\displaystyle \text{Exercse - 18(A)}

\displaystyle \textbf{Question 1: }\text{State, which of the following variables are continuous and which are discrete:}
\displaystyle \text{(a) number of children in your class.}\qquad\text{(b) distance travelled by a car.}
\displaystyle \text{(c) sizes of shoes.}\qquad\text{(d) time.}\qquad\text{(e) number of patients in a hospital.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Number of children in your class: Discrete variable.}
\displaystyle \text{(b) Distance travelled by a car: Continuous variable.}
\displaystyle \text{(c) Sizes of shoes: Discrete variable.}
\displaystyle \text{(d) Time: Continuous variable.}
\displaystyle \text{(e) Number of patients in a hospital: Discrete variable.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Given below are the marks obtained by }30\text{ students in an examination:}
\displaystyle 08,\ 17,\ 33,\ 41,\ 47,\ 23,\ 20,\ 34,\ 09,\ 18,\ 42,\ 14,\ 30,\ 19,\ 29,\ 11,
\displaystyle 36,\ 48,\ 40,\ 24,\ 22,\ 02,\ 16,\ 21,\ 15,\ 32,\ 47,\ 44,\ 33,\ 01.
\displaystyle \text{Taking class intervals }1-10,\ 11-20,\ldots,\ 41-50,\text{ make a frequency table for the}
\displaystyle \text{above distribution.}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Marks}&\text{Tally Marks}&\text{Frequency}\\ \hline 1-10&||||&4\\ \hline 11-20&|||||\ |||&8\\ \hline 21-30&|||||\ |&6\\ \hline 31-40&|||||\ |&6\\ \hline 41-50&|||||\ |&6\\ \hline \text{Total}&&30\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The marks of }24\text{ candidates in Mathematics are given below:}
\displaystyle 45,\ 48,\ 15,\ 23,\ 30,\ 35,\ 40,\ 11,\ 29,\ 0,\ 3,\ 12,\ 48,\ 50,\ 18,\ 30,
\displaystyle 15,\ 30,\ 11,\ 42,\ 23,\ 2,\ 3,\ 44.
\displaystyle \text{The maximum marks are }50.\text{ Make a frequency distribution taking class}
\displaystyle \text{intervals }0-10,\ 10-20,\ \ldots.
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Marks}&\text{Tally\ Marks}&\text{Frequency}\\ \hline 0-10&||||&4\\ \hline 10-20&|||||\ |&6\\ \hline 20-30&|||||\ |&6\\ \hline 30-40&||&2\\ \hline 40-50&|||||\ |&6\\ \hline \text{Total}&&24\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Fill in the blanks:}
\displaystyle \text{(a) A quantity which can vary from one individual to another is called a }\ldots\ldots\ldots .
\displaystyle \text{(b) Sizes of shoes are }\ldots\ldots\ldots\text{ variables.}
\displaystyle \text{(c) Daily temperature is }\ldots\ldots\ldots\text{ variable.}
\displaystyle \text{(d) The range of the data }7,\ 13,\ 6,\ 25,\ 18,\ 20,\ 16\text{ is }\ldots\ldots\ldots .
\displaystyle \text{(e) In the class interval }35-46,\text{ the lower limit is }\ldots\ldots\ldots\text{ and upper limit is }\ldots\ldots\ldots .
\displaystyle \text{(f) The class mark of class interval }22-29\text{ is }\ldots\ldots\ldots .
\displaystyle \text{Answer:}
\displaystyle \text{(a) Variable}
\displaystyle \text{(b) Discrete}
\displaystyle \text{(c) Continuous}
\displaystyle \text{(d) Range}=25-6=19
\displaystyle \text{(e) Lower limit }=35,\qquad \text{Upper limit }=46.
\displaystyle \text{(f) Class mark}=\frac{22+29}{2}=\frac{51}{2}=25.5.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the actual lower class limits, upper class limits and the mid-values}
\displaystyle \text{of the classes }10-19,\ 20-29,\ 30-39\text{ and }40-49.
\displaystyle \text{Answer:}
