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NCERT Exemplar · Class 9 Mathematics Statistics and Probability

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EXERCISE 14.3 1–10 (part 5 of 7)

  1. Exercise 1

    The blood groups of 30\displaystyle 30 students are recorded as follows: A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B, A, A, O, A, AB, B, A, O, B, A, B, A Prepare a frequency distribution table for the data.

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    \[A:12,\quad B:8,\quad O:6,\quad AB:4 \]
    Blood groupABOAB
    Number of students$\displaystyle 12$$\displaystyle 8$$\displaystyle 6$$\displaystyle 4$
    Answer: frequency table above; \(\displaystyle 12+8+6+4=30\).

  2. Exercise 2

    The value of π\displaystyle \pi upto 35\displaystyle 35 decimal places is given below: 3.14159265358979323846264338327950288\displaystyle 3.14159265358979323846264338327950288 Make a frequency distribution of the digits 0\displaystyle 0 to 9\displaystyle 9 after the decimal point.

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    NCERT’s answer
    Digit$\displaystyle 0$$\displaystyle 1$$\displaystyle 2$$\displaystyle 3$$\displaystyle 4$$\displaystyle 5$$\displaystyle 6$$\displaystyle 7$$\displaystyle 8$$\displaystyle 9$
    Frequency$\displaystyle 1$$\displaystyle 2$$\displaystyle 5$$\displaystyle 6$$\displaystyle 3$$\displaystyle 4$$\displaystyle 3$$\displaystyle 2$$\displaystyle 5$$\displaystyle 4$
    \[0:1,\quad 1:2,\quad 2:5,\quad 3:6,\quad 4:3,\quad 5:4,\quad 6:3,\quad 7:2,\quad 8:5,\quad 9:4 \]
    Digit$\displaystyle 0$$\displaystyle 1$$\displaystyle 2$$\displaystyle 3$$\displaystyle 4$$\displaystyle 5$$\displaystyle 6$$\displaystyle 7$$\displaystyle 8$$\displaystyle 9$
    Frequency$\displaystyle 1$$\displaystyle 2$$\displaystyle 5$$\displaystyle 6$$\displaystyle 3$$\displaystyle 4$$\displaystyle 3$$\displaystyle 2$$\displaystyle 5$$\displaystyle 4$
    Answer: table above; \(\displaystyle 1+2+5+6+3+4+3+2+5+4=35\) digits.

  3. Exercise 3

    The scores (out of 100\displaystyle 100) obtained by 33\displaystyle 33 students in a mathematics test are as follows: 69\displaystyle 69, 48\displaystyle 48, 84\displaystyle 84, 58\displaystyle 58, 48\displaystyle 48, 73\displaystyle 73, 83\displaystyle 83, 48\displaystyle 48, 66\displaystyle 66, 58\displaystyle 58, 84\displaystyle 84 66\displaystyle 66, 64\displaystyle 64, 71\displaystyle 71, 64\displaystyle 64, 66\displaystyle 66, 69\displaystyle 69, 66\displaystyle 66, 83\displaystyle 83, 66\displaystyle 66, 69\displaystyle 69, 71\displaystyle 71 81\displaystyle 81, 71\displaystyle 71, 73\displaystyle 73, 69\displaystyle 69, 66\displaystyle 66, 66\displaystyle 66, 64\displaystyle 64, 58\displaystyle 58, 64\displaystyle 64, 69\displaystyle 69, 69\displaystyle 69 Represent this data in the form of a frequency distribution.

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    NCERT’s answer
    Scores$\displaystyle 48$$\displaystyle 58$$\displaystyle 64$$\displaystyle 66$$\displaystyle 69$$\displaystyle 71$$\displaystyle 73$$\displaystyle 81$$\displaystyle 83$$\displaystyle 84$
    Frequency$\displaystyle 3$$\displaystyle 3$$\displaystyle 4$$\displaystyle 7$$\displaystyle 6$$\displaystyle 3$$\displaystyle 2$$\displaystyle 1$$\displaystyle 2$$\displaystyle 2$
    \[48:3,\quad 58:3,\quad 64:4,\quad 66:7,\quad 69:6,\quad 71:3,\quad 73:2,\quad 81:1,\quad 83:2,\quad 84:2 \]
    Score$\displaystyle 48$$\displaystyle 58$$\displaystyle 64$$\displaystyle 66$$\displaystyle 69$$\displaystyle 71$$\displaystyle 73$$\displaystyle 81$$\displaystyle 83$$\displaystyle 84$
    Frequency$\displaystyle 3$$\displaystyle 3$$\displaystyle 4$$\displaystyle 7$$\displaystyle 6$$\displaystyle 3$$\displaystyle 2$$\displaystyle 1$$\displaystyle 2$$\displaystyle 2$
    Answer: table above; \(\displaystyle 3+3+4+7+6+3+2+1+2+2=33\) students.

