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Measures of Central Tendency Test - 10

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Measures of Central Tendency Test - 10
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  • Question 1
    1 / -0
    The average of daily wages for the workers of the two factories combined is 
    No.of wage earnersFactory A 250Factory B 200
    Average daily wage Rs.2.00Rs.2.50
    Solution
    Average wage$$ = \cfrac{x_1w_1+x_2w_2}{w_1+w_2}$$

                             $$=\cfrac{250\times 2+200\times 2.5}{250+200}$$

                             $$=\cfrac{1000}{450}=2.22$$
  • Question 2
    1 / -0
    Find the median of the given data.
    $$30, 32, 24, 34, 26, 28, 30, 35, 33, 25$$
    Solution
    Arranging the data in ascending order:
    $$24,25,26,28,30,30,32,33,34,35$$
    Mean $$=\dfrac {\text { sum of all numbers} }{ \text {count of all numbers }} = \dfrac { 24+25+26+28+30+30+32+33+34+35}{10}  =\dfrac { 297 }{10 } =29.7  $$
    Median of even number of data is the average of the middle two terms that is here the average of $$5^{th}$$ and $$6^{th}$$ terms.
    Median$$ =\dfrac { 30+30 }{ 2 } =30
    $$
  • Question 3
    1 / -0
    Frequency distribution of daily commission received by 100 salesmen is given.
    Daily commission (in Rs.)$$100 - 120$$$$120 - 140$$$$140 - 160$$$$160 - 180$$$$180 - 200$$
    No. of salesman$$20$$$$45$$$$22$$$$9$$$$4$$
    Find mean daily commission received by a salesman.
    Solution

    Rs               No.of salesman        $$x$$           $${ f }_{ x }$$
    $$100-120$$             $$ 20$$                  $$110$$      $$2200$$
    $$120-140 $$             $$ 45$$                  $$130$$        $$5850$$
    $$140-160$$              $$22$$                 $$150$$      $$3300$$
    $$160-180$$              $$9$$                    $$170$$      $$1530$$
    $$180-200$$              $$ 4$$                    $$190$$       $$760$$
                                  $$100$$                              $$13640$$
    $$\bar { x }$$ =$$\cfrac { \sum { fx }  }{ \sum { f }  } =\cfrac { 13640 }{ 100 } =Rs136.4$$
  • Question 4
    1 / -0
    Find the mean of first ten odd natural numbers.
    Solution

    $${\textbf{Step -1: Listing first 10 odd natural numbers .}}$$

                      $${\text{First 10 odd natural numbers  =  1, 3, 5, 7, 9, 11, 13, 15, 17, 19}}$$

    $${\textbf{Step -2: Calculating mean}}$$

                      $${\text{We know that, mean of n numbers  =  }}\dfrac{{{\text{Sum of n numbers}}}}{{\text{n}}}$$

                      $$\therefore {\text{ Mean of first 10 odd natural numbers,}}$$

                      $${\text{M  =  }}\dfrac{{{\text{1  +  3  +  5  +  7  +  9  +  11  +  13  +  15  +  17  +  19}}}}{{{\text{10}}}}$$

                      $$ \Rightarrow {\text{ M  =  }}\dfrac{{{\text{100}}}}{{{\text{10}}}}{\text{  =  10}}$$

    $${\textbf{Thus, the mean of first 10 odd natural numbers is 10 .}}$$

  • Question 5
    1 / -0
    Find the median of:
    $$25, 16, 26, 16, 32, 31, 19, 28$$ and $$35$$.
    Solution
    Arranging the series in ascending order:
    $$16, 16, 19, 25, 26, 28, 31, 32, 35$$
    The series has $$9$$ terms, the median is the middle term.Hence, $$26$$ is the median of the series.
  • Question 6
    1 / -0
    Find the median of:
    $$241, 243, 347, 350, 327, 299, 261, 292, 271, 258$$ and $$257.$$
    Solution
    Arrange the series in ascending order:
    $$241, 243, 257, 258, 261, 271, 292, 299, 327, 347, 350$$
    The series has $$11$$ terms. Hence, the median will be the middle term, i.e. $$6^{th}$$ term.
    Hence, the median is $$271$$
  • Question 7
    1 / -0
    Find the median of the following data.
    $$150, 147, 140, 120, 135, 139, 148, 148$$
    Solution
    The data points are: $$150, 147, 140, 120, 135, 139, 148, 148$$
    Arranging in ascending order: $$120, 135, 139, 140, 147, 148, 148, 150$$
    The number of observations are $$8$$. Since, the observations are even, the median will be the mean of $$\cfrac{n}{2}$$ and $$\cfrac{n+2}{2}$$ observations
    median $$=$$ mean of $$\cfrac{8}{2}$$ and $$\cfrac{8+2}{2}$$ observations
    median $$=$$ mean of 4th and 5th observations
    median $$= \cfrac{140+ 147}{2}$$
    median $$= 143.5$$
  • Question 8
    1 / -0
    Find the median of the data: $$ 35, 48, 92, 76, 64. 52, 51, 63$$ and $$71$$
    Solution
    Arranging the data in ascending order:
    $$35,48,51,52,63,64,71,76,92$$
    Mean $$=\dfrac {\text { sum of all numbers }}{ \text {count of all numbers }} = \dfrac { 35+48+51+52+63+64+71+76+92 }{ 9 }  =\dfrac { 552 }{ 9 } =61.33$$
    Since the number of data is odd the median will be the middlemost term.
    So, Median $$=63$$  of the numbers would become:
    $$ 35,48,52,63,64,66,71,76,92$$ 
    The new Median is $$64$$.
  • Question 9
    1 / -0
    The marks obtained by twenty students are given as follows.
    $$27, 22, 24, 17, 16, 15, 14, 22, 17, 15, 22, 26, 19, 22, 20, 22, 16, 15, 22, 27$$
    Find the modal marks.
    Solution
    Marks obtained by $$20$$ students : $$27, 22, 24, 17, 16, 15, 14, 22, 17, 15, 22, 26, 19, 22, 20, 22, 16, 15, 22, 27$$
    Arranging them in ascending order: $$14, 15, 15,15, 16,16, 17, 17, 19, 20,  22, 22, 22, 22, 22, 22, 24, 26, 27, 27$$
    $$22$$ occurs $$6$$ and maximum number of times in the data. Hence, $$22$$ is the modal marks.
  • Question 10
    1 / -0
    A cricket player scored runs in different matches as follows.
    $$36, 41, 57, 89, 105, 103, 17$$. Find the median.
    Solution
    The data points are: $$36, 41, 57, 89, 105, 103, 17$$
    Arranging in ascending order: $$17, 36, 41, 57, 89, 103, 105$$
    The number of observations are $$7$$. Since, the observations are odd, the median will be the $$\cfrac{n+1}{2}$$ observations
    median $$= \cfrac{7+1}{2}$$ observations
    median $$= 4^{th}$$ observations
    median $$= 57$$
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