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NCERT Exemplar · Class 9 Mathematics Statistics and Probability

72 questions · 72 still being checked

EXERCISE 14.4 1–12 (part 7 of 7)

  1. Exercise 1

    The following are the marks (out of 100\displaystyle 100 ) of 60\displaystyle 60 students in mathematics. 16,13,5,80,86,7,51,48,24,56,70,19,61,17,16,36,34,42,34,35,72,55,75,31,52,28,72,97,74,45,62,68,86,35,85,36,81,75,55,26,95,31,7,78,92,62,52,56,15,63,25,36,54,44,47,27,72,17,4,30.\begin{aligned} & 16,13,5,80,86,7,51,48,24,56,70,19,61,17,16,36,34,42,34,35,72,55,75, \\ & 31,52,28,72,97,74,45,62,68,86,35,85,36,81,75,55,26,95,31,7,78,92,62, \\ & 52,56,15,63,25,36,54,44,47,27,72,17,4,30 . \end{aligned} Construct a grouped frequency distribution table with width 10\displaystyle 10 of each class starting from 0\displaystyle 0-9.

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    Tally each of the $\displaystyle 60$ marks into a class of width 10.\[\begin{array}{c|c} \text{Class} & \text{Frequency} \\ \hline 0\text{-}9 & 4 \\ 10\text{-}19 & 7 \\ 20\text{-}29 & 5 \\ 30\text{-}39 & 10 \\ 40\text{-}49 & 5 \\ 50\text{-}59 & 8 \\ 60\text{-}69 & 5 \\ 70\text{-}79 & 8 \\ 80\text{-}89 & 5 \\ 90\text{-}99 & 3 \\ \hline \text{Total} & 60 \end{array} \]Answer: frequencies $\displaystyle 4$, $\displaystyle 7$, $\displaystyle 5$, $\displaystyle 10$, $\displaystyle 5$, $\displaystyle 8$, $\displaystyle 5$, $\displaystyle 8$, $\displaystyle 5$, $\displaystyle 3$ for classes $\displaystyle 0$-$\displaystyle 9$ through $\displaystyle 90$-$\displaystyle 99$ (table above).NCERT prints: frequencies $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 5$, $\displaystyle 6$, $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 3$, $\displaystyle 2$, $\displaystyle 5$, $\displaystyle 4$ — a misprint: they total $\displaystyle 35$, but the question lists $\displaystyle 60$ marks, and counting them class by class gives $\displaystyle 4$, $\displaystyle 7$, $\displaystyle 5$, $\displaystyle 10$, $\displaystyle 5$, $\displaystyle 8$, $\displaystyle 5$, $\displaystyle 8$, $\displaystyle 5$, 3.

  2. Exercise 2

    Refer to Q1 above. Construct a grouped frequency distribution table with width 10\displaystyle 10 of each class, in such a way that one of the classes is 10\displaystyle 10-20\displaystyle 20 (20\displaystyle 20 not included).

