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Statistics Test - 16

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Statistics Test - 16
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  • Question 1
    1 / -0
    Given median $$= 99.6$$ and mean $$= 101.2$$, then the mode is 96.4
    If true then enter $$1$$ and if false then enter $$0$$
    Solution
    Given, Median = $$99.6$$ and Mean = $$101.2$$
    $$Mode = 3 Median - 2 Mean$$
    $$Mode = 3 \times 99.6 - 2 \times 101.2$$
    $$Mode = 96.4$$
  • Question 2
    1 / -0
     Variable $$4$$$$8$$  $$9$$ $$10$$$$11$$ $$12$$ $$13$$ 
     Frequency $$3$$$$4$$$$5$$ $$8$$ $$4$$ $$3$$ $$2$$ 
    Mode of the distribution 


    Solution
    We know mode of a distribution is the term corresponding to maximum frequency.
    Thus in the given distribution $$8$$ is the maximum frequency and corresponding mode is $$10$$. 
  • Question 3
    1 / -0
    Calculate the median for the following data.
    Marks (out of 60.)$$32$$$$27$$$$26$$$$24$$$$23$$$$21$$
    Number of students $$6$$$$4$$$$7$$$$9$$$$16$$$$2$$
     
    Solution
    Total number of students $$= 6 + 4 + 7 + 9 + 16 + 2 = 44$$

    Since, the number of students is even, the median will be mean of $${(\cfrac{n}{2})}^{th}$$ and $${(\cfrac{n+2}{2})}^{th}$$ number 

    Median = Mean of $$(\cfrac{44}{2})^{th}$$ and $$(\cfrac{44+2}{2})^{th}$$

    Here, $$22^{nd}$$ and $$23^{rd}$$ number is $$24$$ Kg if arranged in ascending order.

    Hence, median = $$\cfrac{24 + 24}{2}= 24$$ kg
  • Question 4
    1 / -0
    The median for the following frequency distribution is:
    $$x:$$123456789
    $$f:$$810111620251596

    Solution
    here,
    $$\Sigma f =120$$
    $$N=120 \Rightarrow even$$
    $$\dfrac N2=\dfrac {120} {2}=60$$
    In cumulative frequency we select $$65$$ as it is closer to $$60$$.
    Corresponding $$x $$ value is $$5$$
    hence option $$B$$ is correct.

  • Question 5
    1 / -0
    Determine the mode of the following data.
    Marks1016121913201114
    Number of Students33426152
    Solution
    maximum number of students got $$ 13$$ marks. Hence, $$13$$  is the mode.
  • Question 6
    1 / -0
    Find the mode of:
    Height ($$cm$$)
    $$37$$
    $$38$$
    $$39$$
    $$40$$
    $$41$$
    Number of plants$$46$$
    $$89$$
    $$93$$$$90$$
    $$153$$
    Solution
    mode $$=41$$ (highest frequency $$= 153$$)
  • Question 7
    1 / -0
    The mode of the following discrete series is:
    $$x_i$$13561215
    $$f_i$$573865
    Solution
    Mode is that data which has highest frequency
    We observe that the value $$6$$ has maximum frequency. So mode $$= 6$$.
  • Question 8
    1 / -0
    Find the mode for the following data :
    Term
    18
    22
    26
    30
    34
    38
    Frequency
     3
      5
     10
     2
      8
     2

    Solution
    Mode= $$26$$ (highest frequency =$$10$$ )
  • Question 9
    1 / -0
    While computing mean of grouped data, we assume that the frequencies are:
    Solution
    While computing mean of grouped data, we assume that the frequencies are centred at the class marks of the class.
    Hence, option B is correct.
  • Question 10
    1 / -0
    The modal class for the following frequency distribution is 
    Marks$$0-10$$ $$10-20$$$$20-40$$$$40-50$$$$50-60$$$$60-70$$$$70-90$$$$90-100$$
    Number of students$$4$$$$6$$$$14$$$$16$$$$14$$$$8$$$$14$$$$5$$

    Solution

    Since $$40-50$$ is the class of maximum frequency, so, it is the modal class.
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