\displaystyle \text{Exercise - 19(A)}

\displaystyle \textbf{Question 1: }\text{The weights of }7\text{ boys in a group are }52\text{ kg},57\text{ kg},55\text{ kg},60\text{ kg},
\displaystyle \text{54 kg},59\text{ kg and }55\text{ kg. Find the mean weight of the group.}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }n=7
\displaystyle \text{Sum of weights}=52+57+55+60+54+59+55
\displaystyle =392\text{ kg}
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{392}{7}=56\text{ kg}
\displaystyle \therefore \text{The mean weight of the group is }56\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The marks obtained by }7\text{ students in a group are }340,180,260,164,
\displaystyle 56,275\text{ and }307\text{ respectively. Find the mean marks per student.}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }n=7
\displaystyle \text{Sum of marks}=340+180+260+164+56+275+307
\displaystyle =1582
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{1582}{7}=226\text{ marks}
\displaystyle \therefore \text{The mean marks per student are }226\text{ marks.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the mean of the first six prime numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{The first six prime numbers are }2,3,5,7,11,13.
\displaystyle n=6
\displaystyle \text{Sum of the prime numbers}=2+3+5+7+11+13=41
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{41}{6}=6.83\text{ (approx.).}
\displaystyle \therefore \text{The mean of the first six prime numbers is }6.83\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the mean of the first }10\text{ odd numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{The first }10\text{ odd numbers are }1,3,5,7,9,11,13,15,17,19.
\displaystyle n=10
\displaystyle \text{Sum of the odd numbers}=1+3+5+7+9+11+13+15+17+19=100
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{100}{10}=10
\displaystyle \therefore \text{The mean of the first }10\text{ odd numbers is }10.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the mean of all the factors of }20.
\displaystyle \text{Answer:}
\displaystyle \text{The factors of }20\text{ are }1,2,4,5,10,20.
\displaystyle n=6
\displaystyle \text{Sum of the factors}=1+2+4+5+10+20=42
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{42}{6}=7
\displaystyle \therefore \text{The mean of all the factors of }20\text{ is }7.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The daily minimum temperatures (in )} \text{F recorded at a place during}
\displaystyle \text{a week are shown below. Find the mean temperature.}
\displaystyle \text{Monday }35.2,\ \text{Tuesday }31.1,\ \text{Wednesday }27.6,\ \text{Thursday }31.8,
\displaystyle \text{Friday }29.3,\ \text{Saturday }23.8
\displaystyle \text{Answer:}
\displaystyle n=6
\displaystyle \text{Sum of temperatures}=35.2+31.1+27.6+31.8+29.3+23.8=178.8^\circ\mathrm{F}
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{178.8}{6}=29.8^\circ\mathrm{F}
\displaystyle \therefore \text{The mean temperature is }29.8^\circ\mathrm{F}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If the mean of }6,8,9,x,13\text{ is }10,\text{ find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{Mean}=10,\qquad n=5
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle 10=\frac{6+8+9+x+13}{5}
\displaystyle 10=\frac{36+x}{5}
\displaystyle 50=36+x
\displaystyle x=50-36=14
\displaystyle \therefore x=14.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The mean height of }6\text{ girls is }148\text{ cm. If the individual heights of}
\displaystyle \text{five of them are }142\text{ cm},154\text{ cm},146\text{ cm},145\text{ cm and }150\text{ cm, find the}
\displaystyle \text{height of the sixth girl.}
\displaystyle \text{Answer:}
\displaystyle \text{Mean height of }6\text{ girls}=148\text{ cm}
