaystyle \text{Total number of families}=200
\displaystyle \Rightarrow 24+40+33+x+30+22+16+7=200
\displaystyle \Rightarrow x=28

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Monthly Expenditure} & \text{Number of families }(f_i) & \text{Mid-value }(x_i) & d_i=x_i-A & f_id_i & cf\\ \hline 1000-1500 & 24 & 1250 & -1500 & -36000 & 24\\ \hline 1500-2000 & 40 & 1750 & -1000 & -40000 & 64\\ \hline 2000-2500 & 33 & 2250 & -500 & -16500 & 97\\ \hline 2500-3000 & 28 & 2750 & 0 & 0 & 125\\ \hline 3000-3500 & 30 & 3250 & 500 & 15000 & 155\\ \hline 3500-4000 & 22 & 3750 & 1000 & 22000 & 177\\ \hline 4000-4500 & 16 & 4250 & 1500 & 24000 & 193\\ \hline 4500-5000 & 7 & 4750 & 2000 & 14000 & 200\\ \hline & \sum f_i=200 & & & \sum f_id_i=-17500 & \\ \hline \end{array}

\displaystyle \therefore \text{Mean}=A+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =2750-\frac{17500}{200}= \text{Rs. } 2662.50
\displaystyle \frac{N}{2}=100,\ cf\text{ just greater than }100\text{ is }125\text{ and its corresponding class is }2500-3000.
\displaystyle \text{Median class}=2500-3000
\displaystyle l=2500,\ cf=97,\ f=28
\displaystyle h=500,\ N=200
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h
\displaystyle =2500+\frac{100-97}{28}\times500
\displaystyle =2500+53.57= \text{Rs. } 2553.57
\\

\displaystyle \textbf{Question 12. }\text{A survey regarding the heights (in cm) of }50\text{ girls of class X of a school was }
\displaystyle \text{conducted and the following data was obtained:} \hspace{0.2cm}\text{[CBSE 2023(C)]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Height\ (in\ cm)} & 120\text{-}130 & 130\text{-}140 & 140\text{-}150 & 150\text{-}160 & 160\text{-}170 & \text{Total}\\ \hline \text{Number\ of\ girls} & 2 & 8 & 12 & 20 & 8 & 50\\ \hline \end{array}

\displaystyle \text{Find the mean and mode of the above data.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{C.I.} & f_i & x_i & d_i=x_i-A & f_id_i\\ \hline 120-130 & 2 & 125 & -20 & -40\\ \hline 130-140 & 8 & 135 & -10 & -80\\ \hline 140-150 & 12 & 145=A & 0 & 0\\ \hline 150-160 & 20 & 155 & 10 & 200\\ \hline 160-170 & 8 & 165 & 20 & 160\\ \hline \text{Total} & \sum f_i=50 & & & \sum f_id_i=240\\ \hline \end{array}

\displaystyle \text{Mean,}
\displaystyle \bar{x}=A+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =145+\frac{240}{50}
\displaystyle =149.8\text{ cm}
\displaystyle \text{Now, modal class is }150-160,\text{ as maximum frequency is }20.
\displaystyle f_{1}=20,\ f_{0}=12,\ f_{2}=8,\ l=150,\ h=10
\displaystyle \text{Mode}=l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h
\displaystyle =150+\frac{20-12}{2\times20-12-8}\times10
\displaystyle =154\text{ cm}
\\

\displaystyle \textbf{Question 13. }\text{The mean of the following frequency distribution is }25. \\ \text{Find the value of }f. \hspace{0.2cm}\text{[CBSE 2022]}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class} & 0\text{-}10 & 10\text{-}20 & 20\text{-}30 & 30\text{-}40 & 40\text{-}50\\ \hline \text{Frequency} & 5 & 18 & 15 & f & 6\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|c|}\hline \text{C.I.} & f_i & x_i & f_ix_i\\ \hline 0-10 & 5 & 5 & 25\\ \hline 10-20 & 18 & 15 & 270\\ \hline 20-30 & 15 & 25 & 375\\ \hline 30-40 & f & 35 & 35f\\ \hline 40-50 & 6 & 45 & 270\\ \hline & \sum f_i=f+44 & & \sum f_ix_i=940+35f\\ \hline \end{array}