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    NCERT’s answer
    Class intervalsFrequency
    $\displaystyle 0$-$\displaystyle 10$$\displaystyle 4$
    $\displaystyle 10$-$\displaystyle 20$$\displaystyle 7$
    $\displaystyle 20$-$\displaystyle 30$$\displaystyle 5$
    $\displaystyle 30$-$\displaystyle 40$$\displaystyle 10$
    $\displaystyle 40$-$\displaystyle 50$$\displaystyle 5$
    $\displaystyle 50$-$\displaystyle 60$$\displaystyle 8$
    $\displaystyle 60$-$\displaystyle 70$$\displaystyle 5$
    $\displaystyle 70$-$\displaystyle 80$$\displaystyle 8$
    $\displaystyle 80$-$\displaystyle 90$$\displaystyle 5$
    $\displaystyle 90$-$\displaystyle 100$$\displaystyle 3$
    Since the marks are integers, each still falls in the same decade as with classes $\displaystyle 0$-$\displaystyle 9$, $\displaystyle 10$-$\displaystyle 19$, …, $\displaystyle 90$-$\displaystyle 99$, so the tally is unchanged.\[\begin{array}{c|c} \text{Class intervals} & \text{Frequency} \\ \hline 0\text{-}10 & 4 \\ 10\text{-}20 & 7 \\ 20\text{-}30 & 5 \\ 30\text{-}40 & 10 \\ 40\text{-}50 & 5 \\ 50\text{-}60 & 8 \\ 60\text{-}70 & 5 \\ 70\text{-}80 & 8 \\ 80\text{-}90 & 5 \\ 90\text{-}100 & 3 \\ \hline \text{Total} & 60 \end{array} \]Answer: frequencies $\displaystyle 4$, $\displaystyle 7$, $\displaystyle 5$, $\displaystyle 10$, $\displaystyle 5$, $\displaystyle 8$, $\displaystyle 5$, $\displaystyle 8$, $\displaystyle 5$, $\displaystyle 3$ for classes $\displaystyle 0$-$\displaystyle 10$ through $\displaystyle 90$-$\displaystyle 100$ (table above).

  3. Exercise 3

    Draw a histogram of the following distribution :
    Heights (in cm)Number of students
    150\displaystyle 150-153\displaystyle 1537\displaystyle 7
    153\displaystyle 153-156\displaystyle 1568\displaystyle 8
    156\displaystyle 156-159\displaystyle 15914\displaystyle 14
    159\displaystyle 159-162\displaystyle 16210\displaystyle 10
    162\displaystyle 162-165\displaystyle 1656\displaystyle 6
    165\displaystyle 165-168\displaystyle 1685\displaystyle 5

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    The classes all have width $\displaystyle 3$ and are already continuous, so no adjustment is needed — each bar's height is just its class frequency.NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-4_Q3Answer: heights $\displaystyle 7$, $\displaystyle 8$, $\displaystyle 14$, $\displaystyle 10$, $\displaystyle 6$, $\displaystyle 5$ over $\displaystyle 150$-$\displaystyle 153$, $\displaystyle 153$-$\displaystyle 156$, $\displaystyle 156$-$\displaystyle 159$, $\displaystyle 159$-$\displaystyle 162$, $\displaystyle 162$-$\displaystyle 165$, $\displaystyle 165$-168.

  4. Exercise 4

    Draw a histogram to represent the following grouped frequency distribution :
    Ages (in years)Number of teachers
    20\displaystyle 20-24\displaystyle 2410\displaystyle 10
    25\displaystyle 25-29\displaystyle 2928\displaystyle 28
    30\displaystyle 30-34\displaystyle 3432\displaystyle 32
    35\displaystyle 35-39\displaystyle 3948\displaystyle 48
    40\displaystyle 40-44\displaystyle 4450\displaystyle 50
    45\displaystyle 45-49\displaystyle 4935\displaystyle 35
    50\displaystyle 50-54\displaystyle 5412\displaystyle 12

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    The classes $\displaystyle 20$-$\displaystyle 24$, $\displaystyle 25$-$\displaystyle 29$, … are discontinuous, with a gap \[25-24 = 1 \] so each boundary shifts by \[\tfrac{1}{2} = 0.5 \] before drawing. \[20\text{-}24 \to 19.5\text{-}24.5,\ \ldots,\ 50\text{-}54 \to 49.5\text{-}54.5 \]NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-4_Q4Answer: continuous classes $\displaystyle 19.5$-$\displaystyle 24.5$, $\displaystyle 24.5$-$\displaystyle 29.5$, $\displaystyle 29.5$-$\displaystyle 34.5$, $\displaystyle 34.5$-$\displaystyle 39.5$, $\displaystyle 39.5$-$\displaystyle 44.5$, $\displaystyle 44.5$-$\displaystyle 49.5$, $\displaystyle 49.5$-$\displaystyle 54.5$ with heights $\displaystyle 10$, $\displaystyle 28$, $\displaystyle 32$, $\displaystyle 48$, $\displaystyle 50$, $\displaystyle 35$, 12.