\displaystyle \text{Total height of }6\text{ girls}=148\times6=888\text{ cm}
\displaystyle \text{Total height of the given }5\text{ girls}=142+154+146+145+150
\displaystyle =737\text{ cm}
\displaystyle \text{Height of the sixth girl}=888-737=151\text{ cm}
\displaystyle \therefore \text{The height of the sixth girl is }151\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The following table shows the weights (in kg) of }15\text{ workers in a}
\displaystyle \text{factory. Calculate the mean weight.}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Weight (kg)}&60&63&66&72&75\\\hline \text{Number of workers}&4&5&3&1&2\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Weight }(x)&\text{Number of workers }(f)&fx\\\hline 60&4&240\\63&5&315\\66&3&198\\72&1&72\\75&2&150\\\hline \text{Total}&15&975\\\hline\end{array}
\displaystyle \sum f=15,\qquad \sum fx=975
\displaystyle \text{Mean }(\overline{x})=\frac{\sum fx}{\sum f}
\displaystyle =\frac{975}{15}=65\text{ kg}
\displaystyle \therefore \text{The mean weight of the workers is }65\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the mean daily wages of }60\text{ workers in a factory from the}
\displaystyle \text{following data.}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Daily wages (Rs.)}&90&110&120&130&150\\\hline \text{No. of workers}&12&14&13&11&10\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Daily wages }(x)&\text{No. of workers }(f)&fx\\\hline 90&12&1080\\110&14&1540\\120&13&1560\\130&11&1430\\150&10&1500\\\hline \text{Total}&60&7110\\\hline\end{array}
\displaystyle \sum f=60,\qquad \sum fx=7110
\displaystyle \text{Mean wages}=\frac{\sum fx}{\sum f}
\displaystyle =\frac{7110}{60}=\frac{711}{6}=118.50
\displaystyle \therefore \text{The mean daily wage of the workers is Rs. }118.50.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The heights (in cm) of }90\text{ plants in a garden are given below.}
\displaystyle \text{Find the mean height of the plants.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Height (cm)}&58&60&62&64&66&74\\\hline \text{Number of plants}&20&25&15&8&12&10\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Height }(x)&\text{Number of plants }(f)&fx\\\hline 58&20&1160\\60&25&1500\\62&15&930\\64&8&512\\66&12&792\\74&10&740\\\hline \text{Total}&90&5634\\\hline\end{array}
\displaystyle \sum f=90,\qquad \sum fx=5634
\displaystyle \text{Mean }(\overline{x})=\frac{\sum fx}{\sum f}
\displaystyle =\frac{5634}{90}=62.6\text{ cm}
\displaystyle \therefore \text{The mean height of the plants is }62.6\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The mean of the following data is }21.6.\text{ Find the value of }p.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline x_i&6&12&18&24&30&36\\\hline f_i&5&4&p&6&4&6\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\\hline 6&5&30\\12&4&48\\18&p&18p\\24&6&144\\30&4&120\\36&6&216\\\hline \text{Total}&25+p&558+18p\\\hline\end{array}
\displaystyle \text{Given mean}=21.6
\displaystyle \therefore 21.6=\frac{558+18p}{25+p}
\displaystyle 21.6(25+p)=558+18p
\displaystyle 540+21.6p=558+18p
\displaystyle 3.6p=18
\displaystyle p=\frac{18}{3.6}=5
\displaystyle \therefore p=5.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If the mean of the following data is }18.75,\text{ find the value of }p.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline x_i&10&15&p&25&30\\\hline f_i&5&10&7&8&2\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline x_i&f_i&f_ix_i\\\hline 10&5&50\\15&10&150\\p&7&7p\\25&8&200\\30&2&60\\\hline \text{Total}&32&460+7p\\\hline\end{array}
\displaystyle \text{Given mean}=18.75
\displaystyle \therefore 18.75=\frac{460+7p}{32}
\displaystyle 18.75\times32=460+7p
\displaystyle 600=460+7p
\displaystyle 7p=140
\displaystyle p=\frac{140}{7}=20
