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Statistics Test - 28

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Statistics Test - 28
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  • Question 1
    1 / -0
    The coefficient of mean deviation from median of observations  40,62,54,90,68,7640, 62, 54, 90, 68, 76  is
    Solution
    Arrange the given observations in ascending order
    40,54,62,68,76,9040,54,62,68,76,90
    Here, number of terms n=6(even)n=6 (even)
    \displaystyle \therefore Median (M) =(n2)thterm+(n2+1)thterm2=62+682=65\displaystyle =\frac{\left ( \frac{n}{2} \right )th\:term+\left ( \frac{n}{2}+1 \right )th\:term}{2}=\frac{62+68}{2}=65

    ΣxiM=25+11+3+3+11+25=78\Sigma \left | x_{i}-M \right |=25+11+3+3+11+25=78
    Mean deviation from median =ΣxiMn=786=13\displaystyle =\frac{\Sigma \left | x_{i}-M \right |}{n}=\frac{78}{6}=13
    \therefore Coefficient of M.D.==M.D.median=1365=0.2\displaystyle =\frac{M.D.}{median}=\frac{13}{65}=0.2
  • Question 2
    1 / -0
    The arithmetic mean of first ten natural numbers is
    Solution
    Formula used:
    Arithmetic mean=Sum of given numbersTotal numbers\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}

    So, by using the above formula we get,
    A.M=1+2+3+...+1010A.M=\dfrac{1+2+3+...+10}{10}
               =5510=\dfrac{55}{10}
               =5.5=5.5

    Hence, option AA is correct.
  • Question 3
    1 / -0
    The sum of the squares of deviation of 10 observations from their mean 50 is 250, then coefficient of variation is
    Solution
    Given,   (xxˉ)2=250,n=10,xˉ=50\sum (x-\bar{x})^2 = 250, n = 10, \bar{x} =50
    Thus standard deviation =(xxˉ)2n=25=5 = \sqrt{\cfrac{\sum (x-\bar{x})^2}{n}}=\sqrt{25}=5
    \therefore Coefficient of variation =σxˉ×100=550×100=\cfrac{\sigma}{\bar{x}}\times 100 =\cfrac{5}{50}\times 100 % =10= 10%

  • Question 4
    1 / -0
    The S.D. of the following freq. dist.
    Class 0-1010-2020-3030-40
    fif_i1342
    Solution
    Class     xix_i   fif_i   ui=xiAhu_i=\dfrac{x_i-A}{h}  ui2u_i^2        fiuif_iu_i     fiui2f_iu_i^2
    0-1051-24-24
    10-20153-11-33
    20-302540000
    30-403521122
    Total10-39
    S.D.S.D.=hfiui2 fi (fiuifi  ) 2=h\sqrt { \cfrac { \sum { { f }_{ i }{ u }_{ i }^{ 2 } }  }{ \sum { { f }_{ i } }  } -{ \left( \cfrac { \sum { f } _{ i }{ u }_{ i } }{ \sum { { f }_{ i } }  }  \right)  }^{ 2 } }
    S.D.=10910(310 ) 2=9S.D.=10\sqrt { \cfrac { 9 }{ 10 } -{ \left( \cfrac { -3 }{ 10 }  \right)  }^{ 2 } } =9
  • Question 5
    1 / -0
    The mean of a dist. is 44. if its coefficient of variation is 58%58\%. Then the S.D. of the dist. is
    Solution
    We know if a distribution having mean xˉ\bar{x} and standard deviation σ\sigma
    then coefficient of variation =σxˉ×100=\cfrac{\sigma}{\bar{x}}\times 100
    σ4×100=58σ=5825=2.32\therefore \cfrac{\sigma}{4}\times 100=58\Rightarrow \sigma = \cfrac{58}{25}=2.32
    Hence required standard deviation is =2.32=2.32
  • Question 6
    1 / -0
    The mean deviation from mean of observations 5, 10, 15, 20, ......85 is
    Solution
    Given series is 5,10,15,....,855,10,15,....,85, which is an AP.
    Last term 85=a+(n1)d85=a+(n-1)d
    n=17\Rightarrow n=17
    Sum of nn terms of AP =172(5+85)=\dfrac{17}{2}(5+85)
    xi=765\sum x_i=765

    xˉ=xin\bar x=\dfrac{\sum x_i}{n}

    xˉ=76517=45\bar x=\dfrac{765}{17}=45

    So, deviations from mean are 40,35,25,20,15,10,5,0,5,10,15,20,25,30,35,4040,35,25,20,15,10,5,0,5,10,15,20,25,30,35,40
    Sum of deviations =360=360

