\displaystyle \textbf{Question 1: }\text{Following are the lives in hours of }15\text{ pieces of the}
\displaystyle \text{components of aircraft engine. Find the median:}
\displaystyle 715,\ 724,\ 725,\ 710,\ 729,\ 745,\ 694,\ 699,\ 696,\ 712,\ 734,\ 728,\ 716,\ 705,\ 719.
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the observations in ascending order, we get}
\displaystyle 694,\ 696,\ 699,\ 705,\ 710,\ 712,\ 715,\ 716,\ 719,\ 724,\ 725,\ 728,\ 729,\ 734,\ 745
\displaystyle \text{Here, }n=15\text{, which is an odd number.}
\displaystyle \therefore\ \text{Median}=\text{value of }\left(\frac{n+1}{2}\right)^{\text{th}}\text{ observation}
\displaystyle =\text{value of }\left(\frac{15+1}{2}\right)^{\text{th}}\text{ observation}
\displaystyle =\text{value of }8^{\text{th}}\text{ observation}=716
\displaystyle \therefore\ \text{Median}=716\text{ hours.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Calculate the median salary of the following data giving}
\displaystyle \text{salaries of }280\text{ persons:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{Salary (in thousands)} & 5-10 & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 & 40-45 & 45-50 \\ \hline \text{No. of persons} & 49 & 133 & 63 & 15 & 6 & 7 & 4 & 2 & 1 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Preparing the cumulative frequency table, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Salary} & \text{Frequency }(f) & \text{Cumulative frequency }(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 \end{array}
\displaystyle N=280\quad\Rightarrow\quad\frac{N}{2}=\frac{280}{2}=140
\displaystyle \text{The cumulative frequency just greater than }140\text{ is }182.
\displaystyle \therefore\ 10-15\text{ is the median class.}
\displaystyle l=10,\quad f=133,\quad F=49,\quad h=5
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =10+\frac{140-49}{133}\times5
\displaystyle =10+\frac{455}{133}=13.42\text{ (approximately)}
\displaystyle \therefore\ \text{Median salary}=Rs.\ 13.42\text{ thousand.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Calculate the missing frequency from the following distribution,}
\displaystyle \text{it being given that the median of the distribution is }24.
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Age in years} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{No. of persons} & 5 & 25 & x & 18 & 7 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Let the missing frequency be }x.
\displaystyle \begin{array}{|c|c|c|} \hline \text{Age in years} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 0-10 & 5 & 5 \\ \hline 10-20 & 25 & 30 \\ \hline 20-30 & x & 30+x \\ \hline 30-40 & 18 & 48+x \\ \hline 40-50 & 7 & 55+x \\ \hline \end{array}
\displaystyle N=55+x
\displaystyle \text{Since the median is }24,\text{ the median class is }20-30.
\displaystyle l=20,\quad f=x,\quad F=30,\quad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle 24=20+\frac{\frac{55+x}{2}-30}{x}\times10
\displaystyle 4=\frac{55+x-60}{2x}\times10
\displaystyle 4=\frac{5(x-5)}{x}
\displaystyle 4x=5x-25
\displaystyle \therefore\ x=25
\displaystyle \therefore\ \text{The missing frequency is }25.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The following table gives the distribution of the life time of}
\displaystyle 400\text{ neon lamps. Find the median life.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Life time (in hours)} & 1500 & 2000 & 2500 & 3000 & 3500 & 4000 & 4500 \\ & -2000 & -2500 & -3000 & -3500 & -4000 & -4500 & -5000 \\ \hline \text{Number of lamps} & 14 & 56 & 60 & 86 & 74 & 62 & 48 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Preparing the cumulative frequency table, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Life time (in hours)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 1500-2000 & 14 & 14 \\ \hline 2000-2500 & 56 & 70 \\ \hline 2500-3000 & 60 & 130 \\ \hline 3000-3500 & 86 & 216 \\ \hline 3500-4000 & 74 & 290 \\ \hline 4000-4500 & 62 & 352 \\ \hline 4500-5000 & 48 & 400 \\ \hline \end{array}
\displaystyle N=400\quad\Rightarrow\quad\frac{N}{2}=\frac{400}{2}=200
\displaystyle \text{The cumulative frequency just greater than }200\text{ is }216.