\displaystyle \text{Difference between the upper limit of one class and lower limit of the next class}=20-19=1.
\displaystyle \text{Adjustment factor}=\frac{1}{2}=0.5.
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Class Interval}&\text{Actual Lower Limit}&\text{Actual Upper Limit}&\text{Mid-value}\\ \hline 10-19&9.5&19.5&14.5\\ \hline 20-29&19.5&29.5&24.5\\ \hline 30-39&29.5&39.5&34.5\\ \hline 40-49&39.5&49.5&44.5\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the actual lower and upper class limits and also the class marks}
\displaystyle \text{of the classes }1.1-2.0,\ 2.1-3.0\text{ and }3.1-4.0.
\displaystyle \text{Answer:}
\displaystyle \text{Difference between successive class limits}=2.1-2.0=0.1.
\displaystyle \text{Adjustment factor}=\frac{0.1}{2}=0.05.
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Class Interval}&\text{Actual Lower Limit}&\text{Actual Upper Limit}&\text{Class Mark}\\ \hline 1.1-2.0&1.05&2.05&1.55\\ \hline 2.1-3.0&2.05&3.05&2.55\\ \hline 3.1-4.0&3.05&4.05&3.55\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Use the table given below to find:}
\displaystyle \text{(a) the actual class limits of the fourth class.}
\displaystyle \text{(b) the class boundaries of the sixth class.}
\displaystyle \text{(c) the class mark of the third class.}
\displaystyle \text{(d) the upper and lower limits of the fifth class.}
\displaystyle \text{(e) the size of the third class.}
\displaystyle \begin{array}{|c|c|}\hline \text{Class Interval}&\text{Frequency}\\ \hline 30-34&7\\ \hline 35-39&10\\ \hline 40-44&12\\ \hline 45-49&13\\ \hline 50-54&8\\ \hline 55-59&4\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Difference between successive class limits}=35-34=1.
\displaystyle \text{Adjustment factor}=\frac{1}{2}=0.5.
\displaystyle \text{(a) Fourth class}=45-49.
\displaystyle \therefore \text{Actual lower limit}=45-0.5=44.5,\qquad \text{actual upper limit}=49+0.5=49.5.
\displaystyle \text{(b) Sixth class}=55-59.
\displaystyle \therefore \text{Class boundaries are }54.5\text{ and }59.5.
\displaystyle \text{(c) Third class}=40-44.
\displaystyle \therefore \text{Class mark}=\frac{40+44}{2}=42.
\displaystyle \text{(d) Fifth class}=50-54.
\displaystyle \therefore \text{Lower limit}=50,\qquad \text{upper limit}=54.
\displaystyle \text{(e) Actual class limits of the third class are }39.5\text{ and }44.5.
\displaystyle \therefore \text{Size of the third class}=44.5-39.5=5.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Construct a cumulative frequency distribution table from the frequency}
\displaystyle \text{table given below:}
\displaystyle \text{(i) }\begin{array}{|c|c|}\hline \text{Class Interval}&\text{Frequency}\\ \hline 0-8&9\\ \hline 8-16&13\\ \hline 16-24&12\\ \hline 24-32&7\\ \hline 32-40&15\\ \hline \end{array}
\displaystyle \text{(ii) }\begin{array}{|c|c|}\hline \text{Class Interval}&\text{Frequency}\\ \hline 1-10&12\\ \hline 11-20&18\\ \hline 21-30&23\\ \hline 31-40&15\\ \hline 41-50&10\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Cumulative frequencies are obtained by successively adding the frequencies.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Frequency}&\text{Cumulative Frequency}\\ \hline 0-8&9&9\\ \hline 8-16&13&9+13=22\\ \hline 16-24&12&22+12=34\\ \hline 24-32&7&34+7=41\\ \hline 32-40&15&41+15=56\\ \hline \end{array}