  5. Exercise 5

    The lengths of 62\displaystyle 62 leaves of a plant are measured in millimetres and the data is represented in the following table :
    Length (in mm)Number of leaves
    118\displaystyle 118-126\displaystyle 1268\displaystyle 8
    127\displaystyle 127-135\displaystyle 13510\displaystyle 10
    136\displaystyle 136-144\displaystyle 14412\displaystyle 12
    145\displaystyle 145-153\displaystyle 15317\displaystyle 17
    154\displaystyle 154-162\displaystyle 1627\displaystyle 7
    163\displaystyle 163-171\displaystyle 1715\displaystyle 5
    172\displaystyle 172-180\displaystyle 1803\displaystyle 3
    Draw a histogram to represent the data above.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    The classes $\displaystyle 118$-$\displaystyle 126$, $\displaystyle 127$-$\displaystyle 135$, … are discontinuous; subtract $\displaystyle 0.5$ from each lower limit and add $\displaystyle 0.5$ to each upper limit. \[118\text{-}126 \to 117.5\text{-}126.5,\ \ldots,\ 172\text{-}180 \to 171.5\text{-}180.5 \] \[8+10+12+17+7+5+3=62 \]NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-4_Q5Answer: continuous classes $\displaystyle 117.5$-$\displaystyle 126.5$, $\displaystyle 126.5$-$\displaystyle 135.5$, $\displaystyle 135.5$-$\displaystyle 144.5$, $\displaystyle 144.5$-$\displaystyle 153.5$, $\displaystyle 153.5$-$\displaystyle 162.5$, $\displaystyle 162.5$-$\displaystyle 171.5$, $\displaystyle 171.5$-$\displaystyle 180.5$ with heights $\displaystyle 8$, $\displaystyle 10$, $\displaystyle 12$, $\displaystyle 17$, $\displaystyle 7$, $\displaystyle 5$, 3.

  6. Exercise 6

    The marks obtained (out of 100\displaystyle 100 ) by a class of 80\displaystyle 80 students are given below :
    MarksNumber of students
    10\displaystyle 10-20\displaystyle 206\displaystyle 6
    20\displaystyle 20-30\displaystyle 3017\displaystyle 17
    30\displaystyle 30-50\displaystyle 5015\displaystyle 15
    50\displaystyle 50-70\displaystyle 7016\displaystyle 16
    70\displaystyle 70-100\displaystyle 10026\displaystyle 26
    Construct a histogram to represent the data above.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\text{minimum class width} = 10 \] \[\text{adjusted height} = \frac{\text{frequency}}{\text{class width}}\times 10 \] Scaling every class to width $\displaystyle 10$ keeps each bar's area proportional to its frequency. \[10\text{-}20:\ \frac{6}{10}\times10=6 \qquad 20\text{-}30:\ \frac{17}{10}\times10=17 \] \[30\text{-}50:\ \frac{15}{20}\times10=7.5 \qquad 50\text{-}70:\ \frac{16}{20}\times10=8 \qquad 70\text{-}100:\ \frac{26}{30}\times10\approx8.67 \] NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-4_Q6 Answer: adjusted heights \(\displaystyle 6,\ 17,\ 7.5,\ 8,\ 8.67\) for classes \(\displaystyle 10\text{-}20,\ 20\text{-}30,\ 30\text{-}50,\ 50\text{-}70,\ 70\text{-}100\).