\displaystyle \therefore\ 3000-3500\text{ is the median class.}
\displaystyle l=3000,\quad f=86,\quad F=130,\quad h=500
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =3000+\frac{200-130}{86}\times500
\displaystyle =3000+\frac{35000}{86}=3406.98\text{ (approximately)}
\displaystyle \therefore\ \text{Median life}=3406.98\text{ hours.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The distribution below gives the weight of }30\text{ students in a}
\displaystyle \text{class. Find the median weight of students.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Weight (in kg)} & 40-45 & 45-50 & 50-55 & 55-60 & 60-65 & 65-70 & 70-75 \\ \hline \text{No. of students} & 2 & 3 & 8 & 6 & 6 & 3 & 2 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Preparing the cumulative frequency table, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Weight (in kg)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 40-45 & 2 & 2 \\ \hline 45-50 & 3 & 5 \\ \hline 50-55 & 8 & 13 \\ \hline 55-60 & 6 & 19 \\ \hline 60-65 & 6 & 25 \\ \hline 65-70 & 3 & 28 \\ \hline 70-75 & 2 & 30 \\ \hline \end{array}
\displaystyle N=30\quad\Rightarrow\quad\frac{N}{2}=\frac{30}{2}=15
\displaystyle \text{The cumulative frequency just greater than }15\text{ is }19.
\displaystyle \therefore\ 55-60\text{ is the median class.}
\displaystyle l=55,\quad f=6,\quad F=13,\quad h=5
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =55+\frac{15-13}{6}\times5
\displaystyle =55+\frac{10}{6}=56.67\text{ (approximately)}
\displaystyle \therefore\ \text{Median weight}=56.67\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A survey regarding the height (in cm) of }51\text{ girls of class X of a}
\displaystyle \text{school was conducted and the following data was obtained. Find the median height.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Less than} & 140 & 145 & 150 & 155 & 160 & 165 \\ \hline \text{No. of girls} & 4 & 11 & 29 & 40 & 46 & 51 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Converting the cumulative frequencies into ordinary frequencies, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Height (in cm)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 135-140 & 4 & 4 \\ \hline 140-145 & 11-4=7 & 11 \\ \hline 145-150 & 29-11=18 & 29 \\ \hline 150-155 & 40-29=11 & 40 \\ \hline 155-160 & 46-40=6 & 46 \\ \hline 160-165 & 51-46=5 & 51 \\ \hline \end{array}
\displaystyle N=51\quad\Rightarrow\quad\frac{N}{2}=\frac{51}{2}=25.5
\displaystyle \text{The cumulative frequency just greater than }25.5\text{ is }29.
\displaystyle \therefore\ 145-150\text{ is the median class.}
\displaystyle l=145,\quad f=18,\quad F=11,\quad h=5
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =145+\frac{25.5-11}{18}\times5
\displaystyle =145+\frac{14.5}{18}\times5=149.03\text{ (approximately)}
\displaystyle \therefore\ \text{Median height}=149.03\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A life insurance agent found the following data for distribution of}
\displaystyle \text{ages of }100\text{ policy holders. Calculate the median age, if policies are only given}
\displaystyle \text{to persons having age }18\text{ years onwards but less than }60\text{ years.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{Age (in years) below} & 20 & 25 & 30 & 35 & 40 & 45 & 50 & 55 & 60 \\ \hline \text{Number of policy holders} & 2 & 6 & 24 & 45 & 78 & 89 & 92 & 98 & 100 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Converting the cumulative frequencies into ordinary frequencies, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Age (in years)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 18-20 & 2 & 2 \\ \hline 20-25 & 6-2=4 & 6 \\ \hline 25-30 & 24-6=18 & 24 \\ \hline 30-35 & 45-24=21 & 45 \\ \hline 35-40 & 78-45=33 & 78 \\ \hline 40-45 & 89-78=11 & 89 \\ \hline 45-50 & 92-89=3 & 92 \\ \hline 50-55 & 98-92=6 & 98 \\ \hline 55-60 & 100-98=2 & 100 \\ \hline \end{array}
\displaystyle N=100\quad\Rightarrow\quad\frac{N}{2}=\frac{100}{2}=50
\displaystyle \text{The cumulative frequency just greater than }50\text{ is }78.