\displaystyle \text{(ii) Cumulative frequencies are obtained by successively adding the frequencies.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Frequency}&\text{Cumulative Frequency}\\ \hline 1-10&12&12\\ \hline 11-20&18&12+18=30\\ \hline 21-30&23&30+23=53\\ \hline 31-40&15&53+15=68\\ \hline 41-50&10&68+10=78\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Construct a frequency distribution table from the following cumulative}
\displaystyle \text{frequency distribution:}
\displaystyle \text{(i) }\begin{array}{|c|c|}\hline \text{Class Interval}&\text{Cumulative Frequency}\\ \hline 10-19&8\\ \hline 20-29&19\\ \hline 30-39&23\\ \hline 40-49&30\\ \hline \end{array}
\displaystyle \text{(ii) }\begin{array}{|c|c|}\hline \text{Class Interval}&\text{Cumulative Frequency}\\ \hline 5-10&18\\ \hline 10-15&30\\ \hline 15-20&46\\ \hline 20-25&73\\ \hline 25-30&90\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(i) First frequency}=8.
\displaystyle \text{Remaining frequencies}=19-8=11,\quad23-19=4,\quad30-23=7.
\displaystyle \begin{array}{|c|c|}\hline \text{Class Interval}&\text{Frequency}\\ \hline 10-19&8\\ \hline 20-29&11\\ \hline 30-39&4\\ \hline 40-49&7\\ \hline \text{Total}&30\\ \hline \end{array}

\displaystyle \text{(ii) First frequency}=18.
\displaystyle \text{Remaining frequencies}=30-18=12,\quad46-30=16,\quad73-46=27,\quad90-73=17.
\displaystyle \begin{array}{|c|c|}\hline \text{Class Interval}&\text{Frequency}\\ \hline 5-10&18\\ \hline 10-15&12\\ \hline 15-20&16\\ \hline 20-25&27\\ \hline 25-30&17\\ \hline \text{Total}&90\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Construct a frequency table from the following data:}
\displaystyle \begin{array}{|c|c|}\hline \text{Marks}&\text{No. of students}\\ \hline \text{less than }10&6\\ \hline \text{less than }20&15\\ \hline \text{less than }30&30\\ \hline \text{less than }40&39\\ \hline \text{less than }50&53\\ \hline \text{less than }60&70\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{For }0-10,\quad f=6.
\displaystyle \text{For }10-20,\quad f=15-6=9.
\displaystyle \text{For }20-30,\quad f=30-15=15.
\displaystyle \text{For }30-40,\quad f=39-30=9.
\displaystyle \text{For }40-50,\quad f=53-39=14.
\displaystyle \text{For }50-60,\quad f=70-53=17.
\displaystyle \begin{array}{|c|c|}\hline \text{Marks}&\text{Frequency}\\ \hline 0-10&6\\ \hline 10-20&9\\ \hline 20-30&15\\ \hline 30-40&9\\ \hline 40-50&14\\ \hline 50-60&17\\ \hline \text{Total}&70\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Construct the frequency distribution table from the following cumulative}
\displaystyle \text{frequency table:}
\displaystyle \begin{array}{|c|c|}\hline \text{Ages}&\text{No. of students}\\ \hline \text{Below }4&0\\ \hline \text{Below }7&85\\ \hline \text{Below }10&140\\ \hline \text{Below }13&243\\ \hline \text{Below }16&300\\ \hline \end{array}
\displaystyle \text{(i) State the number of students in the age group }10-13.
\displaystyle \text{(ii) State the age-group which has the least number of students.}
\displaystyle \text{Answer:}
\displaystyle \text{For }0-4,\quad f=0.
\displaystyle \text{For }4-7,\quad f=85-0=85.
\displaystyle \text{For }7-10,\quad f=140-85=55.
\displaystyle \text{For }10-13,\quad f=243-140=103.
\displaystyle \text{For }13-16,\quad f=300-243=57.
\displaystyle \begin{array}{|c|c|}\hline \text{Age Group}&\text{Frequency}\\ \hline 0-4&0\\ \hline 4-7&85\\ \hline 7-10&55\\ \hline 10-13&103\\ \hline 13-16&57\\ \hline \text{Total}&300\\ \hline \end{array}

\displaystyle \text{(i) Number of students in the age group }10-13=243-140=103.

