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Statistics Test - 18

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Statistics Test - 18
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  • Question 1
    1 / -0
    In an examination, $$10$$ students scored the following marks in Mathematics: $$35, 19, 28, 32, 63, 02, 47, 31, 13, 98$$. Its range is
    Solution
    Range $$=$$ Maximum value of the variable $$-$$ Minimum value of the variable
    $$= 98 - 2 = 96$$
  • Question 2
    1 / -0
    The mean of $$x + 3, x + 5, x + 7, x + 9$$ and $$x + 11$$ is
    Solution
    The observations are : $$x+3,x+5,x+7,x+9, x+11$$
    Mean = $$\dfrac{sum}{number \quad of \quad observations}$$ = $$\dfrac{x+3,x+5,x+7,x+9, x+11}{5}$$ = $$\dfrac{5x + 35}{5} = x + 7$$
  • Question 3
    1 / -0
    A histrogram consists of
    Solution
    A histogram is an area diagram. It can be defined as a set of rectangles with bases along with the intervals between class boundaries and with areas proportional to frequencies in the corresponding classes.
  • Question 4
    1 / -0
    Which  class-interval has the  maximum frequency?

    Solution

    We will form table from above histogram:
     $$CI$$$$Frequency$$ 
     $$0-10$$$$15$$ 
     $$10-20$$$$10$$ 
     $$20-30$$$$20$$ 
     $$30-40$$$$25$$ 
     $$40-50$$$$10$$ 
     $$50-60$$$$5$$ 
    $$\Rightarrow$$  From above table table we can see, $$30-40$$ has frequency $$25$$ which is maximum.
    $$\therefore$$  The class-interval has the maximum frequency is $$30-40$$

  • Question 5
    1 / -0
    What is frequency of the class- interval $$20-30$$ ?

    Solution

    $$\Rightarrow$$  In the above histogram we can see that, frequency of the class interval $$20-30$$ is $$20$$

  • Question 6
    1 / -0
    Lower class limit of $$15 -18$$ is _____
    Solution
    We need to find lower class limit of $$15-18$$.
    Thus, $$15$$ is lesser in $$(15-18) $$, so answer is $$15$$.
  • Question 7
    1 / -0
    Construct a frequency distribution table for the data, taking class-intervals 4-6, 6-8, ............., 14-16.
    11.5
     4.5
     7.4
     9.8
     4.6
    14.2
      6.6
    15.5
    6.3
      6
    5.3
    8.25
     6.4
    15.3
      4.3
    14.4
    7.8
    8.3
    8.4
    6.5
    8.9
    11.7
     4.7
    12.2
    9.2
    12.5
    15.2
    5.8
    10.8
    9.9
    9.4
    7.7
    10.5
    15.8
      8.9
    10.5
    12.7
     8.8
    10.1
      5.5

    Solution
    Crete class interval in $$4 - 6$$, $$6 - 8$$ and so on and count the number of terms in between these intervals:
    C.I.       
    Tally Marks.       Frequency
     4-6
     6-8
     8-10
    10-12
    12-14
    14-16
    $${||||}\! \! \! \! \! \! {\diagdown}\:   ||$$
    $${||||}\! \! \! \! \! \! {\diagdown}\:   |||$$
    $${||||}\! \! \! \! \! \! {\diagdown}\:   {||||}\! \! \! \! \! \! {\diagdown}$$
    $${||||}\! \! \! \! \! \! {\diagdown}\:  |$$
    |||
    $${||||}\! \! \! \! \! \! {\diagdown}\:  |$$
     7
     8
    10
     6  
     3
     6
  • Question 8
    1 / -0
    How many class-intervals have  equal frequency ?

    Solution

    $$\Rightarrow$$  In above histogram we can see, class intervals $$10-20$$ and $$40-50$$ have same frequency which is $$10$$.
    $$\therefore$$  The class intervals have equal frequency are $$2$$.

  • Question 9
    1 / -0
    The difference between the greatest and the least value of the observations is known as 
    Solution
    The difference between the greatest and the least value of the observations is defined as $$Range$$.
  • Question 10
    1 / -0
    Find the mean of first ten odd natural numbers.
    Solution
    First 10 odd natural numbers are: $$1,3,5,7,9,11,13,15,17,19$$
    Mean $$= \cfrac{\text{Sum}}{\text{Number of observations}} =\cfrac{1 + 3 + 5+ 7+9+11+13+15+17 + 19}{10} =10$$
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