  4. Exercise 4

    Prepare a continuous grouped frequency distribution from the following data:
    Mid-pointFrequency
    5\displaystyle 54\displaystyle 4
    15\displaystyle 158\displaystyle 8
    25\displaystyle 2513\displaystyle 13
    35\displaystyle 3512\displaystyle 12
    45\displaystyle 456\displaystyle 6
    Also find the size of class intervals.

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    NCERT’s answer
    Class$\displaystyle 0$-$\displaystyle 10$$\displaystyle 10$-$\displaystyle 20$$\displaystyle 20$-$\displaystyle 30$$\displaystyle 30$-$\displaystyle 40$$\displaystyle 40$-$\displaystyle 50$
    Frequency$\displaystyle 4$$\displaystyle 8$$\displaystyle 13$$\displaystyle 12$$\displaystyle 6$
    Class size $\displaystyle =10$
    \[h = 15-5 = 10 \]\[\text{class interval} = \left(m-\tfrac h2,\ m+\tfrac h2\right) \]
    Class interval$\displaystyle 0$-$\displaystyle 10$$\displaystyle 10$-$\displaystyle 20$$\displaystyle 20$-$\displaystyle 30$$\displaystyle 30$-$\displaystyle 40$$\displaystyle 40$-$\displaystyle 50$
    Frequency$\displaystyle 4$$\displaystyle 8$$\displaystyle 13$$\displaystyle 12$$\displaystyle 6$
    Answer: class size \(\displaystyle 10\); table above.

  5. Exercise 5

    Convert the given frequency distribution into a continuous grouped frequency distribution:
    Class intervalFrequency
    150\displaystyle 150-153\displaystyle 1537\displaystyle 7
    154\displaystyle 154-157\displaystyle 1577\displaystyle 7
    158\displaystyle 158-161\displaystyle 16115\displaystyle 15
    162\displaystyle 162-165\displaystyle 16510\displaystyle 10
    166\displaystyle 166-169\displaystyle 1695\displaystyle 5
    170\displaystyle 170-173\displaystyle 1736\displaystyle 6
    In which intervals would 153.5\displaystyle 153.5 and 157.5\displaystyle 157.5 be included?

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    NCERT’s answer
    Class intervalsFrequency
    $\displaystyle 149.5$-$\displaystyle 153.5$$\displaystyle 7$
    $\displaystyle 153.5$-$\displaystyle 157.5$$\displaystyle 7$
    $\displaystyle 157.5$-$\displaystyle 161.5$$\displaystyle 15$
    $\displaystyle 161.5$-$\displaystyle 165.5$$\displaystyle 10$
    $\displaystyle 165.5$-$\displaystyle 169.5$$\displaystyle 5$
    $\displaystyle 169.5$-$\displaystyle 173.5$$\displaystyle 6$
    $\displaystyle 153.5$ is included in the class interval $\displaystyle 153.5$-$\displaystyle 157.5$ and $\displaystyle 157.5$ in $\displaystyle 157.5$-161.5.
    \[d = \frac{154-153}{2} = 0.5 \]
    Class interval$\displaystyle 149.5$-$\displaystyle 153.5$$\displaystyle 153.5$-$\displaystyle 157.5$$\displaystyle 157.5$-$\displaystyle 161.5$$\displaystyle 161.5$-$\displaystyle 165.5$$\displaystyle 165.5$-$\displaystyle 169.5$$\displaystyle 169.5$-$\displaystyle 173.5$
    Frequency$\displaystyle 7$$\displaystyle 7$$\displaystyle 15$$\displaystyle 10$$\displaystyle 5$$\displaystyle 6$
    Each class runs \(\displaystyle [\text{lower},\text{upper})\), so a boundary value opens the next class: \(\displaystyle 153.5\) starts \(\displaystyle 153.5\)-\(\displaystyle 157.5\); \(\displaystyle 157.5\) starts \(\displaystyle 157.5\)-\(\displaystyle 161.5\).Answer: \(\displaystyle 153.5\) in \(\displaystyle 153.5\)-\(\displaystyle 157.5\); \(\displaystyle 157.5\) in \(\displaystyle 157.5\)-\(\displaystyle 161.5\).

  6. Exercise 6

    The expenditure of a family on different heads in a month is given below:
    HeadFoodEducationClothingHouse RentOthersSavings
    Expenditure (in Rs)4000\displaystyle 40002500\displaystyle 25001000\displaystyle 10003500\displaystyle 35002500\displaystyle 25001500\displaystyle 1500
    Draw a bar graph to represent the data above.