\displaystyle \therefore\ 35-40\text{ is the median class.}
\displaystyle l=35,\quad f=33,\quad F=45,\quad h=5
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =35+\frac{50-45}{33}\times5
\displaystyle =35+\frac{25}{33}=35.76\text{ (approximately)}
\displaystyle \therefore\ \text{Median age}=35.76\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The lengths of }40\text{ leaves of a plant are measured correct to the}
\displaystyle \text{nearest millimetre, and the data obtained is represented in the following table.}
\displaystyle \text{Find the mean length of leaf.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Length (in mm)} & 118-126 & 127-135 & 136-144 & 145-153 & 154-162 & 163-171 & 172-180 \\ \hline \text{No. of leaves} & 3 & 5 & 9 & 12 & 5 & 4 & 2 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks }x_i\text{ and }f_i x_i\text{ are calculated as follows:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Length (in mm)} & f_i & x_i & f_i x_i \\ \hline 118-126 & 3 & 122 & 366 \\ \hline 127-135 & 5 & 131 & 655 \\ \hline 136-144 & 9 & 140 & 1260 \\ \hline 145-153 & 12 & 149 & 1788 \\ \hline 154-162 & 5 & 158 & 790 \\ \hline 163-171 & 4 & 167 & 668 \\ \hline 172-180 & 2 & 176 & 352 \\ \hline \text{Total} & 40 & & 5879 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}
\displaystyle =\frac{5879}{40}=146.975
\displaystyle \therefore\ \text{Mean length of a leaf}=146.975\text{ mm}\approx146.98\text{ mm.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The frequency distribution given below shows the weight of }40
\displaystyle \text{students of a class. Find the median weight of the students.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Weight (in kg)} & 40-45 & 45-50 & 50-55 & 55-60 & 60-65 & 65-70 \\ \hline \text{No. of students} & 9 & 5 & 8 & 9 & 6 & 3 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Preparing the cumulative frequency table, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Weight (in kg)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 40-45 & 9 & 9 \\ \hline 45-50 & 5 & 14 \\ \hline 50-55 & 8 & 22 \\ \hline 55-60 & 9 & 31 \\ \hline 60-65 & 6 & 37 \\ \hline 65-70 & 3 & 40 \\ \hline \end{array}
\displaystyle N=40\quad\Rightarrow\quad\frac{N}{2}=\frac{40}{2}=20
\displaystyle \text{The cumulative frequency just greater than }20\text{ is }22.
\displaystyle \therefore\ 50-55\text{ is the median class.}
\displaystyle l=50,\quad f=8,\quad F=14,\quad h=5
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =50+\frac{20-14}{8}\times5
\displaystyle =50+\frac{30}{8}=53.75
\displaystyle \therefore\ \text{Median weight}=53.75\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{An incomplete distribution is given below:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Variable} & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \text{Frequency} & 12 & 30 & x & 65 & y & 25 & 18 \\ \hline \end{array}
\displaystyle \text{You are given that the median value is }46\text{ and the total number of items is }230.
\displaystyle \text{(i) Using the median formula fill up missing frequencies.}
\displaystyle \text{(ii) Calculate the AM of the completed distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the missing frequencies be }x\text{ and }y.
\displaystyle 12+30+x+65+y+25+18=230
\displaystyle \Rightarrow\ x+y=80\qquad ...(1)
\displaystyle \text{Since the median is }46,\text{ the median class is }40-50.
\displaystyle l=40,\quad f=65,\quad F=12+30+x=42+x,\quad h=10,\quad N=230
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle 46=40+\frac{115-(42+x)}{65}\times10
\displaystyle 6=\frac{73-x}{65}\times10
\displaystyle 390=730-10x
\displaystyle 10x=340\quad\Rightarrow\quad x=34
\displaystyle \text{Using }(1),\quad 34+y=80\quad\Rightarrow\quad y=46
\displaystyle \therefore\ \text{The missing frequencies are }34\text{ and }46.
\displaystyle \text{(ii) The completed distribution and class marks are:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Class interval} & f_i & x_i & f_i x_i \\ \hline 10-20 & 12 & 15 & 180 \\ \hline 20-30 & 30 & 25 & 750 \\ \hline 30-40 & 34 & 35 & 1190 \\ \hline 40-50 & 65 & 45 & 2925 \\ \hline 50-60 & 46 & 55 & 2530 \\ \hline 60-70 & 25 & 65 & 1625 \\ \hline 70-80 & 18 & 75 & 1350 \\ \hline \text{Total} & 230 & & 10550 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}
\displaystyle =\frac{10550}{230}=45.87\text{ (approximately)}
\displaystyle \therefore\ \text{Arithmetic mean}=45.87.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The median of the distribution given below is }14.4.
