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Data Handling Test - 10

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Data Handling Test - 10
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  • Question 1
    1 / -0
    Mode of $$2, 3, 4, 5, 0, 1, 3, 3, 4, 3\ is\ 3.$$
    Solution
    Mode is the term which appears maximum number of times.

    The terms are: 2, 3, 4, 5, 0, 1, 3, 3, 4, 3
    Arranging them in ascending order: 0,1,2,3,3,3,3,4,4,5
    3 is occurring maximum number of times. hence mode is 3.
  • Question 2
    1 / -0
    A student got marks in $$5$$ subjects in a monthly test is given below: $$2, 3, 4, 5, 6$$.
    In these obtained marks, $$4$$ is the
    Solution
    Marks obtained $$= 2,3,4,5,6$$

    Median is the middle term $$= 4$$.
    Mean $$= \dfrac{2+3+4+5+6}{5} = 4$$.
  • Question 3
    1 / -0
    The modal value is the value of the variate which divides the total frequency into two equal parts.
    Solution
    False. Modal value is the value which occurs maximum number of times in the data.
  • Question 4
    1 / -0
    State true or false:
    The mode is the most frequently occurring observation.
    Solution
    The observation occurring the most number of times or which has highest frequency is called the mode.
    Thus, the given statement is true.
  • Question 5
    1 / -0
    The marks of $$20$$ students in a test were as follows:
    $$5, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20$$
    The mode is
    Solution
    Terms are: $$5, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20$$.
    $$15$$ occurs most number of times $$(3$$ times$$)$$. 
    Hence, $$15$$ is the mode of the marks.
  • Question 6
    1 / -0
    If the probability of an event of a random experiment is $$P(E)=0$$, then the event is called an impossible event is 
    Solution
    If $$P(E) = 0$$, this event will never happen, hence it is an impossible event.
  • Question 7
    1 / -0
    Calculate the mode for $$17, 12, 19, 11, 20, 11, 20, 19, 10, 25,19$$.
    Solution
    Mode is the value which occurs most often in the data set of values.
    Given data set is $$17,12,19,11,20,11,20,19,10,25,19$$ 
    In the above data set, value  $$19$$ has occurred many times i.e., 3 times.
    Therefore the mode of the given data set is $$19$$
  • Question 8
    1 / -0
    Median of $$15, 28, 72, 56, 44, 32, 31, 43\ and\ 51\ is\ 43.$$
    Solution
    The terms are: 15, 28, 72, 56, 44, 32, 31, 43 and 51.
    Arranging them in ascending order: 15, 28, 31, 32, 43, 44, 51, 56, 72

    Since the total number of terms is odd that is 9, therefore the median will be the middle term that is the 5th term which is 43.
  • Question 9
    1 / -0
    Mean of $$41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 49, 42, 52, 60 \ is\ 54.8$$
    Solution
    $$Mean$$ $$=$$ $$\dfrac{sum of all numbers}{no. of numbers}$$
    $$\therefore$$ $$mean$$ $$=$$ $$\dfrac{41+ 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 49 + 42 + 52 + 60}{15}$$ $$=$$ $$55.4$$
  • Question 10
    1 / -0
    The mean of $$8, 7, 9, 10, 12, x$$ and $$14$$ is $$12$$, then find the value of $$x$$.
    Solution
    Given mean is $$12$$.
    Therefore, mean $$=\dfrac {8+7+9+10+12+x+14}{7}$$
    $$\Rightarrow 12=\dfrac {60+x}{7}$$
    $$\Rightarrow 84=60+x$$
    $$\Rightarrow x=24$$
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