  7. Exercise 7

    Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way :
    Class interval (km/h)Frequency
    30\displaystyle 30-40\displaystyle 403\displaystyle 3
    40\displaystyle 40-50\displaystyle 506\displaystyle 6
    50\displaystyle 50-60\displaystyle 6025\displaystyle 25
    60\displaystyle 60-70\displaystyle 7065\displaystyle 65
    70\displaystyle 70-80\displaystyle 8050\displaystyle 50
    80\displaystyle 80-90\displaystyle 9028\displaystyle 28
    90\displaystyle 90-100\displaystyle 10014\displaystyle 14
    Draw a histogram and frequency polygon representing the data above.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Each class has width $\displaystyle 10$, so histogram bars are adjacent rectangles of that width with heights equal to the frequencies.\[\text{class mark} = \frac{\text{lower limit}+\text{upper limit}}{2} \]\[\frac{30+40}{2}=35,\quad \frac{40+50}{2}=45,\quad \frac{50+60}{2}=55,\quad \frac{60+70}{2}=65,\quad \frac{70+80}{2}=75,\quad \frac{80+90}{2}=85,\quad \frac{90+100}{2}=95 \]The frequency polygon closes at the zero-frequency marks of the adjacent classes \(\displaystyle 20\text{-}30\) and \(\displaystyle 100\text{-}110\), i.e. \(\displaystyle (25,0)\) and \(\displaystyle (105,0)\).NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-4_Q7Answer: histogram (solid) and frequency polygon (dashed) as shown in the figure.

  8. Exercise 8

    Refer to Q. 7\displaystyle 7 : Draw the frequency polygon representing the above data without drawing the histogram.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\text{class mark} = \frac{\text{lower limit}+\text{upper limit}}{2} \]\[35,\ 45,\ 55,\ 65,\ 75,\ 85,\ 95 \]\[(35,3),\ (45,6),\ (55,25),\ (65,65),\ (75,50),\ (85,28),\ (95,14) \]Closed at \(\displaystyle (25,0)\) and \(\displaystyle (105,0)\).NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-4_Q8Answer: frequency polygon as shown in the figure.

  9. Exercise 9

    Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them.
    Section ASection B
    MarksFrequencyMarksFrequency
    0\displaystyle 0-15\displaystyle 155\displaystyle 50\displaystyle 0-15\displaystyle 153\displaystyle 3
    15\displaystyle 15-30\displaystyle 3012\displaystyle 1215\displaystyle 15-30\displaystyle 3016\displaystyle 16
    30\displaystyle 30-45\displaystyle 4528\displaystyle 2830\displaystyle 30-45\displaystyle 4525\displaystyle 25
    45\displaystyle 45-60\displaystyle 6030\displaystyle 3045\displaystyle 45-60\displaystyle 6027\displaystyle 27
    60\displaystyle 60-75\displaystyle 7535\displaystyle 3560\displaystyle 60-75\displaystyle 7540\displaystyle 40
    75\displaystyle 75-90\displaystyle 9013\displaystyle 1375\displaystyle 75-90\displaystyle 9010\displaystyle 10
    Represent the marks of the students of both the sections on the same graph by two frequency polygons. What do you observe?

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\text{class mark} = \frac{\text{lower limit}+\text{upper limit}}{2} \]\[\frac{0+15}{2}=7.5,\ \frac{15+30}{2}=22.5,\ \frac{30+45}{2}=37.5,\ \frac{45+60}{2}=52.5,\ \frac{60+75}{2}=67.5,\ \frac{75+90}{2}=82.5 \]Section A's frequencies \(\displaystyle 5,12,28,30,35,13\) and Section B's \(\displaystyle 3,16,25,27,40,10\) are plotted against these marks, each polygon closed at \(\displaystyle (-7.5,0)\) and \(\displaystyle (97.5,0)\).NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-4_Q9Section A leads in four classes (\(\displaystyle 0\text{-}15,\ 30\text{-}45,\ 45\text{-}60,\ 75\text{-}90\)); Section B leads in \(\displaystyle 15\text{-}30\) (\(\displaystyle 16\) against \(\displaystyle 12\)) and in \(\displaystyle 60\text{-}75\) (\(\displaystyle 40\) against \(\displaystyle 35\)). Both polygons peak at \(\displaystyle 60\text{-}75\). The totals, \(\displaystyle 123\) for A and \(\displaystyle 121\) for B, are close.Answer: two frequency polygons as shown (Section A solid, Section B dashed); Section A leads in four classes, Section B leads in \(\displaystyle 15\text{-}30\) and \(\displaystyle 60\text{-}75\); totals close (\(\displaystyle 123\) vs \(\displaystyle 121\)).

