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NCERT Exemplar · Class 9 Mathematics Statistics and Probability

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EXERCISE 14.1 1–10 (part 1 of 7)

  1. Write the correct answer in each of the following :

    Exercise 1

    The class mark of the class 90\displaystyle 90-120\displaystyle 120 is : (A) 90\displaystyle 90 (B) 105\displaystyle 105 (C) 115\displaystyle 115 (D) 120\displaystyle 120

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 105\) \[\text{Class mark} = \frac{\text{lower limit} + \text{upper limit}}{2} = \frac{90+120}{2} = 105 \]

  2. Exercise 2

    The range of the data : 25,18,20,22,16,6,17,15,12,30,32,10,19,8,11,20\displaystyle 25,18,20,22,16,6,17,15,12,30,32,10,19,8,11,20 is (A) 10\displaystyle 10 (B) 15\displaystyle 15 (C) 18\displaystyle 18 (D) 26\displaystyle 26

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 26\) \[\text{Range} = \max(x_i) - \min(x_i) = 32 - 6 = 26 \]

  3. Exercise 3

    In a frequency distribution, the mid value of a class is 10\displaystyle 10 and the width of the class is 6\displaystyle 6 . The lower limit of the class is : (A) 6\displaystyle 6 (B) 7\displaystyle 7 (C) 8\displaystyle 8 (D) 12\displaystyle 12

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 7\) \[\text{Lower limit} = m - \frac{h}{2} = 10 - \frac{6}{2} = 7 \]

  4. Exercise 4

    The width of each of five continuous classes in a frequency distribution is 5\displaystyle 5 and the lower class-limit of the lowest class is 10\displaystyle 10 . The upper class-limit of the highest class is: (A) 15\displaystyle 15 (B) 25\displaystyle 25 (C) 35\displaystyle 35 (D) 40\displaystyle 40

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 35\) \[\text{Upper limit of highest class} = 10 + 5\times 5 = 35 \]

  5. Exercise 5

    Let m\displaystyle m be the mid-point and l\displaystyle l be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is : (A) 2m+l\displaystyle 2 m+l (B) 2m−l\displaystyle 2 m-l (C) m−l\displaystyle m-l (D) m−2l\displaystyle m-2 l

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 2m-l\) \[m = \frac{a+l}{2} \quad\Rightarrow\quad a = 2m-l \]

  6. Exercise 6

    The class marks of a frequency distribution are given as follows : 15\displaystyle 15, 20\displaystyle 20, 25\displaystyle 25, ... The class corresponding to the class mark 20\displaystyle 20 is : (A) 12.5–17.5\displaystyle 12.5\text{–}17.5 (B) 17.5–22.5\displaystyle 17.5\text{–}22.5 (C) 18.5–21.5\displaystyle 18.5\text{–}21.5 (D) 19.5–20.5\displaystyle 19.5\text{–}20.5

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 17.5\text{–}22.5\) \[h = 20-15 = 5 \] \[\text{Class} = \left(20-\tfrac{h}{2},\, 20+\tfrac{h}{2}\right) = (17.5,\ 22.5) \]
  7. The class corresponding to the class mark $\displaystyle 20$ is :

    Exercise 7

    In the class intervals 10\displaystyle 10-20\displaystyle 20, 20\displaystyle 20-30\displaystyle 30, the number 20\displaystyle 20 is included in : (A) 10−20\displaystyle 10-20 (B) 20−30\displaystyle 20-30 (C) both the intervals (D) none of these intervals

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 20\text{–}30\) The upper limit of a class is excluded, the lower limit of the next class is included. \[20 \in [20,\,30) \]

  8. Exercise 8

    A grouped frequency table with class intervals of equal sizes using 250\displaystyle 250-270\displaystyle 270 (270\displaystyle 270 not included in this interval) as one of the class interval is constructed for the following data : 268\displaystyle 268, 220\displaystyle 220, 368\displaystyle 368, 258\displaystyle 258, 242\displaystyle 242, 310\displaystyle 310, 272\displaystyle 272, 342\displaystyle 342, 310\displaystyle 310, 290\displaystyle 290, 300\displaystyle 300, 320\displaystyle 320, 319\displaystyle 319, 304\displaystyle 304, 402\displaystyle 402, 318\displaystyle 318, 406\displaystyle 406, 292\displaystyle 292, 354\displaystyle 354, 278\displaystyle 278, 210\displaystyle 210, 240\displaystyle 240, 330\displaystyle 330, 316\displaystyle 316, 406\displaystyle 406, 215\displaystyle 215, 258\displaystyle 258, 236. The frequency of the class 310\displaystyle 310-330\displaystyle 330 is: (A) 4\displaystyle 4 (B) 5\displaystyle 5 (C) 6\displaystyle 6 (D) 7\displaystyle 7

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 6\) \[h = 270-250 = 20 \] \[\{310,310,320,319,318,316\} \subset [310,330) \] \[\text{Frequency} = 6 \]Answer: \(\displaystyle 6\).
  9. The frequency of the class $\displaystyle 310$-$\displaystyle 330$ is:

    Exercise 9

    A grouped frequency distribution table with classes of equal sizes using 63\displaystyle 63-72\displaystyle 72 (72\displaystyle 72 included) as one of the class is constructed for the following data : 30\displaystyle 30, 32\displaystyle 32, 45\displaystyle 45, 54\displaystyle 54, 74\displaystyle 74, 78\displaystyle 78, 108\displaystyle 108, 112\displaystyle 112, 66\displaystyle 66, 76\displaystyle 76, 88\displaystyle 88, 40\displaystyle 40, 14\displaystyle 14, 20\displaystyle 20, 15\displaystyle 15, 35\displaystyle 35, 44\displaystyle 44, 66\displaystyle 66, 75\displaystyle 75, 84\displaystyle 84, 95\displaystyle 95, 96\displaystyle 96, 102\displaystyle 102, 110\displaystyle 110, 88\displaystyle 88, 74\displaystyle 74, 112\displaystyle 112, 14\displaystyle 14, 34\displaystyle 34, 44. The number of classes in the distribution will be : (A) 9\displaystyle 9 (B) 10\displaystyle 10 (C) 11\displaystyle 11 (D) 12\displaystyle 12

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 10\) \[h = 72-63+1 = 10 \] \[13\text{–}22,\ 23\text{–}32,\ \dots,\ 103\text{–}112 \] \[\text{Number of classes} = 10 \]Answer: \(\displaystyle 10\).
  10. The number of classes in the distribution will be :

    Exercise 10

    To draw a histogram to represent the following frequency distribution :
    Class interval5\displaystyle 5-10\displaystyle 1010\displaystyle 10-15\displaystyle 1515\displaystyle 15-25\displaystyle 2525\displaystyle 25-45\displaystyle 4545\displaystyle 45-75\displaystyle 75
    Frequency6\displaystyle 612\displaystyle 1210\displaystyle 108\displaystyle 815\displaystyle 15
    the adjusted frequency for the class 25\displaystyle 25-45\displaystyle 45 is : (A) 6\displaystyle 6 (B) 5\displaystyle 5 (C) 3\displaystyle 3 (D) 2\displaystyle 2

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 2\) \[\text{Adjusted frequency} = \text{frequency} \times \frac{\text{minimum class width}}{\text{class width}} \] \[= 8 \times \frac{5}{20} = 2 \]