\displaystyle \text{(ii) The age group }0-4\text{ has the least number of students, i.e. }0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Fill in the blanks in the following table:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Frequency}&\text{Cumulative Frequency}\\ \hline 25-34&\ldots\ldots&15\\ \hline 35-44&\ldots\ldots&28\\ \hline 45-54&21&\ldots\ldots\\ \hline 55-64&16&\ldots\ldots\\ \hline 65-74&\ldots\ldots&73\\ \hline 75-84&12&\ldots\ldots\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Frequency of }25-34=15.
\displaystyle \text{Frequency of }35-44=28-15=13.
\displaystyle \text{Cumulative frequency of }45-54=28+21=49.
\displaystyle \text{Cumulative frequency of }55-64=49+16=65.
\displaystyle \text{Frequency of }65-74=73-65=8.
\displaystyle \text{Cumulative frequency of }75-84=73+12=85.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Frequency}&\text{Cumulative Frequency}\\ \hline 25-34&15&15\\ \hline 35-44&13&28\\ \hline 45-54&21&49\\ \hline 55-64&16&65\\ \hline 65-74&8&73\\ \hline 75-84&12&85\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The value of }\pi\text{ up to }50\text{ decimal places is:}
\displaystyle 3.14159265358979323846264338327950288419716939937510
\displaystyle \text{(i) Make a frequency distribution table of the digits from }0\text{ to }9\text{ after the decimal place.}
\displaystyle \text{(ii) Which are the most and the least occurring digits?}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Counting the }50\text{ digits after the decimal point, we get:}
\displaystyle \begin{array}{|c|c|}\hline \text{Digit}&\text{Frequency}\\ \hline 0&2\\ \hline 1&5\\ \hline 2&5\\ \hline 3&8\\ \hline 4&4\\ \hline 5&5\\ \hline 6&4\\ \hline 7&4\\ \hline 8&5\\ \hline 9&8\\ \hline \text{Total}&50\\ \hline \end{array}

\displaystyle \text{(ii) The highest frequency is }8.
\displaystyle \therefore \text{The most occurring digits are }3\text{ and }9.
\displaystyle \text{The lowest frequency is }2.
\displaystyle \therefore \text{The least occurring digit is }0.
\displaystyle \\

\displaystyle \text{Exercse - 18(B)}

\displaystyle \textbf{Question 1: }\text{Construct a frequency polygon for the following distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class Intervals}&0-4&4-8&8-12&12-16&16-20&20-24\\ \hline \text{Frequency}&4&7&10&15&11&6\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Class mark}=\frac{\text{Lower limit}+\text{Upper limit}}{2}.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Class Mark}&\text{Frequency}\\ \hline -4-0&-2&0\\ \hline 0-4&2&4\\ \hline 4-8&6&7\\ \hline 8-12&10&10\\ \hline 12-16&14&15\\ \hline 16-20&18&11\\ \hline 20-24&22&6\\ \hline 24-28&26&0\\ \hline \end{array}
\displaystyle \therefore \text{The resulting graph is the required frequency polygon.}\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Construct a combined histogram and frequency polygon for the following}
\displaystyle \text{frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class Intervals}&10-20&20-30&30-40&40-50&50-60\\ \hline \text{Frequency}&3&5&6&4&2\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Draw the histogram with class intervals on the }x\text{-axis and frequencies on the }y\text{-axis.}
\displaystyle \text{Class mark}=\frac{\text{Lower limit}+\text{Upper limit}}{2}.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Class Mark}&\text{Frequency}\\ \hline 0-10&5&0\\ \hline 10-20&15&3\\ \hline 20-30&25&5\\ \hline 30-40&35&6\\ \hline 40-50&45&4\\ \hline 50-60&55&2\\ \hline 60-70&65&0\\ \hline \end{array}
\displaystyle \therefore \text{The resulting graph is the required frequency polygon.}