\displaystyle \text{Mean}=25
\displaystyle \frac{\sum f_ix_i}{\sum f_i}=25
\displaystyle \Rightarrow \frac{940+35f}{f+44}=25
\displaystyle \Rightarrow 35f+940=25f+1100
\displaystyle \Rightarrow 10f=160
\displaystyle \Rightarrow f=16
\\

\displaystyle \textbf{Question 14. }\text{Find the mean of the following data using assumed mean method:} \hspace{0.2cm}\text{[CBSE 2022]}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class} & 0\text{-}5 & 5\text{-}10 & 10\text{-}15 & 15\text{-}20 & 20\text{-}25\\ \hline \text{Frequency} & 8 & 7 & 10 & 13 & 12\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Class} & \text{Class marks }(x_i) & \text{Frequency }(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=A & 10 & 0 & 0\\ \hline 15-20 & 17.5 & 13 & 5 & 65\\ \hline 20-25 & 22.5 & 12 & 10 & 120\\ \hline & & \sum f_i=50 & & \sum f_id_i=70\\ \hline \end{array}

\displaystyle \text{Mean}=A+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =12.5+\frac{70}{50}
\displaystyle =13.9
\

\displaystyle \textbf{Question 15. }\text{If mode of the following frequency distribution is }55, \\ \text{then find the value of }x. \hspace{0.2cm}\text{[CBSE 2022]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class} & 0\text{-}15 & 15\text{-}30 & 30\text{-}45 & 45\text{-}60 & 60\text{-}75 & 75\text{-}90\\ \hline \text{Frequency} & 10 & 7 & x & 15 & 10 & 12\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|}\hline \text{C.I.} & \text{Frequency}\\ \hline 0-15 & 10\\ \hline 15-30 & 7\\ \hline 30-45 & x\\ \hline 45-60 & 15\\ \hline 60-75 & 10\\ \hline 75-90 & 12\\ \hline \end{array}

\displaystyle \text{As mode is }55,\text{ so modal class is }45-60.
\displaystyle l=45,\ f_{1}=15,\ f_{0}=x,\ f_{2}=10,\ h=15
\displaystyle \text{Mode}=55
\displaystyle \Rightarrow l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h=55
\displaystyle \Rightarrow 45+\left(\frac{15-x}{30-x-10}\right)\times15=55
\displaystyle \Rightarrow 3(15-x)=2(20-x)
\displaystyle \Rightarrow x=5
\\

\displaystyle \textbf{Question 16. }\text{If the mean of the first }n\text{ natural numbers is }15,\text{ then find }n. \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Mean}=15,\quad \frac{1+2+\cdots+n}{n}=15
\displaystyle \Rightarrow \frac{1+2+3+\cdots+n}{n}=15
\displaystyle \Rightarrow \frac{n(1+n)}{2n}=15
\displaystyle \Rightarrow 1+n=30
\displaystyle \Rightarrow n=29
\\

\displaystyle \textbf{Question 17. }\text{Find the mean of the following distribution:} \hspace{0.2cm}\text{[CBSE 2020]}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class} & 3\text{-}5 & 5\text{-}7 & 7\text{-}9 & 9\text{-}11 & 11\text{-}13\\ \hline \text{Frequency} & 5 & 10 & 10 & 7 & 8\\ \hline \end{array}

\displaystyle \text{Answer:}
\displaystyle \text{Table for mean: Let }A=8,\text{ here }h=2

\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{C.I.} & x & f & u=\frac{x-8}{2} & fu\\ \hline 3-5 & 4 & 5 & -2 & -10\\ \hline 5-7 & 6 & 10 & -1 & -10\\ \hline 7-9 & 8 & 10 & 0 & 0\\ \hline 9-11 & 10 & 7 & 1 & 7\\ \hline 11-13 & 12 & 8 & 2 & 16\\ \hline & & \sum f=40 & & \sum fu=3\\ \hline \end{array}

\displaystyle \text{Mean}=A+\frac{\sum fu}{\sum f}\times h=8+\frac{3}{40}\times2=8+0.15=8.15
\\

