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Measures of Central Tendency Test - 41

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Measures of Central Tendency Test - 41
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  • Question 1
    1 / -0
    The average (arithmetic mean) of tt and yy is 1515, and the average of ww and xx is 1515. What is the average of t,w,xt, w, x and yy
    Solution
    Given, Average of tt and yy == 1515
    t + y2\Rightarrow \dfrac {t \space + \space y}{2} == 1515
    t\Rightarrow t ++ yy == 1515 ×\times 22
    t\Rightarrow t ++ yy == 3030

    Average ofww and xx == 1515
    w + x2\Rightarrow \dfrac {w \space + \space x}{2} == 1515
    w\rightarrow w ++ xx == 1515 ×\times 22
    w\Rightarrow w ++ xx == 3030
    Now, average of tt, yy, ww and xx == (t + y + w + x)4\dfrac {(t \space + \space y \space + \space w \space + \space x)}{4}
    == (t + y) + (w + x)4\dfrac {(t \space + \space y) \space + \space (w \space + \space x)}{4}
    == 30 + 304\dfrac {30 \space + \space 30}{4}
    == 604\dfrac {60}{4}
    == 1515
    Therefore, average of tt, yy, ww and xx is 15'15'.
  • Question 2
    1 / -0
    The mode of the observations 5,4,4,3,5,x,3,4,3,5,4,3,5, 4, 4, 3, 5, x, 3, 4, 3, 5, 4, 3, and 55 is 33. What is the median?
    Solution
    Since frequency of mode (3)(3) should be maximum x=3x=3
    Now, arranging data in ascending order we get
    3,3,3,3,3,4,4,4,4,,5,5,5,5
    Median = Middle term =4=4
  • Question 3
    1 / -0
    The mean of 48,27,36,24,x48, 27, 36, 24, x, and 2x2x is 37.x=37.x =
    Solution
    Mean == sumofno.totalno\dfrac{sum \quad of \quad no.}{total \quad no}
    37=48+27+36+24+x+2x6\Rightarrow 37=\dfrac{48+27+36+24+x+2x}{6}
    135+3x=222\Rightarrow 135+3x=222
    3x=222135\Rightarrow 3x=222-135
    3x=87\Rightarrow 3x=87
    x=873=29\Rightarrow x=\dfrac{87}{3}=29
  • Question 4
    1 / -0
    Given, 10,18,4,15,3,21,x10, 18, 4, 15, 3, 21, x
     If xx is the median of the 77 numbers listed above, which of the following could be the value of xx?
    Solution
    Given numbers
    10,10, 18,18, 4,4, 15,15, 3,3, 21,21, xx
    Median is the middle value in the list of numbers. To find the median, the numbers have to be in the numerical order.

    As xx is the given median, xx should be placed in the middle of the list in numerical order. The list contains 77 numbers. So, xx should be placed in the 4th{4}^{th} position.

    Rewriting the list in increasing order,
    3,3, 4,4, 10,10, x,x, 15,15, 18,18, 2121

    So, xx should lie between 1010 and 1515.
    Possible values of xx, 1010 \leq xx \leq 1515
    From the given options, 
    Only 1414 satisfies the above condition.

    Therefore, the value of xx is 1414.
  • Question 5
    1 / -0
    The average of six numbers is 4. If the average of two of those numbers is 2, what is the average of the other four numbers?
    Solution

  • Question 6
    1 / -0
    The median of 2,7,4,8,9,12, 7, 4, 8, 9, 1 is
    Solution
    1,2,4,7,8,91,2,4,7,8,9
    No. of terms =6=6
    Median =6+12=3.5=\dfrac{6+1}{2}=3.5 term
    (4)th Mean = Median  4+72=5.5\dfrac{4+7}{2}=5.5
  • Question 7
    1 / -0
    For a collection of 1111 items, sum of all items is 132132, then the arithmetic mean is.
    Solution

  • Question 8
    1 / -0
    The mode of the data 5,5,5,5,5,1,2,2,3,3,3,4,4,4,45, 5, 5, 5, 5, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 is
    Solution
    Step 1: Find the mode.\textbf{Step 1: Find the mode.}

                    The given observations are : 5,5,5,5,5,1,2,2,3,3,3,4,4,4,4\text{The given observations are : } 5, 5, 5, 5, 5, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 

                    In any series of observations, mode is the observation which appears the maximum\text{In any series of observations, mode is the observation which appears the maximum} 

                    number of times.\text{number of times.}  

                    Given data is 5,5,5,5,5,1,2,2,3,3,3,4,4,4,4 \text{Given data is 5,5,5,5,5,1,2,2,3,3,3,4,4,4,4 }

                    In the given observations, 5 appears the maximum number of times. \text{In the given observations, 5 appears the maximum number of times. }  

                    So, the mode is 5.\text{So, the mode is 5.}  

    Hence, the correct option is D.\textbf{Hence, the correct option is D.}
  • Question 9
    1 / -0
    A random sample of 20 people is classified in the following table according to their ages
    AgeFrequency
    15-252
    25-354
    35-456
    45-555
    55-653
    What is the mean age of this group of people?
    Solution
    Mean= (frequency×MidPoint)i(frequency)iMean=\ \dfrac{\sum { (frequency\times Mid-Point)_i }}{\sum{(frequency)_i}}
    Mean= 83020=41.5Mean=\ \dfrac{830}{20}=41.5

  • Question 10
    1 / -0
    The following observations have been arranged in the ascending order. If the median of the data 29,32,48,50,x,x+2,72,78,84,9529, 32, 48, 50, x, x+2, 72, 78, 84, 95 is 6363, then the value of xx is ___________.
    Solution
    Given data is 29,32,48,50,x,x+2,72,78, 84,9529, 32, 48, 50, x, x+2, 72, 78,  84, 95
    Also given median of data =63=63.
    Since, total number of observation are even. 
    So, median=x+(x+2)2\text{median}=\dfrac{x+(x+2)}{2}
    =2x+22=x+1=\dfrac{2x+2}{2}=x+1
    Thus 63=x+163=x+1
    x=62\Rightarrow x=62
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