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SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Mathematics Statistics and Probability

96 questions · 96 still being checked

EXERCISE 13.1 1–10 (part 1 of 11)

  1. Choose the correct answer from the given four options:

    Exercise 1

    In the formula xˉ=a+∑fidi∑fi,\bar{x}=a+\frac{\sum f_i d_i}{\sum f_i}, for finding the mean of grouped data di\displaystyle d_i's are deviations from a\displaystyle a of
    (A)
    lower limits of the classes
    (B)
    upper limits of the classes
    (C)
    mid points of the classes
    (D)
    frequencies of the class marks

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    NCERT’s answer
    (C)
    (C) mid points of the classes.In the assumed-mean method the deviation of each class from the assumed mean \(\displaystyle a\) is \[d_i = x_i - a \] where \(\displaystyle x_i\) is the class mark (mid point) of the \(\displaystyle i\)-th class.

  2. Exercise 2

    While computing mean of grouped data, we assume that the frequencies are
    (A)
    evenly distributed over all the classes
    (B)
    centred at the classmarks of the classes
    (C)
    centred at the upper limits of the classes
    (D)
    centred at the lower limits of the classes

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    NCERT’s answer
    (B)
    (B) centred at the classmarks of the classes.Each class's \(\displaystyle f_i\) observations are assumed equal to its mid-point \(\displaystyle x_i\), since only class limits and frequencies are known, not the individual observations.

  3. Exercise 3

    If xi\displaystyle x_i's are the mid points of the class intervals of grouped data, fi\displaystyle f_i's are the corresponding frequencies and xˉ\displaystyle \bar{x} is the mean, then ∑(fixi−xˉ)\displaystyle \sum\left(f_i x_i-\bar{x}\right) is equal to
    (A)
    0\displaystyle 0 (B) -1\displaystyle 1 (C) 1\displaystyle 1 (D) 2\displaystyle 2

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    (A)
    (A) \(\displaystyle 0\), reading the expression as \(\displaystyle \sum f_i(x_i-\bar{x})\).\[\bar{x} = \frac{\sum f_i x_i}{\sum f_i} \] \[\sum f_i(x_i-\bar{x}) = \sum f_i x_i - \bar{x}\sum f_i \] \[= \sum f_i x_i - \sum f_i x_i = 0 \]

  4. Exercise 4

    In the formula xˉ=a+h∑fiui∑fi\displaystyle \bar{x}=a+h \quad \frac{\sum f_i u_i}{\sum f_i} \quad, for finding the mean of grouped frequency distribution, ui=\displaystyle u_i=
    (A)
    xi+ah\displaystyle \frac{x_i+a}{h}
    (B)
    h(xi−a)\displaystyle h\left(x_i-a\right)
    (C)
    xi−ah\displaystyle \frac{x_i-a}{h}
    (D)
    a−xih\displaystyle \frac{a-x_i}{h}

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    (C)
    (C) \(\displaystyle \frac{x_i-a}{h}\).The step-deviation method divides the assumed-mean deviation by the class size \(\displaystyle h\). \[d_i = x_i - a \] \[u_i = \frac{d_i}{h} = \frac{x_i - a}{h} \]

  5. Exercise 5

    The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
    (A)
    mean
    (B)
    median
    (C)
    mode
    (D)
    all the three above

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    NCERT’s answer
    (B)
    (B) median.The less-than and more-than ogives meet where their cumulative frequencies are equal, i.e. at \(\displaystyle \frac{N}{2}\) observations -- the point that defines the median.

  6. Exercise 6

    For the following distribution :
    Class0\displaystyle 0-5\displaystyle 55\displaystyle 5-10\displaystyle 1010\displaystyle 10-15\displaystyle 1515\displaystyle 15-20\displaystyle 2020\displaystyle 20-25\displaystyle 25
    Frequency10\displaystyle 1015\displaystyle 1512\displaystyle 1220\displaystyle 209\displaystyle 9
    the sum of lower limits of the median class and modal class is
    (A)
    15\displaystyle 15
    (B)
    25\displaystyle 25
    (C)
    30\displaystyle 30
    (D)
    35\displaystyle 35

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    NCERT’s answer
    (B)
    (B) 25.\[N = 10+15+12+20+9 = 66, \qquad \frac{N}{2}=33 \] Cumulative frequencies: \(\displaystyle 10,\ 25,\ 37,\ 57,\ 66\). \[\text{Median class } 10\text{-}15 \quad (\text{cf}=37 \ge 33) \] \[\text{Modal class } 15\text{-}20 \quad (f=20 \text{ is the highest frequency}) \] \[10+15=25 \]