\displaystyle \textbf{Question 18. }\text{Find the mode of the following data:} \hspace{0.2cm}\text{[CBSE 2020]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Class} & 0\text{-}20 & 20\text{-}40 & 40\text{-}60 & 60\text{-}80 & 80\text{-}100 & 100\text{-}120 & 120\text{-}140\\ \hline \text{Frequency} & 6 & 8 & 10 & 12 & 6 & 5 & 3\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \text{Table for mode:}

\displaystyle \begin{array}{|c|c|}\hline \text{C.I.} & f\\ \hline 0-20 & 6\\ \hline 20-40 & 8\\ \hline 40-60 & 10\\ \hline 60-80 & 12\\ \hline 80-100 & 6\\ \hline 100-120 & 5\\ \hline 120-140 & 3\\ \hline \end{array}

\displaystyle \text{Maximum frequency }(f_{1})=12
\displaystyle \text{Modal class is }60-80
\displaystyle f_{1}=12,\ f_{0}=10,\ f_{2}=6,\ h=20,\ l=60
\displaystyle \therefore \text{Mode}=l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h
\displaystyle =60+\frac{12-10}{24-10-6}\times20
\displaystyle =60+\frac{2}{8}\times20=65
\\

\displaystyle \textbf{Question 19. }\text{The median of the following data is }525. \text{Find the values of }x\text{ and }y,
\displaystyle \text{if total frequency is }100: \hspace{0.2cm}\text{[CBSE 2020]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline \text{Class} & 0\text{-}100 & 100\text{-}200 & 200\text{-}300 & 300\text{-}400 & 400\text{-}500 & 500\text{-}600 & 600\text{-}700 & 700\text{-}800 & 800\text{-}900 & 900\text{-}1000\\ \hline \text{Frequency} & 2 & 5 & x & 12 & 17 & 20 & y & 9 & 7 & 4\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \text{We have, median}=525

\displaystyle \begin{array}{|c|c|c|}\hline \text{C.I.} & f & c.f\\ \hline 0-100 & 2 & 2\\ \hline 100-200 & 5 & 7\\ \hline 200-300 & x & 7+x\\ \hline 300-400 & 12 & 19+x\\ \hline 400-500 & 17 & 36+x\\ \hline 500-600 & 20 & 56+x\\ \hline 600-700 & y & 56+x+y\\ \hline 700-800 & 9 & 65+x+y\\ \hline 800-900 & 7 & 72+x+y\\ \hline 900-1000 & 4 & 76+x+y\\ \hline & 100 & \\ \hline \end{array}

\displaystyle \text{We have }76+x+y=100
\displaystyle \Rightarrow x+y=24\qquad (i)
\displaystyle \text{As median is }525,\text{ therefore median class is }500-600.
\displaystyle l=500,\ \frac{N}{2}=\frac{100}{2}=50,\ cf=36+x,\ f=20,\ h=100
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h
\displaystyle \Rightarrow 525=500+\frac{50-(36+x)}{20}\times100
\displaystyle \Rightarrow 25=(50-36-x)5
\displaystyle \Rightarrow 25=70-5x
\displaystyle \Rightarrow 5x=45\Rightarrow x=9
\displaystyle \text{From }(i),\text{ we get}
\displaystyle 9+y=24\Rightarrow y=15
\displaystyle \therefore x=9,\ y=15
\\

\displaystyle \textbf{Question 20. }\text{The arithmetic mean of the following frequency distribution is }53.
\displaystyle \text{Find the value of }k. \hspace{0.2cm}\text{[CBSE 2019]}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class} & 0\text{-}20 & 20\text{-}40 & 40\text{-}60 & 60\text{-}80 & 80\text{-}100\\ \hline \text{Frequency} & 12 & 15 & 32 & k & 13\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{C.I.} & f_i & x_i & u_i=\frac{x_i-A}{h} & f_iu_i\\ \hline 0-20 & 12 & 10 & -2 & -24\\ \hline 20-40 & 15 & 30 & -1 & -15\\ \hline 40-60 & 32 & 50=A & 0 & 0\\ \hline 60-80 & k & 70 & 1 & k\\ \hline 80-100 & 13 & 90 & 2 & 26\\ \hline & \sum f_i=k+72 & & & \sum f_iu_i=-13+k\\ \hline \end{array}

