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Measures of Central Tendency Test - 26

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Measures of Central Tendency Test - 26
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  • Question 1
    1 / -0
    The mean of the following distribution is
    class
    Frequency
    0-5
    4
    5-10
    5
    10-15
    7
    15-20
    12
    20-25
    7
    25-30
    5

    Solution
    Mean == sum of (Classmark×frequency)Sum of frequency\dfrac{sum \ of \ (Class mark \times frequency)}{Sum \ of \ frequency}

    Mean == 2.5×4+7.5×5+12.5× 7+17.5×12+22.5× 7+27.5×54+5+7+12+7+5\dfrac{2.5\times 4 + 7.5\times5 + 12.5\times 7+ 17.5\times12 + 22.5\times 7 + 27.5\times 5}{4 +5+7+12+7 +5}
    Mean = 64040=16\dfrac{640}{40} = 16
  • Question 2
    1 / -0
    The numbers 4,6  and  8 4, 6\  and\  8  are the frequencies of (x+ 2),x and  (x1)(x +  2), x \ and\  (x - 1) respectively . If their arithmetic mean is 8,8,  the value of  xx  is
    Solution
    8=8 = 4×(x+2)+6x+8×(x1)4+6+8\dfrac{4\times \left ( x+2 \right )+6x+8\times \left ( x-1 \right )} {4+6+8}
    =4x+8+6x+8x818=\dfrac{4x+8+6x+8x-8}{18}
    \therefore x=8x = 8
  • Question 3
    1 / -0
    The examination marks of 250250 pupils are recorded below. Find the mean.
    Class interval
    0-9
    10-19
    20-29
    30-39
    40-49
    Frequency
    0
    2
    6
    24
    36
    Class interval
    50-59
    60-69
    70-79
    80-89
    90-99
    Frequency
    47
    55
    40
    27
    13
    Solution
    Mean=Total (mid value × frequency)TotalfrequencyMean = \displaystyle \frac{Total  (mid  value  \times  frequency)}{Total frequency}
    Class interval
      Mid- value
     Fequency.  
    Mid-value ×\times frequency
    0-9
     4.5
    0
    0
    10-19
    14.5
    2
    29
    20-29
    24.5
    6
    147
    30-39
    34.5
    24
    828
    40-49
    44.5
    36
    1602
    50-59
    54.5
    47
    2561.5
    60-69
    64.5
    55
    3547.5
    70-79
    74.5
    40
    2980
    80-89
    84.5
    27
    2281.5
    99-99
    94.5
    13
    1228.5


    Total - 250
    Total 15205
    Mean=Total (mid value × frequency)Total frequenct=15205250=60.82Mean = \displaystyle \frac{Total  (mid  value  \times  frequency)}{Total  frequenct} = \frac{15205}{250}= 60.82
  • Question 4
    1 / -0
    A contractor employed 1818  labourers at Rs. 1212 per day, 10 10 labourers at Rs.13.50 13.50 per  day,5 5 labourers at Rs.25 25 per  day and 22 labourers at Rs.42 42 per day. The average  wage of a labourer per day is 
    Solution
    Mean = TotalSumTotalCount\dfrac{Total Sum}{Total Count}