\displaystyle \therefore p=20.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The mean age of a group of }40\text{ students is }17.45\text{ years. Find the}
\displaystyle \text{missing frequencies.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Age (years)}&15&16&17&18&19&20\\\hline \text{Number of students}&3&?&9&11&?&3\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Let the missing frequencies corresponding to ages }16\text{ and }19\text{ be }x\text{ and }y.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Age }(x_i)&\text{Number of students }(f_i)&f_ix_i\\\hline 15&3&45\\16&x&16x\\17&9&153\\18&11&198\\19&y&19y\\20&3&60\\\hline \text{Total}&26+x+y&456+16x+19y\\\hline\end{array}
\displaystyle \text{Since the total number of students is }40,
\displaystyle 26+x+y=40
\displaystyle x+y=14\qquad\ldots\text{(i)}
\displaystyle \text{Given mean}=17.45
\displaystyle \text{Mean }(\overline{x})=\frac{\sum f_ix_i}{\sum f_i}
\displaystyle 17.45=\frac{456+16x+19y}{40}
\displaystyle 698=456+16x+19y
\displaystyle 16x+19y=242\qquad\ldots\text{(ii)}
\displaystyle \text{From (i), }x=14-y
\displaystyle \text{Substituting in (ii),}
\displaystyle 16(14-y)+19y=242
\displaystyle 224-16y+19y=242
\displaystyle 3y=18
\displaystyle y=6
\displaystyle \therefore x=14-6=8
\displaystyle \therefore \text{The missing frequencies are }8\text{ and }6.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Using the assumed mean method, calculate the mean weekly wage from the}
\displaystyle \text{following frequency distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Weekly wages (Rs.)}&950&1000&1050&1100&1250&1500&1600\\\hline \text{Number of workers}&24&18&13&15&20&11&9\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Let the assumed mean }A=1100.
\displaystyle d_i=x_i-A
\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&d_i=x_i-A&f_id_i\\\hline 950&24&-150&-3600\\1000&18&-100&-1800\\1050&13&-50&-650\\1100&15&0&0\\1250&20&150&3000\\1500&11&400&4400\\1600&9&500&4500\\\hline \text{Total}&110&&5850\\\hline\end{array}
\displaystyle \sum f_i=110,\qquad \sum f_id_i=5850
\displaystyle \text{Mean }(\overline{x})=A+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =1100+\frac{5850}{110}
\displaystyle =1100+53.18=1153.18
\displaystyle \therefore \text{The mean weekly wage is Rs. }1153.18.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Using the step-deviation method, find the mean from the following data.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline x_i&18&19&20&21&22&23&24\\\hline f_i&170&320&530&700&230&140&110\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Let the assumed mean }A=21\text{ and }h=1.
\displaystyle u_i=\frac{x_i-A}{h}=\frac{x_i-21}{1}
\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&u_i=\frac{x_i-21}{1}&f_iu_i\\\hline 18&170&-3&-510\\19&320&-2&-640\\20&530&-1&-530\\21&700&0&0\\22&230&1&230\\23&140&2&280\\24&110&3&330\\\hline \text{Total}&2200&&-840\\\hline\end{array}
\displaystyle \sum f_i=2200,\qquad \sum f_iu_i=-840
\displaystyle \text{Mean }(\overline{x})=A+h\frac{\sum f_iu_i}{\sum f_i}
\displaystyle =21+1\left(\frac{-840}{2200}\right)
\displaystyle =21-0.3818=20.6182
\displaystyle \therefore \text{The mean is }20.62\text{ (approx.).}
\displaystyle \\

\displaystyle \text{Exercise - 19(B)}

\displaystyle \textbf{Question 1: }\text{Find the median of:}

\displaystyle \text{(i) }15,6,16,8,22,21,9,18,25
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the data in ascending order:}
\displaystyle 6,8,9,15,16,18,21,22,25
\displaystyle n=9,\text{ which is odd.}
\displaystyle \text{Median}=\left(\frac{n+1}{2}\right)^{\text{th}}\text{ term}
\displaystyle =\left(\frac{9+1}{2}\right)^{\text{th}}\text{ term}=5^{\text{th}}\text{ term}
\displaystyle =16
\displaystyle \therefore \text{Median}=16.

