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Statistics Test - 32

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Statistics Test - 32
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  • Question 1
    1 / -0
    The distance (in km) of $$40$$ engineers from their residence to their place of work were found as follows:
    $$5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8,$$
    $$3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12$$
    How many engineers travelled distance more than $$14$$ and less than $$26$$?
    Solution
    Distance of engineers from their residence to their work place is given by 
    $$5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8,$$
    $$3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12$$
    Now, the distances less than more than $$15$$ and less than $$25$$ are $$15,16,17,18,19,20,21,22,23,24,25$$
    The distance$$(>14\ and\ <26)$$ travelled by engineers are $$20,25,19,17,18,17,16,15,18,15,15$$
    $$\therefore$$ The number of engineers who travelled distance more than $$14$$ and less than $$26$$ is $$11$$.
    Hence, option C is correct.
  • Question 2
    1 / -0

    Directions For Questions

    Use the table given below to find:
    Class Interval
    Frequency
    $$30-34$$
    $$7$$
    $$35-39$$
    $$10$$
    $$40-44$$
    $$12$$
    $$45-49$$
    $$13$$
    $$50-54$$
    $$8$$
    $$55-59$$
    $$4$$

    ...view full instructions

    The class mark of the third class is:
    Solution
    The third class is $$40 -44$$
    Class Mark $$= \cfrac{\text{Lower limit} + \text{upper limit}}{2}$$
    Class Mark $$= \cfrac{40 + 44}{2}$$
    Class Mark $$= 42$$
  • Question 3
    1 / -0

    Directions For Questions

    The table, given below, shows the frequency distribution of the weekly wages of the employees of a company:
    Weekly wages (in Rs.)Number of empolyees
    $$800 - 899$$
    $$900 - 999$$
    $$1000 - 1099$$
    $$1100 - 1199$$
    $$1200 - 1299$$
    $$22$$
    $$27$$
    $$23$$
    $$18$$
    $$15$$

    ...view full instructions

    Find the class mark of the first class.
    Solution
    The first class has the interval: $$800 - 899$$
    The class mark will be the mean of the lower limit and the upper limit of the class.
    Class mark $$= \cfrac{\text{lower limit} + \text{upper limit}}{2}$$
    Class mark $$= \cfrac{800 + 899}{2}$$
    Class mark $$= 849.5$$
  • Question 4
    1 / -0
    The blood groups of 30 students of class VIII are recorded as follows:
    $$A,B,O,O,AB,O,A,O,B,A,O,B,A,O,O,$$
    $$A,AB,O,A,A,O,O,AB,B,A,O,B,A,B,O.$$
    Which is the most common and which is the rarest blood group among these students?
    Answer is in form of  (Rarest, Common)
    Solution
    Most common is the one which has occurreded very frequently and the rarest the one which has occurred very few times.
    From the given data set $$O$$ has occurred red very frequently and $$AB$$ has occurred very few times.
    Therefore, the rarest is $$AB$$ and common is $$O$$.
    Hence, $$(\text{Rarest, Common})=(AB,O)$$
  • Question 5
    1 / -0
    The A.M. of the first ten odd numbers is
    Solution
    First ten odd numbers are $$1, 3, 5, 7, 9, 11, 13, 15, 17, 19$$ respectively.
    $$A.M. (\overline {x}) = \dfrac{\text {sum of all the terms}}{\text {total number of terms}}$$
     
    $$A.M. \left(\overline { x } \right) = \displaystyle\frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10}$$
                       
                       $$= \displaystyle\frac{100}{10}$$
                             
                       $$=10$$

    $$\therefore$$ the $$A.M$$ of first ten odd numbers is $$10$$ 
  • Question 6
    1 / -0
    The horizontal line in a bar graph is called:
    Solution
    In the Cartesian coordinate system, the horizontal reference line is called $$x$$-axis.
    Therefore, the horizontal line in a bar graph is called $$x$$-axis.
    Hence, option $$A$$ is correct.
  • Question 7
    1 / -0
    The arithmetic mean of first ten natural numbers is
    Solution
    Formula used:
    $$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$$

    So, by using the above formula we get,
    $$A.M=\dfrac{1+2+3+...+10}{10}$$
               $$=\dfrac{55}{10}$$
               $$=5.5$$

    Hence, option $$A$$ is correct.
  • Question 8
    1 / -0

    Directions For Questions

    Study the following frequency table and choose the appropriate answer :
    Class IntervalsFrequency
    $$0-10$$$$6$$
    $$10-20$$$$2$$
    $$20-30$$$$13$$
    $$30-40$$$$17$$
    $$40-50$$$$11$$
    $$50-60$$$$4$$
    $$60-70$$$$8$$
    $$70-80$$$$7$$

    ...view full instructions

    The class-mark of class $$50-60$$ is:
    Solution
    The number in the middle of the class is called class mark.
    i.e, $$55$$
  • Question 9
    1 / -0
    The range of $$15, 14, x, 25, 30, 35$$ is $$23$$. Find the least possible value of $$x$$.
    Solution
    Let $$x$$ be the least value and $$35$$ be the Highest value.
    Since, Range $$=$$ Highest value $$-$$ Lowest value
    $$\Rightarrow  23 = 35 - x $$
    $$\Rightarrow  x = 12$$
    $$\therefore $$ Option $$B$$ is correct.
  • Question 10
    1 / -0
    The following marks were obtained by the students in a test
    $$81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62$$.
    What is the range of the marks obtained?
    Solution
    The given observation are: $$81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62$$.
    The minimum marks obtained are: $$62$$
    Maximum marks obtained are: $$95$$
    Thus, range $$=$$ maximum marks $$-$$ minimum marks $$= 95 - 62 = 33$$
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