\displaystyle \text{We have,}
\displaystyle A=50,\ h=20
\displaystyle \Rightarrow A+\frac{\sum f_iu_i}{\sum f_i}\times h=53
\displaystyle \Rightarrow 50+\frac{k-13}{k+72}\times20=53
\displaystyle \Rightarrow 20k-260=3k+216
\displaystyle \Rightarrow k=\frac{476}{17}=28
\\

\displaystyle \textbf{Question 21. }\text{The table below shows the daily expenditure on food of }25
\displaystyle \text{ households in a locality.} \text{Find the mean daily expenditure on food.} \hspace{0.2cm}\text{[CBSE 2019]}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Daily\ expenditure\ (in\ Rs)} & 100\text{-}150 & 150\text{-}200 & 200\text{-}250 & 250\text{-}300 & 300\text{-}350\\ \hline \text{Number\ of\ households} & 4 & 5 & 12 & 2 & 2\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{C.I.} & f_i & x_i & u_i=\frac{x_i-A}{h} & f_iu_i\\ \hline 100-150 & 4 & 125 & -2 & -8\\ \hline 150-200 & 5 & 175 & -1 & -5\\ \hline 200-250 & 12 & 225=A & 0 & 0\\ \hline 250-300 & 2 & 275 & 1 & 2\\ \hline 300-350 & 2 & 325 & 2 & 4\\ \hline & \sum f_i=25 & & & \sum f_iu_i=-7\\ \hline \end{array}

\displaystyle \text{Mean}=A+\frac{\sum f_iu_i}{\sum f_i}\times h
\displaystyle =225-\frac{7}{25}\times50
\displaystyle =225-14=211
\displaystyle \therefore \text{Mean daily expenditure is Rs. }211.
\\

\displaystyle \textbf{Question 22. }\text{Find the mode of the following frequency distribution.} \hspace{0.2cm}\text{[CBSE 2019]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Class} & 0\text{-}10 & 10\text{-}20 & 20\text{-}30 & 30\text{-}40 & 40\text{-}50 & 50\text{-}60 & 60\text{-}70\\ \hline \text{Frequency} & 8 & 10 & 10 & 16 & 12 & 6 & 7\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|}\hline \text{Classes} & f\\ \hline 0-10 & 8\\ \hline 10-20 & 10\\ \hline 20-30 & 10\\ \hline 30-40 & 16\\ \hline 40-50 & 12\\ \hline 50-60 & 6\\ \hline 60-70 & 7\\ \hline \end{array}

\displaystyle f_{0}=10,\quad f_{1}=16,\quad f_{2}=12
\displaystyle \text{Mode}=l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h
\displaystyle =30+\left(\frac{16-10}{32-10-12}\right)\times10
\displaystyle =30+\frac{6}{10}\times10
\displaystyle =30+6=36
\\

\displaystyle \textbf{Question 23. }\text{The mean of the following distribution is }18. \text{Find the frequency }f
\displaystyle \text{of the class }19\text{-}21. \hspace{0.2cm}\text{[CBSE 2018]}

\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Class} & 11\text{-}13 & 13\text{-}15 & 15\text{-}17 & 17\text{-}19 & 19\text{-}21 & 21\text{-}23 & 23\text{-}25\\ \hline \text{Frequency} & 3 & 6 & 9 & 13 & f & 5 & 4\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Class (C.I.)} & \text{Frequency }(f_i) & x_i & f_ix_i\\ \hline 11-13 & 3 & 12 & 36\\ \hline 13-15 & 6 & 14 & 84\\ \hline 15-17 & 9 & 16 & 144\\ \hline 17-19 & 13 & 18 & 234\\ \hline 19-21 & f & 20 & 20f\\ \hline 21-23 & 5 & 22 & 110\\ \hline 23-25 & 4 & 24 & 96\\ \hline \text{Total} & \sum f_i=40+f & & \sum f_ix_i=704+20f\\ \hline \end{array}