\displaystyle \text{Find the values of }x\text{ and }y,\text{ if the total frequency is }20.
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Class interval} & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 \\ \hline \text{Frequency} & 4 & x & 5 & y & 1 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Since the total frequency is }20,
\displaystyle 4+x+5+y+1=20
\displaystyle \Rightarrow x+y=10\qquad ...(1)
\displaystyle \text{Since the median is }14.4,\text{ the median class is }12-18.
\displaystyle l=12,\quad f=5,\quad F=4+x,\quad h=6,\quad N=20
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle 14.4=12+\frac{10-(4+x)}{5}\times6
\displaystyle 2.4=\frac{6-x}{5}\times6
\displaystyle 12=6(6-x)
\displaystyle 2=6-x\quad\Rightarrow\quad x=4
\displaystyle \text{Using }(1),\quad 4+y=10\quad\Rightarrow\quad y=6
\displaystyle \therefore\ x=4\text{ and }y=6.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The median of the following data is }50.\text{ Find the values of }p
\displaystyle \text{and }q,\text{ if the sum of all the frequencies is }90.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Marks} & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 & 80-90 \\ \hline \text{Frequency} & p & 15 & 25 & 20 & q & 8 & 10 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Since the sum of all the frequencies is }90,
\displaystyle p+15+25+20+q+8+10=90
\displaystyle \Rightarrow p+q=12\qquad ...(1)
\displaystyle \text{Here, }N=90\quad\Rightarrow\quad\frac{N}{2}=45.
\displaystyle \text{Taking }50-60\text{ as the median class, we have}
\displaystyle l=50,\quad f=20,\quad F=p+15+25=p+40,\quad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle 50=50+\frac{45-(p+40)}{20}\times10
\displaystyle 0=\frac{5-p}{20}\times10
\displaystyle \therefore\ p=5
\displaystyle \text{Using }(1),\quad 5+q=12\quad\Rightarrow\quad q=7
\displaystyle \therefore\ p=5\text{ and }q=7.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Height of }50\text{ students of class X of a school are recorded}
\displaystyle \text{and the following data is obtained. Find the median height.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Height (in cm)} & 130-135 & 135-140 & 140-145 & 145-150 & 150-155 & 155-160 \\ \hline \text{No. of students} & 4 & 11 & 12 & 7 & 10 & 6 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Preparing the cumulative frequency table, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Height (in cm)} & \text{Frequency }(f) & \text{Cumulative frequency }(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=50\quad\Rightarrow\quad\frac{N}{2}=\frac{50}{2}=25
\displaystyle \text{The cumulative frequency just greater than }25\text{ is }27.
\displaystyle \therefore\ 140-145\text{ is the median class.}
\displaystyle l=140,\quad f=12,\quad F=15,\quad h=5
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =140+\frac{25-15}{12}\times5
\displaystyle =140+\frac{50}{12}=144.17\text{ (approximately)}
\displaystyle \therefore\ \text{Median height}=144.17\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The monthly expenditures on milk in }200\text{ families of a housing}
\displaystyle \text{society are given below. Find the value of }x\text{ and also find the median and}
\displaystyle \text{mean expenditure.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Monthly expenditure in Rs.} & 1000 & 1500 & 2000 & 2500 & 3000 & 3500 & 4000 & 4500 \\ & -1500 & -2000 & -2500 & -3000 & -3500 & -4000 & -4500 & -5000 \\ \hline \text{Number of families} & 24 & 40 & 33 & x & 30 & 22 & 16 & 7 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the total number of families is }200,
\displaystyle 24+40+33+x+30+22+16+7=200
\displaystyle 172+x=200\quad\Rightarrow\quad x=28
\displaystyle \therefore\ \text{The missing frequency is }28.
\displaystyle \text{The cumulative frequencies of the completed distribution are:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Monthly expenditure} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 1000-1500 & 24 & 24 \\ \hline 1500-2000 & 40 & 64 \\ \hline 2000-2500 & 33 & 97 \\ \hline 2500-3000 & 28 & 125 \\ \hline 3000-3500 & 30 & 155 \\ \hline 3500-4000 & 22 & 177 \\ \hline 4000-4500 & 16 & 193 \\ \hline 4500-5000 & 7 & 200 \\ \hline \end{array}
\displaystyle N=200\quad\Rightarrow\quad\frac{N}{2}=\frac{200}{2}=100
\displaystyle \text{The cumulative frequency just greater than }100\text{ is }125.