\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Construct a frequency polygon for the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class Intervals}&10-14&15-19&20-24&25-29&30-34\\ \hline \text{Frequency}&5&8&12&9&4\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The given class intervals are inclusive. Convert them into continuous class intervals.}
\displaystyle \text{Adjustment factor}=\frac{15-14}{2}=0.5.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Class Mark}&\text{Frequency}\\ \hline 5-9&7&0\\ \hline 10-14&12&5\\ \hline 15-19&17&8\\ \hline 20-24&22&12\\ \hline 25-29&27&9\\ \hline 30-34&32&4\\ \hline 35-39&37&0\\ \hline \end{array}
\displaystyle \therefore \text{The resulting graph is the required frequency polygon.}\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The daily wages in a factory are distributed as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Daily wages (in }\text{Rs.}\text{)}&125-175&175-225&225-275&275-325&325-375\\ \hline \text{Number of workers}&4&20&22&10&6\\ \hline \end{array}
\displaystyle \text{Draw a frequency polygon for this distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Class mark}=\frac{\text{Lower limit}+\text{Upper limit}}{2}.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Class Mark}&\text{Frequency}\\ \hline 75-125&100&0\\ \hline 125-175&150&4\\ \hline 175-225&200&20\\ \hline 225-275&250&22\\ \hline 275-325&300&10\\ \hline 325-375&350&6\\ \hline 375-425&400&0\\ \hline \end{array}
\displaystyle \therefore \text{The resulting graph is the required frequency polygon.}\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Draw frequency polygons for each of the following frequency distributions:}
\displaystyle \text{(a) using histogram}\qquad\text{(b) without using histogram.}
\displaystyle \text{(i) }\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{C.I.}&10-30&30-50&50-70&70-90&90-110&110-130&130-150\\ \hline f&4&7&5&9&5&6&4\\ \hline \end{array}
\displaystyle \text{(ii) }\begin{array}{|c|c|c|c|c|c|c|}\hline \text{C.I.}&5-15&15-25&25-35&35-45&45-55&55-65\\ \hline f&8&16&18&14&8&2\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Class mark}=\frac{\text{Lower limit}+\text{Upper limit}}{2}.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Class Mark}&f\\ \hline -10-10&0&0\\ \hline 10-30&20&4\\ \hline 30-50&40&7\\ \hline 50-70&60&5\\ \hline 70-90&80&9\\ \hline 90-110&100&5\\ \hline 110-130&120&6\\ \hline 130-150&140&4\\ \hline 150-170&160&0\\ \hline \end{array}

\displaystyle \text{(a) Using histogram: Draw the histogram for the given distribution.}
\displaystyle \text{(b) Without using histogram: Mark the class marks on the }x\text{-axis and frequencies on the }y\text{-axis.}

\displaystyle \text{(ii) Class mark}=\frac{\text{Lower limit}+\text{Upper limit}}{2}.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class Interval}&\text{Class Mark}&f\\ \hline -5-5&0&0\\ \hline 5-15&10&8\\ \hline 15-25&20&16\\ \hline 25-35&30&18\\ \hline 35-45&40&14\\ \hline 45-55&50&8\\ \hline 55-65&60&2\\ \hline 65-75&70&0\\ \hline \end{array}

\displaystyle \text{(a) Using histogram: Draw the histogram for the given distribution.}
\displaystyle \text{(b) Without using histogram: Mark the class marks on the }x\text{-axis and frequencies on the }y\text{-axis.}

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