  7. Exercise 7

    Consider the following frequency distribution :
    Class0\displaystyle 0-5\displaystyle 56\displaystyle 6-11\displaystyle 1112\displaystyle 12-17\displaystyle 1718\displaystyle 18-23\displaystyle 2324\displaystyle 24-29\displaystyle 29
    Frequency13\displaystyle 1310\displaystyle 1015\displaystyle 158\displaystyle 811\displaystyle 11
    The upper limit of the median class is
    (A)
    17\displaystyle 17
    (B)
    17.5\displaystyle 5
    (C)
    18\displaystyle 18
    (D)
    18.5\displaystyle 5

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    NCERT’s answer
    (B)
    (B) 17.5.The classes are inclusive; their continuous (exclusive) limits are \[-0.5\text{-}5.5,\ 5.5\text{-}11.5,\ 11.5\text{-}17.5,\ 17.5\text{-}23.5,\ 23.5\text{-}29.5 \] \[N = 13+10+15+8+11 = 57, \qquad \frac{N}{2}=28.5 \] Cumulative frequencies: \(\displaystyle 13,\ 23,\ 38,\ 46,\ 57\). \[\text{Median class } 11.5\text{-}17.5 \quad (\text{cf}=38 \ge 28.5) \] Upper limit \(\displaystyle =17.5\).

  8. Exercise 8

    For the following distribution :
    MarksNumber of students
    Below 10\displaystyle 103\displaystyle 3
    Below 20\displaystyle 2012\displaystyle 12
    Below 30\displaystyle 3027\displaystyle 27
    Below 40\displaystyle 4057\displaystyle 57
    Below 50\displaystyle 5075\displaystyle 75
    Below 60\displaystyle 6080\displaystyle 80
    the modal class is
    (A)
    10\displaystyle 10-20\displaystyle 20
    (B)
    20\displaystyle 20-30\displaystyle 30
    (C)
    30\displaystyle 30-40\displaystyle 40
    (D)
    50\displaystyle 50-60\displaystyle 60

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    (C)
    (C) \(\displaystyle 30\text{-}40\) \[f_{10\text{-}20}=12-3=9,\ f_{20\text{-}30}=27-12=15,\ f_{30\text{-}40}=57-27=30,\ f_{40\text{-}50}=75-57=18,\ f_{50\text{-}60}=80-75=5 \] The modal class has the highest frequency.

  9. Exercise 9

    Consider the data :
    Class65\displaystyle 65-85\displaystyle 8585\displaystyle 85-105\displaystyle 105105\displaystyle 105-125\displaystyle 125125\displaystyle 125-145\displaystyle 145145\displaystyle 145-165\displaystyle 165165\displaystyle 165-185\displaystyle 185185\displaystyle 185-205\displaystyle 205
    Frequency4\displaystyle 45\displaystyle 513\displaystyle 1320\displaystyle 2014\displaystyle 147\displaystyle 74\displaystyle 4
    The difference of the upper limit of the median class and the lower limit of the modal class is
    (A)
    0\displaystyle 0 (B) 19\displaystyle 19
    (C)
    20\displaystyle 20
    (D)
    38\displaystyle 38

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 20\) \[N=4+5+13+20+14+7+4=67,\quad \frac{N}{2}=33.5 \] \[\text{cf}:\ 4,\ 9,\ 22,\ 42,\ 56,\ 63,\ 67 \] Median class \(\displaystyle 125\text{-}145\) (first cf exceeding \(\displaystyle 33.5\)); upper limit \(\displaystyle 145\). Modal class \(\displaystyle 125\text{-}145\) (largest frequency \(\displaystyle 20\)); lower limit \(\displaystyle 125\). \[145-125=20 \]

  10. Exercise 10

    The times, in seconds, taken by 150\displaystyle 150 atheletes to run a 110\displaystyle 110 m hurdle race are tabulated below :
    Class13.8\displaystyle 13.8-14\displaystyle 1414\displaystyle 14-14.2\displaystyle 14.214.2\displaystyle 14.2-14.4\displaystyle 14.414.4\displaystyle 14.4-14.6\displaystyle 14.614.6\displaystyle 14.6-14.8\displaystyle 14.814.8\displaystyle 14.8-15\displaystyle 15
    Frequency2\displaystyle 24\displaystyle 45\displaystyle 571\displaystyle 7148\displaystyle 4820\displaystyle 20
    The number of atheletes who completed the race in less then 14.6\displaystyle 14.6 seconds is :
    (A)
    11\displaystyle 11
    (B)
    71\displaystyle 71
    (C)
    82\displaystyle 82
    (D)
    130\displaystyle 130

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    (C)
    (C) \(\displaystyle 82\) \[2+4+5+71=82 \] Sum of the frequencies of the classes \(\displaystyle 13.8\text{-}14,\ 14\text{-}14.2,\ 14.2\text{-}14.4,\ 14.4\text{-}14.6\), all ending at or before \(\displaystyle 14.6\text{ s}\).