\displaystyle \therefore\ 2500-3000\text{ is the median class.}
\displaystyle l=2500,\quad f=28,\quad F=97,\quad h=500
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =2500+\frac{100-97}{28}\times500
\displaystyle =2500+\frac{1500}{28}=2553.57\text{ (approximately)}
\displaystyle \therefore\ \text{Median expenditure}=Rs.\ 2553.57.
\displaystyle \text{For calculating the mean, we use the class marks }x_i.
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Monthly expenditure} & f_i & x_i & f_i x_i \\ \hline 1000-1500 & 24 & 1250 & 30000 \\ \hline 1500-2000 & 40 & 1750 & 70000 \\ \hline 2000-2500 & 33 & 2250 & 74250 \\ \hline 2500-3000 & 28 & 2750 & 77000 \\ \hline 3000-3500 & 30 & 3250 & 97500 \\ \hline 3500-4000 & 22 & 3750 & 82500 \\ \hline 4000-4500 & 16 & 4250 & 68000 \\ \hline 4500-5000 & 7 & 4750 & 33250 \\ \hline \text{Total} & 200 & & 532500 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}
\displaystyle =\frac{532500}{200}=2662.5
\displaystyle \therefore\ \text{Mean expenditure}=Rs.\ 2662.50.
\displaystyle \\

\displaystyle \textbf{Question 15: }\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 below.}
\displaystyle \text{Find the mean and median of the following 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 (Periods)} & 7 & 14 & 13 & 12 & 20 & 11 & 15 & 8 \\ \hline \end{array}
\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{Number of cars} & f_i & x_i & f_i x_i \\ \hline 0-10 & 7 & 5 & 35 \\ \hline 10-20 & 14 & 15 & 210 \\ \hline 20-30 & 13 & 25 & 325 \\ \hline 30-40 & 12 & 35 & 420 \\ \hline 40-50 & 20 & 45 & 900 \\ \hline 50-60 & 11 & 55 & 605 \\ \hline 60-70 & 15 & 65 & 975 \\ \hline 70-80 & 8 & 75 & 600 \\ \hline \text{Total} & 100 & & 4070 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{4070}{100}=40.7
\displaystyle \therefore\ \text{Mean number of cars}=40.7.
\displaystyle \text{Now, preparing the cumulative frequency table for finding the median, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Number of cars} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 0-10 & 7 & 7 \\ \hline 10-20 & 14 & 21 \\ \hline 20-30 & 13 & 34 \\ \hline 30-40 & 12 & 46 \\ \hline 40-50 & 20 & 66 \\ \hline 50-60 & 11 & 77 \\ \hline 60-70 & 15 & 92 \\ \hline 70-80 & 8 & 100 \\ \hline \end{array}
\displaystyle N=100\quad\Rightarrow\quad\frac{N}{2}=\frac{100}{2}=50
\displaystyle \text{The cumulative frequency just greater than }50\text{ is }66.
\displaystyle \therefore\ 40-50\text{ is the median class.}
\displaystyle l=40,\quad f=20,\quad F=46,\quad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =40+\frac{50-46}{20}\times10
\displaystyle =40+2=42
\displaystyle \therefore\ \text{Median number of cars}=42.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The lengths of }40\text{ leaves of a plant are measured correct to the}
\displaystyle \text{nearest millimetre, and the data obtained is represented in the following table.}
\displaystyle \text{Find the median length of the leaves.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Length (in mm)} & 118-126 & 127-135 & 136-144 & 145-153 & 154-162 & 163-171 & 172-180 \\ \hline \text{Number of leaves} & 3 & 5 & 9 & 12 & 5 & 4 & 2 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the lengths are measured correct to the nearest millimetre, the classes}
\displaystyle \text{are converted into continuous class intervals by subtracting and adding }0.5.