\displaystyle \text{(ii) }10,75,3,15,9,47,12,48,4,81,17,27
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the data in ascending order:}
\displaystyle 3,4,9,10,12,15,17,27,47,48,75,81
\displaystyle n=12,\text{ which is even.}
\displaystyle \text{Median}=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text{th}}\text{ term}+\left(\frac{n}{2}+1\right)^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}\left\{6^{\text{th}}\text{ term}+7^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}(15+17)=16
\displaystyle \therefore \text{Median}=16.

\displaystyle \text{(iii) }55,60,35,51,29,63,72,91,85,82
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the data in ascending order:}
\displaystyle 29,35,51,55,60,63,72,82,85,91
\displaystyle n=10,\text{ which is even.}
\displaystyle \text{Median}=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text{th}}\text{ term}+\left(\frac{n}{2}+1\right)^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}\left\{5^{\text{th}}\text{ term}+6^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}(60+63)=61.5
\displaystyle \therefore \text{Median}=61.5.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The runs scored by }11\text{ members of a cricket team are }26,38,53,18,66,72,
\displaystyle 0,47,32,7,35.\text{ Find the median score.}
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the data in ascending order:}
\displaystyle 0,7,18,26,32,35,38,47,53,66,72
\displaystyle n=11,\text{ which is odd.}
\displaystyle \text{Median}=\left(\frac{n+1}{2}\right)^{\text{th}}\text{ term}
\displaystyle =\left(\frac{11+1}{2}\right)^{\text{th}}\text{ term}=6^{\text{th}}\text{ term}
\displaystyle =35
\displaystyle \therefore \text{The median score is }35.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The heights (in cm) of }9\text{ girls are }144.2,148.5,152.1,143.7,145,149.6,
\displaystyle 150,146.5,147.3.\text{ Find the median height.}
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the data in ascending order:}
\displaystyle 143.7,144.2,145,146.5,147.3,148.5,149.6,150,152.1
\displaystyle n=9,\text{ which is odd.}
\displaystyle \text{Median}=\left(\frac{n+1}{2}\right)^{\text{th}}\text{ term}
\displaystyle =\left(\frac{9+1}{2}\right)^{\text{th}}\text{ term}=5^{\text{th}}\text{ term}
\displaystyle =147.3\text{ cm}
\displaystyle \therefore \text{The median height is }147.3\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The ages (in years) of }10\text{ teachers in a school are }34,37,53,46,52,43,
\displaystyle 31,36,40,50.\text{ Find the median age.}
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the data in ascending order:}
\displaystyle 31,34,36,37,40,43,46,50,52,53
\displaystyle n=10,\text{ which is even.}
\displaystyle \text{Median}=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text{th}}\text{ term}+\left(\frac{n}{2}+1\right)^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}\left\{5^{\text{th}}\text{ term}+6^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}(40+43)=\frac{83}{2}=41.5\text{ years}
\displaystyle \therefore \text{The median age is }41.5\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The weights (in kg) of }8\text{ children are }13.4,10.6,12.7,17.2,14.3,15,16.5,9.8.
