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Statistics Test - 45

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Statistics Test - 45
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  • Question 1
    1 / -0
    The choices of the fruits of 42 students in a class are as follows:
    A, O, B, M, A, G, B, G, A, G, B, M. A, O, M, A, B, G, M, B, A, O, M, O, G, B, O, M, G, A, A, B, M, O, M, G, B, A, M, O, M, O,
    where A, B, G, M, and O stands for Apple, Banana, Grapes, Mango and Orange respectively.
    which fruit is liked
    by most of the students?
    Solution
    The number of $$A$$ appearing in the list is $$9$$

    The number of $$B$$ appearing in the list is $$8$$

    The number of $$G$$ appearing in the list is $$7$$

    The number of $$M$$ appearing in the list are $$10$$

    The number of $$O$$ appearing in the list are $$8$$

    So, $$M$$ appeared the most, and $$M$$ stands for Mango.

    So, Mango $$M$$ is the fruit that is liked by most of the students.
  • Question 2
    1 / -0
    The median of the data : $$ 3,4,5,6,7,3,4$$ is  
    Solution
    Rearranging the given data into ascending order, we get 
    $$3,3,4,4,5,6,7,$$
    Total number of observations $$=7$$ which is odd
    Hence, the $$4^{th}$$ term is the median.
    $$\therefore$$ Median $$=4$$

    Hence, option (C) is the correct answer.
  • Question 3
    1 / -0
    What is the mode of $$19, 19, 15, 20, 25, 15, 20, 15$$?
    Solution
    Since observation $$15$$ is occuring the most in the data
    $$19,19,15,20,25,15,20,15$$
    19,19,15,20,25,15,20,15

    So, the mode of the given data is $$15$$
    .
    Hence, the correct alternative answer is (A).
  • Question 4
    1 / -0
    If maximum cost of any times is Rs. 500 and minimum Rs 75, the coefficient of range will be - 
    Solution
    Range coefficient 
    $$\dfrac{500 - 75}{500 + 75} = \dfrac{425}{575} = 0.739$$
  • Question 5
    1 / -0
    Coefficient of range of variable series $$ 10, 20, 30, 40, 50, 60 $$ is 
    Solution
    Coefficient of Range 
    $$= \dfrac{60 - 10}{60 + 10} = \dfrac{50}{70} = \dfrac{5}{7}$$
    Thus (D) is correct
  • Question 6
    1 / -0
    Marks obtained by students are $$ 25, 35, 45 $$ and $$55 $$ their mean deviation is -
    Solution
    $$\bar{x} = \dfrac{25 + 35 + 45 + 55}{4} = \dfrac{160}{4} = 40$$ 
    Mean $$ = 40 $$ 
    Mean deviation $$ = \dfrac{\sum |x_{i}-\bar{x}}{N} = \dfrac{40}{4} = 10$$

  • Question 7
    1 / -0
    coefficient of Range can be defined as - 
    Solution
    Thus (C) is correct
  • Question 8
    1 / -0
    In a series $$\sum x^{2} = 100, n = 5 $$ and  $$\sum x = 20 $$ , then standard deviation is 
    Solution
    Standard deviation 
    $$\sigma \sqrt{\dfrac{\sum x^{2}}{n} - \left ( \dfrac{\sum }{n} \right )^{2}}  = \sqrt{\dfrac{100}{5} - \left ( \dfrac{20}{5} \right )^{2}} $$
    $$\sqrt{20-16} = \sqrt{4} = 2 $$
  • Question 9
    1 / -0
    If $$ N = 10 \sum x = 120 $$ and $$ \sigma _{x} = 60 $$ , the variation coefficient is - 
    Solution
    Variation coefficient 
    $$= \dfrac{\sigma _{x}}{Mean} \times 100$$
    $$ = \dfrac{60 \times 10}{120} \times = 500$$
  • Question 10
    1 / -0
    Standard deviation of data $$6, 10, 4, 7, 4, 5$$ is - 
    Solution
    Mean $$(\bar{x}) = \dfrac{6+ 10 + 4 + 7 + 4 + 5}{6} = \dfrac{36}{6} = 6 $$
    Standard deviation $$(\sigma ) = \sqrt{\dfrac{1}{N} \times \sum  (x_{1} - \bar{x})^{2}} $$
    $$=\sqrt{\dfrac{1}{6} \times 26} = \sqrt{\dfrac{13}{3}}$$
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