\displaystyle \begin{array}{|c|c|c|} \hline \text{Length (in mm)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 117.5-126.5 & 3 & 3 \\ \hline 126.5-135.5 & 5 & 8 \\ \hline 135.5-144.5 & 9 & 17 \\ \hline 144.5-153.5 & 12 & 29 \\ \hline 153.5-162.5 & 5 & 34 \\ \hline 162.5-171.5 & 4 & 38 \\ \hline 171.5-180.5 & 2 & 40 \\ \hline \end{array}
\displaystyle N=40\quad\Rightarrow\quad\frac{N}{2}=\frac{40}{2}=20
\displaystyle \text{The cumulative frequency just greater than }20\text{ is }29.
\displaystyle \therefore\ 144.5-153.5\text{ is the median class.}
\displaystyle l=144.5,\quad f=12,\quad F=17,\quad h=9
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =144.5+\frac{20-17}{12}\times9
\displaystyle =144.5+\frac{27}{12}=146.75
\displaystyle \therefore\ \text{Median length of the leaves}=146.75\text{ mm.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the mean and median for the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Classes} & 5-15 & 15-25 & 25-35 & 35-45 & 45-55 & 55-65 & 65-75 \\ \hline \text{Frequency} & 2 & 3 & 5 & 7 & 4 & 2 & 2 \\ \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 5-15 & 2 & 10 & 20 \\ \hline 15-25 & 3 & 20 & 60 \\ \hline 25-35 & 5 & 30 & 150 \\ \hline 35-45 & 7 & 40 & 280 \\ \hline 45-55 & 4 & 50 & 200 \\ \hline 55-65 & 2 & 60 & 120 \\ \hline 65-75 & 2 & 70 & 140 \\ \hline \text{Total} & 25 & & 970 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{970}{25}=38.8
\displaystyle \therefore\ \text{Mean}=38.8.
\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 5-15 & 2 & 2 \\ \hline 15-25 & 3 & 5 \\ \hline 25-35 & 5 & 10 \\ \hline 35-45 & 7 & 17 \\ \hline 45-55 & 4 & 21 \\ \hline 55-65 & 2 & 23 \\ \hline 65-75 & 2 & 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 }17.
\displaystyle \therefore\ 35-45\text{ is the median class.}
\displaystyle l=35,\quad f=7,\quad F=10,\quad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =35+\frac{12.5-10}{7}\times10
\displaystyle =35+\frac{25}{7}=38.57\text{ (approximately)}
\displaystyle \therefore\ \text{Median}=38.57.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Medical check-up was carried out for }35\text{ students of a class}
\displaystyle \text{and their weights were recorded as follows. Find the difference between the mean}
\displaystyle \text{weight and the median weight.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Weight (in kg)} & 38-40 & 40-42 & 42-44 & 44-46 & 46-48 & 48-50 & 50-52 \\ \hline \text{Number of students} & 3 & 2 & 4 & 5 & 14 & 4 & 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{Weight (in kg)} & f_i & x_i & f_i x_i \\ \hline 38-40 & 3 & 39 & 117 \\ \hline 40-42 & 2 & 41 & 82 \\ \hline 42-44 & 4 & 43 & 172 \\ \hline 44-46 & 5 & 45 & 225 \\ \hline 46-48 & 14 & 47 & 658 \\ \hline 48-50 & 4 & 49 & 196 \\ \hline 50-52 & 3 & 51 & 153 \\ \hline \text{Total} & 35 & & 1603 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{1603}{35}=45.8\text{ kg}
\displaystyle \text{Now, preparing the cumulative frequency table for finding the median, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Weight (in kg)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 38-40 & 3 & 3 \\ \hline 40-42 & 2 & 5 \\ \hline 42-44 & 4 & 9 \\ \hline 44-46 & 5 & 14 \\ \hline 46-48 & 14 & 28 \\ \hline 48-50 & 4 & 32 \\ \hline 50-52 & 3 & 35 \\ \hline \end{array}
\displaystyle N=35\quad\Rightarrow\quad\frac{N}{2}=\frac{35}{2}=17.5
\displaystyle \text{The cumulative frequency just greater than }17.5\text{ is }28.