  10. Exercise 10

    The mean of the following distribution is 50.
    x\displaystyle \boldsymbol{x}f\displaystyle f
    10\displaystyle 1017\displaystyle 17
    30\displaystyle 305a+3\displaystyle 5 a+3
    50\displaystyle 5032\displaystyle 32
    70\displaystyle 707a−11\displaystyle 7 a-11
    90\displaystyle 9019\displaystyle 19
    Find the value of a\displaystyle a and hence the frequencies of 30\displaystyle 30 and 70.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle a=5$, frequency of $\displaystyle 30$ is $\displaystyle 28$ and that of $\displaystyle 70$ is $\displaystyle 24$ .
    \[\bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]\[\sum f_i = 17+(5a+3)+32+(7a-11)+19 = 60+12a \]\[\sum f_i x_i = 10(17)+30(5a+3)+50(32)+70(7a-11)+90(19) = 2800+640a \]\[\frac{2800+640a}{60+12a}=50 \]\[2800+640a = 3000+600a \]\[40a=200 \implies a=5 \]\[5a+3 = 28, \qquad 7a-11 = 24 \]Answer: \(\displaystyle a=5\); frequency of \(\displaystyle 30\) is \(\displaystyle 28\), frequency of \(\displaystyle 70\) is \(\displaystyle 24\).

  11. Exercise 11

    The mean marks (out of 100\displaystyle 100) of boys and girls in an examination are 70\displaystyle 70 and 73\displaystyle 73, respectively. If the mean marks of all the students in that examination is 71\displaystyle 71, find the ratio of the number of boys to the number of girls.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 2$ : $\displaystyle 1$
    Let \(\displaystyle b\) and \(\displaystyle g\) be the numbers of boys and girls.\[\frac{70b+73g}{b+g} = 71 \]\[70b+73g = 71b+71g \]\[2g = b \]\[\frac{b}{g} = \frac{2}{1} \]Answer: ratio of boys to girls \(\displaystyle = 2:1\).

  12. Exercise 12

    A total of 25\displaystyle 25 patients admitted to a hospital are tested for levels of blood sugar, (mg/dl)\displaystyle (\mathrm{mg} / \mathrm{dl}) and the results obtained were as follows :
    87\displaystyle 8771\displaystyle 7183\displaystyle 8367\displaystyle 6785\displaystyle 85
    77\displaystyle 7769\displaystyle 6976\displaystyle 7665\displaystyle 6585\displaystyle 85
    85\displaystyle 8554\displaystyle 5470\displaystyle 7068\displaystyle 6880\displaystyle 80
    73\displaystyle 7378\displaystyle 7868\displaystyle 6885\displaystyle 8573\displaystyle 73
    81\displaystyle 8178\displaystyle 7881\displaystyle 8177\displaystyle 7775\displaystyle 75
    Find mean, median and mode (mg/dl) of the above data.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Mean $\displaystyle =75.64$, Median $\displaystyle =77$, Mode $\displaystyle =85$
    \[\sum x_i = 393+372+357+377+392 = 1891 \]\[\text{Mean} = \frac{1891}{25} = 75.64 \]Arranging the \(\displaystyle 25\) values in ascending order:\[54,65,67,68,68,69,70,71,73,73,75,76,\mathbf{77},77,78,78,80,81,81,83,85,85,85,85,87 \]\[\frac{n+1}{2} = \frac{26}{2} = 13\text{th term} \]\[\text{Median} = 77 \]\(\displaystyle 85\) occurs \(\displaystyle 4\) times, more than any other value.\[\text{Mode} = 85 \]Answer: mean \(\displaystyle =75.64\), median \(\displaystyle =77\), mode \(\displaystyle =85\) mg/dl.