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NCERT Exemplar · Class 10 Mathematics Statistics and Probability

96 questions · 96 still being checked

EXERCISE 13.4 11–14 (part 11 of 11)

  1. Exercise 11

    Size of agricultural holdings in a survey of 200\displaystyle 200 families is given in the following Compute median and mode size of the holdings.
    Size of agricultural holdings (in hec)Number of families
    0\displaystyle 0-5\displaystyle 510\displaystyle 10
    5\displaystyle 5-10\displaystyle 1015\displaystyle 15
    10\displaystyle 10-15\displaystyle 1530\displaystyle 30
    15\displaystyle 15-20\displaystyle 2080\displaystyle 80
    20\displaystyle 20-25\displaystyle 2540\displaystyle 40
    25\displaystyle 25-30\displaystyle 3020\displaystyle 20
    30\displaystyle 30-35\displaystyle 355\displaystyle 5

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    \[N=200,\quad \dfrac{N}{2}=100 \] Median class $\displaystyle 15$-$\displaystyle 20$: \(\displaystyle l=15,\ cf=55,\ f=80,\ h=5\). \[\text{Median}=15+\left(\dfrac{100-55}{80}\right)5=15+\dfrac{225}{80}=17.8125 \] Modal class $\displaystyle 15$-$\displaystyle 20$ (highest frequency): \(\displaystyle f_1=80,\ f_0=30,\ f_2=40\). \[\text{Mode}=15+\left(\dfrac{80-30}{2(80)-30-40}\right)5=15+\dfrac{250}{90}=17.78 \] Answer: Median \(\displaystyle \approx17.81\) hectares, Mode \(\displaystyle \approx17.78\) hectares.

  2. Exercise 12

    The annual rainfall record of a city for 66\displaystyle 66 days is given in the following table.
    Rainfall (in cm)0\displaystyle 0-10\displaystyle 1010\displaystyle 10-20\displaystyle 2020\displaystyle 20-30\displaystyle 3030\displaystyle 30-40\displaystyle 4040\displaystyle 40-50\displaystyle 5050\displaystyle 50-60\displaystyle 60
    Number of days22\displaystyle 2210\displaystyle 108\displaystyle 815\displaystyle 155\displaystyle 56\displaystyle 6
    Calculate the median rainfall using ogives (or move than type and of less than type)

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    NCERT’s answer
    Median rainfall \(\displaystyle =21.25\) cm
    \[N=66,\quad \dfrac{N}{2}=33 \] Less than type: \(\displaystyle 10\!:\!22,\ 20\!:\!32,\ 30\!:\!40,\ 40\!:\!55,\ 50\!:\!60,\ 60\!:\!66\). More than type: \(\displaystyle 0\!:\!66,\ 10\!:\!44,\ 20\!:\!34,\ 30\!:\!26,\ 40\!:\!11,\ 50\!:\!6\). NCERT_Solution_Class10_Maths_Exemplar_Ch13_Ex13-4_Q12 Curves meet near \(\displaystyle x=21.25\). Median class $\displaystyle 20$-$\displaystyle 30$: \(\displaystyle l=20,\ cf=32,\ f=8,\ h=10\). \[\text{Median}=20+\left(\dfrac{33-32}{8}\right)10=21.25 \] Answer: Median rainfall \(\displaystyle =21.25\) cm.

  3. Exercise 13

    The following is the frequency distribution of duration for 100\displaystyle 100 calls made on a mobile phone.
    Duration (in s)Number of calls
    95\displaystyle 95-125\displaystyle 12514\displaystyle 14
    125\displaystyle 125-155\displaystyle 15522\displaystyle 22
    155\displaystyle 155-185\displaystyle 18528\displaystyle 28
    185\displaystyle 185-215\displaystyle 21521\displaystyle 21
    215\displaystyle 215-245\displaystyle 24515\displaystyle 15
    [Find the average duration of a call.]

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

    NCERT’s answer
    average \(\displaystyle =170.3\) sec.
    \[\begin{array}{c|ccccc} x & 110 & 140 & 170 & 200 & 230 \\ f & 14 & 22 & 28 & 21 & 15 \\ fx & 1540 & 3080 & 4760 & 4200 & 3450 \end{array} \] \[\sum f=100,\quad \sum fx=17030 \] \[\text{Mean}=\dfrac{\sum fx}{\sum f}=\dfrac{17030}{100}=170.3 \] Answer: Average duration of a call \(\displaystyle =170.3\) s.

  4. Exercise 14

    50\displaystyle 50 students enter for a school javelin throw competition. The distance (in metre) thrown are recorded below
    Distance (in m)0\displaystyle 0-20\displaystyle 2020\displaystyle 20-40\displaystyle 4040\displaystyle 40-60\displaystyle 6060\displaystyle 60-80\displaystyle 8080\displaystyle 80-100\displaystyle 100
    Number of students6\displaystyle 611\displaystyle 1117\displaystyle 1712\displaystyle 124\displaystyle 4
    (i)
    Construct a cumulative frequency table.
    (ii)
    Draw a cumulative frequency curve (less than type) and calculate the median distance drawn by using this curve.
    (iii)
    Calculate the median distance by using the formula for median.
    (iv)
    Are the median distance calculated in (ii) and (iii) same?

    Matches the book, not yet reviewed

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

    NCERT’s answer
    (i)
    Distance (in m)No. of studentsCummulative frequency
    $\displaystyle 0$-$\displaystyle 20$$\displaystyle 6$$\displaystyle 6$
    $\displaystyle 20$-$\displaystyle 40$$\displaystyle 11$$\displaystyle 17$
    $\displaystyle 40$-$\displaystyle 60$$\displaystyle 17$$\displaystyle 34$
    $\displaystyle 60$-$\displaystyle 80$$\displaystyle 12$$\displaystyle 46$
    $\displaystyle 80$-$\displaystyle 100$$\displaystyle 4$$\displaystyle 50$
    (iii)
    49.$\displaystyle 41$ m.
    (i)
    Less than $\displaystyle 20$: $\displaystyle 6$; less than $\displaystyle 40$: $\displaystyle 17$; less than $\displaystyle 60$: $\displaystyle 34$; less than $\displaystyle 80$: $\displaystyle 46$; less than $\displaystyle 100$: 50.
    (ii)
    NCERT_Solution_Class10_Maths_Exemplar_Ch13_Ex13-4_Q14
    \[N=50,\quad \dfrac{N}{2}=25 \]
    Curve meets \(\displaystyle y=25\) near \(\displaystyle x\approx49\)-\(\displaystyle 50\).
    (iii)
    Median class $\displaystyle 40$-$\displaystyle 60$: \(\displaystyle l=40,\ cf=17,\ f=17,\ h=20\).
    \[\text{Median}=40+\left(\dfrac{25-17}{17}\right)20=40+\dfrac{160}{17}\approx49.41 \]
    (iv)
    The two values agree, within the precision of reading a hand-drawn curve.
    Answer: Median \(\displaystyle \approx49.41\) m; (iv) yes.