\displaystyle \text{Find the median weight.}
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the data in ascending order:}
\displaystyle 9.8,10.6,12.7,13.4,14.3,15,16.5,17.2
\displaystyle n=8,\text{ which is even.}
\displaystyle \text{Median}=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text{th}}\text{ term}+\left(\frac{n}{2}+1\right)^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}\left\{4^{\text{th}}\text{ term}+5^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}(13.4+14.3)=\frac{27.7}{2}=13.85\text{ kg}
\displaystyle \therefore \text{The median weight is }13.85\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the median weight for the following data.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Weight (kg)}&45&46&48&50&52&54&55\\\hline \text{Number of boys}&8&5&6&9&7&4&2\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Writing the cumulative frequency table:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Weight (kg)}&\text{Number of boys }(f)&\text{Cumulative frequency }(cf)\\\hline 45&8&8\\46&5&13\\48&6&19\\50&9&28\\52&7&35\\54&4&39\\55&2&41\\\hline\end{array}
\displaystyle n=41,\text{ which is odd.}
\displaystyle \text{Median}=\left(\frac{n+1}{2}\right)^{\text{th}}\text{ term}
\displaystyle =\left(\frac{41+1}{2}\right)^{\text{th}}\text{ term}=21^{\text{st}}\text{ term}
\displaystyle \text{The }21^{\text{st}}\text{ observation corresponds to }50\text{ kg.}
\displaystyle \therefore \text{The median weight is }50\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Calculate the median for the following frequency distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Variate}&3&6&10&12&7&15\\\hline \text{Frequency}&3&4&2&8&13&10\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Arrange the variates in ascending order and prepare the cumulative frequency table.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Variate}&\text{Frequency }(f)&\text{Cumulative frequency }(cf)\\\hline 3&3&3\\6&4&7\\7&13&20\\10&2&22\\12&8&30\\15&10&40\\\hline\end{array}
\displaystyle n=40,\text{ which is even.}
\displaystyle \text{Median}=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text{th}}\text{ term}+\left(\frac{n}{2}+1\right)^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}\left\{20^{\text{th}}\text{ term}+21^{\text{st}}\text{ term}\right\}
\displaystyle =\frac{1}{2}(7+10)=\frac{17}{2}=8.5
\displaystyle \therefore \text{The median is }8.5.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The hearts of }60\text{ patients were examined through X-ray and the observations}
\displaystyle \text{obtained are given below. Find the median.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Diameter of heart (mm)}&120&121&122&123&124&125\\\hline \text{Number of patients}&7&9&15&12&6&11\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Writing the cumulative frequency table:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Diameter of heart (mm)}&\text{Number of patients }(f)&\text{Cumulative frequency }(cf)\\\hline 120&7&7\\121&9&16\\122&15&31\\123&12&43\\124&6&49\\125&11&60\\\hline\end{array}
\displaystyle n=60,\text{ which is even.}
\displaystyle \text{Median}=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text{th}}\text{ term}+\left(\frac{n}{2}+1\right)^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}\left\{30^{\text{th}}\text{ term}+31^{\text{st}}\text{ term}\right\}
\displaystyle \text{The }30^{\text{th}}\text{ and }31^{\text{st}}\text{ observations are both }122\text{ mm.}
\displaystyle \text{Median}=\frac{1}{2}(122+122)=122\text{ mm}
\displaystyle \therefore \text{The median diameter of the heart is }122\text{ mm.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the median for the following data.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{Variate}&23&26&20&30&28&25&18&16\\\hline \text{Frequency}&4&6&13&5&11&4&8&9\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the variates in ascending order and writing the cumulative frequency table:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Variate}&\text{Frequency }(f)&\text{Cumulative frequency }(cf)\\\hline 16&9&9\\18&8&17\\20&13&30\\23&4&34\\25&4&38\\26&6&44\\28&11&55\\30&5&60\\\hline\end{array}
\displaystyle n=60,\text{ which is even.}
\displaystyle \text{Median}=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text{th}}\text{ term}+\left(\frac{n}{2}+1\right)^{\text{th}}\text{ term}\right\}
\displaystyle =\frac{1}{2}\left\{30^{\text{th}}\text{ term}+31^{\text{st}}\text{ term}\right\}
\displaystyle \text{The }30^{\text{th}}\text{ observation is }20\text{ and the }31^{\text{st}}\text{ observation is }23.
\displaystyle \text{Median}=\frac{1}{2}(20+23)=\frac{43}{2}=21.5
\displaystyle \therefore \text{The median is }21.5.
\displaystyle \\

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