    Mean deviation from mean =36017=21.17=\dfrac{360}{17}=21.17
  • Question 7
    1 / -0
    The mean deviation of a+b2  and  ab2\displaystyle \frac{a+b}{2}\: \: and\: \: \frac{a-b}{2} (where a and b >0) is?
    Solution

    The given observations are: a+b2\dfrac{a + b}{2} and ab2\dfrac{a - b}{2}
    Mean of the observations = a+b2+ab22=a2\dfrac{\dfrac{a + b}{2} + \dfrac{a - b}{2}}{2} = \dfrac{a}{2}

    Now, mean deviation = a+b2a2+ab2a22\dfrac{|\dfrac{a + b}{2} - \dfrac{a}{2}| + |\dfrac{a - b}{2} - \dfrac{a}{2}|}{2} = b2\dfrac{b}{2}

  • Question 8
    1 / -0
    Standard deviation of the distribution 1,3,5,.....,131, 3, 5,....., 13 will be
    Solution
    Mean of the given numbers is xˉ=1+3+5+7+9+11+137=497=7\bar x =\dfrac{1+3+5+7+9+11+13}{7}=\dfrac{49}{7}=7

    total number of values n=7n=7

    Standard deviation is given by SD=i=1n(xixˉ)2n1SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n-1}}

        SD=(17)2+(37)2+(57)2+(77)2+(97)2+(117)2+(137)271\implies SD=\sqrt{\dfrac{(1-7)^2+(3-7)^2+(5-7)^2+(7-7)^2+(9-7)^2+(11-7)^2+(13-7)^2}{7-1}}

        SD=36+16+4+0+4+16+366\implies SD=\sqrt{\dfrac{36+16+4+0+4+16+36}{6}}

        SD=1126\implies SD=\sqrt{\dfrac{112}{6}}

        SD=18.6667=4.32\implies SD=\sqrt{18.6667}=4.32

    Therefore the standard deviation for the given values is 4.324.32
  • Question 9
    1 / -0
    In the following table pass percentage of three schools from the year 20012001 to the year 20062006 are given. Which school students performance is more consistent?

    200120012002200220032003200420042005200520062006
    School (X)808089897979838384846565
    School (Y)929294947676757580806363
    School (Z)939397976767636370708585
    Solution
    Average of x=80+89+79+83+84+656=4806=80x=\cfrac { 80+89+79+83+84+65 }{ 6 } =\cfrac { 480 }{ 6 } =80
    Average of y=92+94+76+75+80+636=4806=80y=\cfrac { 92+94+76+75+80+63 }{ 6 } =\cfrac { 480 }{ 6 } =80
    Average of z=93+97+67+63+70+856=79.16z=\cfrac { 93+97+67+63+70+85 }{ 6 } =79.16
    Average of x and y are equal and more than z.
    So,their performance is consistent. 
  • Question 10
    1 / -0
    The mean deviation of first 8 composite numbers is
    Solution
    Mean deviation is the mean of the absolute differences between each value in the data set and the mean of the data set.

    Mean is the sum of the values in the data set divided by the total number of values.

    A composite number is an integer which is not a prime.
    The first 88 composite numbers are 4,6,8,9,10,12,14,154,6,8,9,10,12,14,15

    Mean of the composite numbers is 4+6+8+9+10+12+14+158\dfrac{4+6+8+9+10+12+14+15}{8}

        mean=788=9.75\implies mean =\dfrac{78}{8}=9.75

    The absolute difference between the data values and mean is 9.754,9.756,9.758,9.759,9.7510,9.7512,9.7514,9.7515|9.75-4|,|9.75-6|,|9.75-8|,|9.75-9|,|9.75-10|,|9.75-12|,|9.75-14|,|9.75-15| 

    Therefore, the data  values after finding the absolute differences are 5.75,3.75,1.75,0.75,0.25,2.25,4.25,5.255.75,3.75,1.75,0.75,0.25,2.25,4.25,5.25 

    Therefore, the mean deviation is 5.75+3.75+1.75+0.75+0.25+2.25+4.25+5.258=248=3\dfrac{5.75+3.75+1.75+0.75+0.25+2.25+4.25+5.25}{8}=\dfrac{24}{8}=3

    Therefore, the mean deviation of first eight composite numbers is 33
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