\displaystyle \text{Mean}=\frac{\sum f_ix_i}{\sum f_i}
\displaystyle \Rightarrow 18=\frac{704+20f}{40+f}
\displaystyle \Rightarrow 18(40+f)=704+20f
\displaystyle \Rightarrow 720+18f=704+20f
\displaystyle \Rightarrow 16=2f
\displaystyle \Rightarrow f=\frac{16}{2}=8
\\

\displaystyle \textbf{Question 24. }\text{The average score of boys in the examination of a school is }71\text{ and that}
\displaystyle \text{of the girls is }73. \text{The average score of the school in the examination is }71.8.
\displaystyle \text{Find the ratio of the number of boys to the number of girls who appeared in the examination.} \\ \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}

\displaystyle \text{Let number of boys in the school be }x.
\displaystyle \text{Average score of boys }=71
\displaystyle \text{Total score of boys in the examination of the school }=71\times x=71x
\displaystyle \text{Let number of girls in the school be }y.
\displaystyle \text{Average score of girls }=73
\displaystyle \text{Total score of girls in the examination of the school }=73\times y=73y
\displaystyle \text{Now, average score of the school in examination }=71.8
\displaystyle \therefore \frac{\text{Total score of boys}+\text{Total score of girls}}{\text{Total number of boys and girls}}=71.8
\displaystyle \Rightarrow \frac{71x+73y}{x+y}=71.8
\displaystyle \Rightarrow 71x+73y=71.8x+71.8y
\displaystyle \Rightarrow 1.2y=0.8x
\displaystyle \Rightarrow \frac{1.2}{0.8}=\frac{x}{y}
\displaystyle \Rightarrow x:y=3:2
\\

\displaystyle \textbf{Question 25. }\text{The following data gives the information on the observed life times (in hours) of }
\displaystyle 150\text{ electrical} \text{ components:} \hspace{0.2cm}\text{[CBSE 2016]}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Life\ time\ (in\ hours)} & 0\text{-}20 & 20\text{-}40 & 40\text{-}60 & 60\text{-}80 & 80\text{-}100\\ \hline \text{Frequency} & 15 & 10 & 35 & 50 & 40\\ \hline \end{array}

\displaystyle \text{Find the mode of the distribution.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|}\hline \text{Life time (in hours)} & \text{Frequency}\\ \hline 0-20 & 15\\ \hline 20-40 & 10\\ \hline 40-60 & 35\\ \hline 60-80 & 50\\ \hline 80-100 & 40\\ \hline \end{array}

\displaystyle \text{Modal class is }60-80
\displaystyle l=60,\ f_{0}=35,\ f_{1}=50,\ f_{2}=40,\ h=20
\displaystyle \text{Mode}=l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h
\displaystyle =60+\left(\frac{50-35}{(2\times50)-35-40}\right)\times20
\displaystyle =60+\left(\frac{15}{100-75}\right)\times20
\displaystyle =60+\left(\frac{300}{25}\right)
\displaystyle =60+12=72
\\

\displaystyle \textbf{Question 26. }\text{Find the unknown values in the following table:} \hspace{0.2cm}\text{[CBSE 2016]}

\displaystyle \begin{array}{|c|c|c|}\hline \text{Class interval} & \text{frequency} & \text{Cumulative frequency}\\ \hline 0\text{-}10 & 5 & 5\\ \hline 10\text{-}20 & 7 & x_{1}\\ \hline 20\text{-}30 & x_{2} & 18\\ \hline 30\text{-}40 & 5 & x_{3}\\ \hline 40\text{-}50 & x_{4} & 30\\ \hline \end{array}

\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|}\hline \text{Class interval} & \text{frequency} & \text{Cumulative frequency}\\ \hline 0-10 & 5 & 5\\ \hline 10-20 & 7 & x_{1}\\ \hline 20-30 & x_{2} & 18\\ \hline 30-40 & 5 & x_{3}\\ \hline 40-50 & x_{4} & 30\\ \hline \end{array}

\displaystyle \text{From table:}
\displaystyle x_{1}=7+5=12
\displaystyle 18=x_{1}+x_{2}=12+x_{2}\Rightarrow x_{2}=6
\displaystyle x_{3}=18+5=23
\displaystyle 30=x_{3}+x_{4}\Rightarrow x_{4}=30-x_{3}=30-23=7
\\

