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Statistics Test - 54

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Statistics Test - 54
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  • Question 1
    1 / -0
    In a frequency distributions, mode is $$7.88$$, mean is $$8.32$$, then median is 
    Solution
    Using the Empirical Formula,
    $$\begin{array} { l }  { 3 \text { Median = Mode + 2 Mean } } \\ { \Rightarrow  \text { Median } = \dfrac { \text { Mode } + 2 \text { Mean } } { 3 }  }\end{array}$$

    On substituting the values, we get
    $$\begin{array} { l } { \Rightarrow \dfrac { 7.88 + 2 ( 8.32 ) } { 3 } }\\{ \Rightarrow \dfrac { 7.88 + 16.64 } { 3 } } \\ { \Rightarrow \dfrac { 24.52 } { 3 } } \\ { \Rightarrow 8.173\approx 8.17 } \end{array}$$
  • Question 2
    1 / -0
    The median class of the following distribution is ..............
    x5-1010-2020-3030-4040-5050-60
    f48121695

    Solution
    To find the median class:
    First add the total number of frequencies.
    Here, $$f = 54$$
     $$=$$ $$\dfrac{Total \space\ number\space\ of \space\ frequency }{2}$$
     $$=$$ $$\dfrac{54 }{2} =27$$ 
    Hence, the value 27 lies in the class interval 30-40 from the cummulative frequency table.
  • Question 3
    1 / -0
    Mode of the data $$3,2,3,2,3,5,6,6,5,3,2,5$$ is
  • Question 4
    1 / -0
    The modal class of the given frequency distribution is
    Marks Obtained $$10-20$$$$20-30$$$$30-40$$$$40-50$$
    Cumulative Frequency $$7$$$$35$$$$27$$$$34$$
    Solution
    The modal class is the class with the highest frequency.

    From the table,
    Highest frequency $$=35$$

    The frequency $$35$$ belongs to class $$20-30.$$

    Hence the modal class is $$20-30$$
  • Question 5
    1 / -0
    .

  • Question 6
    1 / -0
    The median wage of the worker in the following table is 
    Wages/Week
    (Rs.)
    no. of workers c.f
    50-59 15 15
    60-69 40 55
    70-79 50 105
    80-89 60 165
    90-99 45 210
    100-109 40 250
    110-119 15 265
  • Question 7
    1 / -0
    Mode of the data $$3,2,5,2,3,5,6,6,5,3,5,2,5$$ is 
    Solution


         $${\textbf{Step -1: Find the number used maximum times.}}$$

          $${\text{Given numbers: 3,2,5,2,3,5,6,6,5,3,5,2,5}} $$

            $$ {x_i}:{\text{2  3  5  6}} $$

            $$  {f_i}:{\text{3  3  5  2  }} $$

            $$\rightarrow {x_i} = {\text{Given numbers}} $$

            $$\rightarrow {f_i} = {\text{how many times Given number is written}} $$

             $$  {\text{Mode of the data is the maximum times a number is repeated in the given data}}{\text{.}} $$

             $${\text{Since 5 is repeated maximum times in the given data, mode is 5.}}$$

          $$  {\textbf{Hence, The correct answer is option D}} $$

     

     

  • Question 8
    1 / -0
    The ages of 40 girls in a class are given below:
    Age (in years)1415161718
    Number of girls5815102
    Find the mean age.
    Solution

  • Question 9
    1 / -0
    $$9898, 8998, 9889, 8899, 9988, 8989, 9998$$.
    If the numbers above are arranged in an ascending or descending order. Which number will appear in the middle place?
    Solution

  • Question 10
    1 / -0
    The median and mode respectively of a frequency distribution are $$26$$ and $$29$$. Then its mean is
    Solution
    Here we can use direct formula,

    $$ 3 \times Median - 2 \times Mean = Mode$$
    $$ 3 \times 26 - 2 \times Mean = 29$$
    $$ 78 - 29 = 2 \times Mean$$
    $$ Mean = \dfrac{49}{2}$$
    $$ Mean = 24.5$$

    Option B  is correct
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