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    Six heads on the horizontal axis, numbered \(\displaystyle 1\)–\(\displaystyle 6\); expenditure (Rs) on the vertical axis. Equal-width bars, equal gaps.NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-3_Q6Key: \(\displaystyle 1=\)Food, \(\displaystyle 2=\)Education, \(\displaystyle 3=\)Clothing, \(\displaystyle 4=\)House Rent, \(\displaystyle 5=\)Others, \(\displaystyle 6=\)Savings.Answer: bar graph, heights \(\displaystyle 4000,\,2500,\,1000,\,3500,\,2500,\,1500\) (Rs) over heads \(\displaystyle 1\)–\(\displaystyle 6\) as keyed.

  7. Exercise 7

    Expenditure on Education of a country during a five year period (2002\displaystyle 2002-2006\displaystyle 2006), in crores of rupees, is given below:
    Elementary education240\displaystyle 240
    Secondary Education120\displaystyle 120
    University Education190\displaystyle 190
    Teacher's Training20\displaystyle 20
    Social Education10\displaystyle 10
    Other Educational Programmes115\displaystyle 115
    Cultural programmes25\displaystyle 25
    Technical Education125\displaystyle 125
    Represent the information above by a bar graph.

    Not cross-checked

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

    Eight programmes on the horizontal axis, numbered \(\displaystyle 1\)–\(\displaystyle 8\); expenditure (Rs crore) on the vertical axis. Equal-width bars, equal gaps.NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-3_Q7Key: \(\displaystyle 1=\)Elementary, \(\displaystyle 2=\)Secondary, \(\displaystyle 3=\)University, \(\displaystyle 4=\)Teacher's Training, \(\displaystyle 5=\)Social, \(\displaystyle 6=\)Other Programmes, \(\displaystyle 7=\)Cultural, \(\displaystyle 8=\)Technical.Answer: bar graph, heights \(\displaystyle 240,\,120,\,190,\,20,\,10,\,115,\,25,\,125\) (Rs crore) over programmes \(\displaystyle 1\)–\(\displaystyle 8\) as keyed.

  8. Exercise 8

    The following table gives the frequencies of most commonly used letters a,e,i,o\displaystyle a, e, i, o, r,t,u\displaystyle r, t, u from a page of a book :
    Lettersa\displaystyle ae\displaystyle ei\displaystyle io\displaystyle or\displaystyle rt\displaystyle tu\displaystyle u
    Frequency75\displaystyle 75125\displaystyle 12580\displaystyle 8070\displaystyle 7080\displaystyle 8095\displaystyle 9575\displaystyle 75
    Represent the information above by a bar graph.

    Not cross-checked

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

    Seven letters on the horizontal axis, numbered \(\displaystyle 1\)–\(\displaystyle 7\); frequency on the vertical axis. Equal-width bars, equal gaps.NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-3_Q8Key: \(\displaystyle 1=a,\ 2=e,\ 3=i,\ 4=o,\ 5=r,\ 6=t,\ 7=u\).Answer: bar graph, heights \(\displaystyle 75,\,125,\,80,\,70,\,80,\,95,\,75\) for letters \(\displaystyle a,e,i,o,r,t,u\) as keyed.

  9. Exercise 9

    If the mean of the following data is 20.2\displaystyle 20.2, find the value of p\displaystyle p :
    x\displaystyle \boldsymbol{x}10\displaystyle 1015\displaystyle 1520\displaystyle 2025\displaystyle 2530\displaystyle 30
    f\displaystyle f6\displaystyle 68\displaystyle 8p\displaystyle p10\displaystyle 106\displaystyle 6

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    NCERT’s answer
    $\displaystyle 20$
    \[\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \] \[\sum f_i x_i = 10(6)+15(8)+20(p)+25(10)+30(6) = 610+20p \] \[\sum f_i = 6+8+p+10+6 = 30+p \] \[\frac{610+20p}{30+p} = 20.2 \] \[610+20p = 606+20.2p \] \[0.2p = 4 \]Answer: \(\displaystyle p = 20\)

  10. Exercise 10

    Obtain the mean of the following distribution:
    FrequencyVariable
    4\displaystyle 44\displaystyle 4
    8\displaystyle 86\displaystyle 6
    14\displaystyle 148\displaystyle 8
    11\displaystyle 1110\displaystyle 10
    3\displaystyle 312\displaystyle 12

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    NCERT’s answer
    8.$\displaystyle 05$
    \[\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \] \[\sum f_i = 4+8+14+11+3 = 40 \] \[\sum f_i x_i = 4(4)+8(6)+14(8)+11(10)+3(12) = 322 \] \[\text{Mean} = \frac{322}{40} = 8.05 \]Answer: Mean \(\displaystyle = 8.05\).