\displaystyle \textbf{Question 27. }\text{In a retail market, fruit vendor were selling mangoes in packing boxes.}
\displaystyle \text{These boxes contained varying number of mangoes. The following was the distribution:} \\ \hspace{0.2cm}\text{[CBSE 2016]}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{No.\ of\ mangoes} & 50\text{-}52 & 53\text{-}55 & 56\text{-}58 & 59\text{-}61 & 62\text{-}64\\ \hline \text{No.\ of\ boxes} & 15 & 110 & 135 & 115 & 25\\ \hline \end{array}

\displaystyle \text{Find the mean and median number of mangoes kept in a packing box.}
\displaystyle \text{Answer:}

\displaystyle \text{Converting into continuous frequency distribution,}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{CI} & f_i & x_i & d_i=x_i-A & f_id_i & c.f\\ \hline 49.5-52.5 & 15 & 51 & -6 & -90 & 15\\ \hline 52.5-55.5 & 110 & 54 & -3 & -330 & 125\\ \hline 55.5-58.5 & 135 & 57=A & 0 & 0 & 260\\ \hline 58.5-61.5 & 115 & 60 & 3 & 345 & 375\\ \hline 61.5-64.5 & 25 & 63 & 6 & 150 & 400\\ \hline & \sum f_i=400 & & & \sum f_id_i=75 & \\ \hline \end{array}

\displaystyle \text{Mean}=A+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =57+\frac{75}{400}=57.1875
\displaystyle \text{For median:}
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)\times h
\displaystyle \text{Here }N=400
\displaystyle \Rightarrow \frac{N}{2}=200,\ cf\text{ just greater than }200\text{ is }260\text{ and its corresponding class is }55.5-58.5,
\displaystyle \therefore \text{Median class is }55.5-58.5.
\displaystyle l=55.5,\ f=135,\ cf=125,\ h=3
\displaystyle \text{Median}=55.5+\left(\frac{200-125}{135}\right)\times3=55.5+\frac{75}{135}\times3
\displaystyle =55.5+\frac{225}{135}=55.5+1.666=57.16
\

\displaystyle \textbf{Question 28. }\text{Some students of Class X donated for the welfare of old age persons. Their }
\displaystyle \text{contributions are shown in the following frequency distribution:} \hspace{0.2cm}\text{[CBSE 2016]}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Amount\ (in\ Rs)} & 0\text{-}20 & 20\text{-}40 & 40\text{-}60 & 60\text{-}80 & 80\text{-}100\\ \hline \text{No.\ of\ students} & 5 & 8 & 12 & 11 & 4\\ \hline \end{array}

\displaystyle \text{Find median and mode for their contribution.}
\displaystyle \text{Answer:}

\displaystyle \begin{array}{|c|c|c|}\hline \text{Amount in (Rs. )}\ (\text{Class interval}) & \text{Numbers of students }(f_i) & c.f.\\ \hline 0-20 & 5 & 5\\ \hline 20-40 & 8 & 13\\ \hline 40-60 & 12 & 25\\ \hline 60-80 & 11 & 36\\ \hline 80-100 & 4 & 40\\ \hline & N=\sum f_i=40 & \\ \hline \end{array}

\displaystyle N=40
\displaystyle \frac{N}{2}=20;\ cf\ 25\text{ is just greater than }20\text{ and its corresponding class is }40-60
\displaystyle \therefore \text{Median class}=40-60
\displaystyle l=40,\ f=12,\text{ and }c.f.=13,\ h=60-40=20
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)\times h
\displaystyle =40+\left(\frac{20-13}{12}\right)\times20=40+\frac{7}{12}\times20=\text{Rs. } \ 51.66
\displaystyle \text{For mode, modal class }40-60,\text{ as maximum frequency is }12.
\displaystyle f_{0}=8,\ f_{1}=12,\ f_{2}=11,\ h=60-40=20
\displaystyle \text{Mode}=l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)\times h=40+\left(\frac{12-8}{24-8-11}\right)\times20=40+\frac{80}{5}=\text{Rs. } \ 56
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