\displaystyle \therefore\ 46-48\text{ is the median class.}
\displaystyle l=46,\quad f=14,\quad F=14,\quad h=2
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =46+\frac{17.5-14}{14}\times2
\displaystyle =46+\frac{7}{14}=46.5\text{ kg}
\displaystyle \therefore\ \text{Difference}=46.5-45.8=0.7\text{ kg}
\displaystyle \therefore\ \text{Difference between the mean and median weights}=0.7\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{During a medical checkup, height of }35\text{ students of a class}
\displaystyle \text{were recorded as follows. Find the difference between the mean height and}
\displaystyle \text{median height.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Height (in cm)} & 90-100 & 100-110 & 110-120 & 120-130 & 130-140 & 140-150 \\ \hline \text{Number of students} & 3 & 2 & 4 & 5 & 14 & 7 \\ \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{Height (in cm)} & f_i & x_i & f_i x_i \\ \hline 90-100 & 3 & 95 & 285 \\ \hline 100-110 & 2 & 105 & 210 \\ \hline 110-120 & 4 & 115 & 460 \\ \hline 120-130 & 5 & 125 & 625 \\ \hline 130-140 & 14 & 135 & 1890 \\ \hline 140-150 & 7 & 145 & 1015 \\ \hline \text{Total} & 35 & & 4485 \\ \hline \end{array}
\displaystyle \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{4485}{35}=128.14\text{ cm (approximately)}
\displaystyle \text{Now, preparing the cumulative frequency table for finding the median, we get}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Height (in cm)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 90-100 & 3 & 3 \\ \hline 100-110 & 2 & 5 \\ \hline 110-120 & 4 & 9 \\ \hline 120-130 & 5 & 14 \\ \hline 130-140 & 14 & 28 \\ \hline 140-150 & 7 & 35 \\ \hline \end{array}
\displaystyle N=35\quad\Rightarrow\quad\frac{N}{2}=\frac{35}{2}=17.5
\displaystyle \text{The cumulative frequency just greater than }17.5\text{ is }28.
\displaystyle \therefore\ 130-140\text{ is the median class.}
\displaystyle l=130,\quad f=14,\quad F=14,\quad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =130+\frac{17.5-14}{14}\times10
\displaystyle =130+\frac{35}{14}=132.5\text{ cm}
\displaystyle \therefore\ \text{Difference}=132.5-128.14=4.36\text{ cm (approximately)}
\displaystyle \therefore\ \text{Difference between mean and median heights}=4.36\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Following data shows the number of family members living in}
\displaystyle \text{different bungalows of a locality. If the median number of members is found to be }5,
\displaystyle \text{find the values of }p\text{ and }q.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Number of members} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 & \text{Total} \\ \hline \text{Number of bungalows} & 10 & p & 60 & q & 5 & 120 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the total number of bungalows is }120,
\displaystyle 10+p+60+q+5=120
\displaystyle \Rightarrow\ p+q=45\qquad ...(1)
\displaystyle \text{Since the median is }5,\text{ the median class is }4-6.
\displaystyle l=4,\quad f=60,\quad F=10+p,\quad h=2,\quad N=120
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle 5=4+\frac{60-(10+p)}{60}\times2
\displaystyle 1=\frac{50-p}{30}
\displaystyle 30=50-p\quad\Rightarrow\quad p=20
\displaystyle \text{Using }(1),\quad 20+q=45\quad\Rightarrow\quad q=25
\displaystyle \therefore\ p=20\text{ and }q=25.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The population of lions was noted in different regions across}
\displaystyle \text{the world in the following table. If the median of the given data is }525,\text{ find}
\displaystyle \text{the values of }x\text{ and }y.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Number of lions} & 0 & 100 & 200 & 300 & 400 & 500 & 600 & 700 & 800 & 900 & \text{Total} \\ & -100 & -200 & -300 & -400 & -500 & -600 & -700 & -800 & -900 & -1000 & \text{Total} \\ \hline \text{Number of regions} & 2 & 5 & 9 & 12 & x & 20 & 15 & 9 & y & 2 & 100 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Since the total frequency is }100,
\displaystyle 2+5+9+12+x+20+15+9+y+2=100
\displaystyle \Rightarrow\ x+y=26\qquad ...(1)
\displaystyle \text{Since the median is }525,\text{ the median class is }500-600.
\displaystyle l=500,\quad f=20,\quad F=28+x,\quad h=100,\quad N=100
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle 525=500+\frac{50-(28+x)}{20}\times100
\displaystyle 25=5(22-x)
\displaystyle 25=110-5x
\displaystyle 5x=85\quad\Rightarrow\quad x=17
\displaystyle \text{Using }(1),\quad 17+y=26\quad\Rightarrow\quad y=9
\displaystyle \therefore\ x=17\text{ and }y=9.
\displaystyle \\

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