    Mean = 18×12+10×13.50+5×25+2×4218+10+5+2\dfrac{18\times12 + 10\times13.50 + 5\times25 + 2\times42}{18+10+5+2}
    = 56035\dfrac{560}{35}
    =16= 16
  • Question 5
    1 / -0
    In a class test in English 1010 students scored 7575 marks, 1212 students scored 6060 marks, 88 scored 4040 marks and 33 scored 3030 marks, the mode  for their score is
    Solution
    Step -1: Defining mode.\textbf{Step -1: Defining mode.}
                      In a given data set, the mode is the value that appears most often among every other values.\text{In a given data set, the mode is the value that appears most often among every other values.}
                      Given data set says, \text{Given data set says, }
                      10 Students scored 75 marks.\text{10 Students scored 75 marks.}
                      12 Students scored 60 marks.\text{12 Students scored 60 marks.}
                      8 Students scored 40 marks.\text{8 Students scored 40 marks.}
                      3 Students scored 30 marks.\text{3 Students scored 30 marks.}
    Step -2: Finding the mode of their score.\textbf{Step -2: Finding the mode of their score.}
                      The mode of their score will be the marks occurring most often.\text{The mode of their score will be the marks occurring most often.}
                      12 Students scored 60 marks.\therefore \text{12 Students scored 60 marks.}
                      Hence 60 is the mode of their score.\text{Hence 60 is the mode of their score.}
    Hence,The mode of their score is 60. (Option C)\textbf{Hence,The mode of their score is 60. (Option C)}
  • Question 6
    1 / -0
    The marks of 2020 students in a test were as follows:
    5,6,8,9,10,11,11,12,13,13,14,14,15,15,15,16,16,18,19,205, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20
    The median is
    Solution
    Here, nn == 2020 which is an even number.
    \therefore Median =(n2)thterm+(n2+1)thterm2= \displaystyle \cfrac{\left ( \dfrac{n}{2} \right )^{th} \text{term} + \left ( \dfrac{n}{2}+1\right )^{th} \text{term}}{2}
    \therefore Median =(202)thterm+(202+1)thterm2= \displaystyle \cfrac{\left ( \dfrac{20}{2} \right )^{th} \text{term} + \left ( \dfrac{20}{2}+1\right )^{th} \text{term}}{2}
    =10thterm+11thterm2\displaystyle =\cfrac{10^{th}\: \text{term} + 11^{th}\: \text{term}}{2}
    =13+142=13.5\displaystyle =\cfrac{13+14}{2}=13.5
  • Question 7
    1 / -0
    The median of variables 5, 8, 10, 13, 17, 21, 27, 30, 33, 42 is
    Solution

  • Question 8
    1 / -0
    Arithmetic mean for the given data is
    Marks
    No. of students
    0 - 10
    10 - 20
    20 - 30
    30 - 40
    40 - 50
    50 - 60
    5
    10
    25
    30
    20
    10

    Solution
    Mean == Sum of(class mark×frequency)Sum of frequency\dfrac{\text {Sum of} (\text {class mark} \times \text {frequency})}{\text {Sum of frequency}}

    Mean == 5×5+15×10+25× 25+35×30+45× 20+55×105+10+25+30+20+10\dfrac{5\times 5 + 15\times10 + 25\times 25+ 35\times30 + 45\times 20 + 55\times 10}{5 +10+25+30+20 +10}

    Mean == 3300100\dfrac{3300}{100} = 3333
  • Question 9
    1 / -0
    The average weight of 1010 men is decreased by 3 kg 3 \space kg  when one of them whose weight is 80kg80 kg  is replaced by a new person. The weight of the new person is
    Solution
    Let the mean of 1010 persons =M M
    Sum of weight of the persons =10M = 10M

    After replacing a person weighing 80 Kg with another weighing sayx x Kg, sum=10M80+xsum = 10M - 80 +x
    New Mean=M3 = M-3
    hence,10(M3)=10M80+x 10(M-3) = 10M - 80 +x
    10M30=10M80+x10M- 30= 10M -80 +x
    x=50Kgx = 50 Kg
  • Question 10
    1 / -0
    In a family, there are 88 men, 77 women and 55 children whose mean ages separately are respectively 24,2024,20 and 66 years. The mean age of the family is
    Solution
    Here we have three collections for which A1=24,n1=8;A2=20,n2=7{ A }_{ 1 }=24,{ n }_{ 1 }=8;{ A }_{ 2 }=20,{ n }_{ 2 }=7 and A3=6,n3=5{ A }_{ 3 }=6,{ n }_{ 3 }=5.
    Their combined mean is the required mean.
    By the formula  A=n1A1+n2A2+n3A3n1+n2+n3=8×24+7×20+5×68+7+5\displaystyle A=\frac { { n }_{ 1 }{ A }_{ 1 }+{ n }_{ 2 }{ A }_{ 2 }+{ n }_{ 3 }{ A }_{ 3 } }{ { n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 } } =\frac { 8\times 24+7\times 20+5\times 6 }{ 8+7+5 }
     =192+140+3020=36220=18.1\displaystyle =\frac { 192+140+30 }{ 20 } =\frac { 362 }{ 20 } =18.1
    \therefore The mean age of the family =18.1=18.1 years
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