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Statistics Test - 24

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Statistics Test - 24
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  • Question 1
    1 / -0
    The median class of the frequency distribution given below is _______.
    Class$$0 - 10$$$$10 - 20$$$$20 - 30$$$$30 - 40$$$$40 - 50$$
    Frequency7$$15$$$$13$$$$17$$$$10$$
    Solution
    $$C.I$$ $$Frequency$$ $$Cumulative\,frequency$$ 
    $$0-10$$$$7$$ $$7$$ 
    $$10-20$$ $$15$$ $$22$$ 
    $$20-30$$ $$13$$ $$35$$ 
    $$30-40$$ $$17$$ $$52$$ 
    $$40-50$$ $$10$$ $$62$$ 
     $$N=62$$  
    We have, $$N = 62$$
    $$\therefore \quad \displaystyle\frac{N}{2} = 31$$
    Cumulative frequency just greater than $$\cfrac N2$$ is $$35$$.
    So the corresponding median class is $$20-30$$
  • Question 2
    1 / -0
    Size$$6$$$$7$$$$8$$$$9$$$$10$$
    No. of Shoes $$4$$$$5$$$$1$$$$2$$$$1$$
    Find the mode.
    Solution
     Size$$6$$ $$7$$ $$8$$ $$9$$ $$10$$ 
    No. of size $$4$$ $$5$$ $$1$$ $$2$$ $$1$$ 
    Mode is the quantity which is higher in frequency.
    Here, $$7$$ occurs for $$5$$ times.
    $$\therefore$$ Mode is $$7$$.
  • Question 3
    1 / -0
    The mode of the following data 8, 7, 9, 8, 7, 8, 10, 7, 10, 6, 7, 6, 10, 8, 6, 8 is ____
    Solution
    Arranging the given number in ascending order,
    $$6,6,6,7,7,7,7,8,8,8,8,8,9,10,10,10$$
    $$\Rightarrow$$  The number which occurs maximum times is called mode.
    $$\Rightarrow$$  So, here we can see number $$8$$ occurs maximum times i.e $$5\,times$$.
    $$\therefore$$   $$Mode=8$$
  • Question 4
    1 / -0
    Find the modal class 
     Test ScoresFrequencyGreater than Cummulative frequency
    $$91-100$$$$9$$$$20$$
    $$81-90$$
    $$6$$$$11$$
    $$71-80$$$$3$$$$5$$
    $$61-70$$$$0$$$$2$$
    $$51-60$$$$2$$$$2$$
    Solution
    $$\Rightarrow$$  We know that modal class is a group that has highest frequency.
    $$\Rightarrow$$  Here, in given frequency table we can see class interval $$91-100$$ has highest frequency i.e. $$9$$
    $$\therefore$$  Modal class $$=91-100$$
  • Question 5
    1 / -0
    Median of the following freq. dist.
    $$x_i$$$$3$$$$6$$$$10$$$$12$$$$7$$$$15$$
    $$f_i$$$$3$$$$4$$$$2$$$$8$$$$13$$$$10$$
    Solution
    $$C.I$$ $$Frequency$$ $$Cumulative\,frequency$$ 
    $$3$$$$3$$ $$3$$ 
    $$6$$ $$4$$ $$7$$ 
    $$10$$ $$2$$ $$9$$ 
    $$12$$ $$8$$ $$17$$ 
    $$7$$ $$13$$ $$30$$ 
     $$15$$$$10$$  $$40$$
    Here $$N=40$$,
     then $$\dfrac{N}{2}=\dfrac{40}{2}=20$$
    $$\dfrac{N}{2}+1=\dfrac{42}{2}=21$$
    As the corresponding median to $$30$$ which is nearest to $$21$$ is $$7$$, so required result is $$7$$.
  • Question 6
    1 / -0
    In a batch of $$13$$ students, $$4$$ have failed. The marks of successful candidates are $$41, 57, 36, 35, 71, 50,$$ and $$40$$. The median marks are 
    Solution
     In question, it is given that their are $$13$$ students in a batch.
    Out of $$13$$, $$4$$ students have failed and $$7$$ passed students marks are  given.
    We have to determine median marks, but we cannot find median marks here, because to find median we have know the marks of  $$13$$ students  , but in question only  $$7$$ students' marks are given.
  • Question 7
    1 / -0
    Find the mean of the data given below, if following are the marks obtained by students in a class test in the mathematics subject.
    Marks:102030405060
    Number of student51232404548
    Solution
    Mean $$=$$ $$ \dfrac{Total\, marks\, obtained \, by \, students}{Total \, students}$$
    Mean $$=$$ $$ \dfrac{50+240+960+1600+2250+2880}{5+12+32+40+45+48}$$
    Mean $$=$$ $$ \dfrac{7980}{182}$$
    Mean $$=$$$$ 43.846 $$
    $$ \boxed{\therefore Mean = 43.846}$$
  • Question 8
    1 / -0
    Mode of the distribution is:

    Marks 4  5  6  7  8 
    No. of students 3 510 6 1
    Solution
    Mode is the class that has the maximum frequency. 

    Here, we can see that the maximum number of students (10 students) have scored 6 marks.

    $$\therefore$$  Mode $$=6$$
  • Question 9
    1 / -0
    The value of median of 
    Income In Rs 100011001200130014001500
    No. of persons142621182815
    Solution
    $$Income$$
    $$(in\,Rs.)$$ 
    $$No.\,of\,persons$$ $$c.f$$ 
    $$1000$$ $$14$$ $$14$$ 
    $$1100$$ $$26$$ $$40$$ 
    $$1200$$ $$21$$ $$61$$ 
    $$1300$$ $$18$$ $$79$$ 
    $$1400$$ $$28$$$$107$$ 
    $$1500$$$$15$$ $$122$$ 
     $$N=122$$  
    $$Median=\dfrac{1}{2}\left[\left(\dfrac{122}{2}\right)^{th}term+\left(\dfrac{122}{2}+1\right)^{th}term\right]$$

                    $$=\dfrac{1}{2}\left[61^{th}term+62^{th}term\right]$$

    $$61^{th}term=1200$$ and $$62^{th}term=1300$$

    $$\therefore$$   $$Median=\dfrac{1}{2}[1200+1300]$$

    $$\therefore$$   $$Median=1250$$
  • Question 10
    1 / -0
    The value of $$p$$, for which the following date
    $$3,5,0,7,5,3,5,6,p,7,6,4,9$$ has mode $$5$$ is
    Solution

    We have,

    $$3,5,0,7,5,3,5,6,p,7,6,4,9$$

    So,

    Ascending order

    $$0,3,3,4,5,5,5,6,6,7,7,9,p$$

    So,

    $$ \text{mode}=5 $$

    $$ p=5 $$

    Hence, this is the answer.

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