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Statistics Test - 34

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Statistics Test - 34
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  • Question 1
    1 / -0
    Suppose for $$40$$ observations, the variance is $$50$$. If all the observations are increased by $$20$$, the variance of these increased observation will be
    Solution
    The variance for $$40$$ observations is $$50$$
    Variance is independent of the number of observations. Therefore, variance for $$60$$ observations is $$50$$
  • Question 2
    1 / -0
    Variance of the first $$11$$ natural numbers is:
    Solution
    Sum of first $$n$$ natural numbers is $$\dfrac {n^2-1}{12}$$
    $$=\dfrac {11^2-1}{12}$$
    $$=\dfrac {120}{12}$$
    $$=10$$
    Thus, variance is $$\sqrt {\dfrac {n^2-1}{12}}$$ $$=\sqrt {10}$$.
  • Question 3
    1 / -0
    If a, b are constants then, $$Var(a+bX)$$ is
    Solution
    $$Var(a+bX)=Var(a)+Var(bX)$$

    $$\implies Var(a+bX)=0+Var(bX)$$ (Since, $$Var(c)=0$$)

    $$\implies Var(a+bX)=b^2Var(X)$$ (Since, $$Var(aX)=a^2Var(X)$$)
  • Question 4
    1 / -0
    If $$X, Y $$ are independent then $$SD(X-Y)$$ is:
    Solution
    $$SD(X-Y)=SD(X)+SD(-Y)$$

    $$\implies SD(X-Y)=SD(X)+SD(Y)$$ (since, $$SD(aX)=|a|SD(X)$$)
  • Question 5
    1 / -0
    If the standard deviation of a population is $$9$$, the population variance is:
    Solution
    We know that variance is square of standard deviation.

    Given that standard deviation is $$9$$

    Therefore, variance $$=9^2=81$$
  • Question 6
    1 / -0
     If total sum of square is $$20$$ and sample variance is $$5$$ then total number of observations are
    Solution
    We know that variance is $$\dfrac{\text{sum of the squares}}{\text{total observations}}$$

    Given that sum of squares is $$20$$ and
    the variance is $$5$$

    Therefore, $$5=\dfrac{20}{\text{total observations}}$$

    $$\implies \text{total observations}=\dfrac{20}5=4$$
  • Question 7
    1 / -0
    Find $$Var(2X+3)$$
    Solution
    $$Var(2X+3)=Var(2X)+Var(3)$$

    $$\implies Var(2X+3)=Var(2X)+0$$ (Since, $$Var(c)=0$$)

    $$\implies Var(2X+3)=2^2Var(X)$$ (Since, $$Var(aX)=a^2Var(X)$$)

    $$\implies Var(2X+3)=4Var(X)$$
  • Question 8
    1 / -0
    If the standard deviation of a set of scores is $$1.2$$ and their mean is $$10$$, then the coefficient of variation of the scores is
    Solution
    Given : standard deviation$$(\sigma)=1.2,$$ mean$$(\overline {X})=10$$.
    Coefficient of variation(C.V.) $$=\dfrac{\sigma}{\overline {X}}\times 100=\dfrac{1.2}{10}\times 100=12$$
    $$\therefore$$ C.V. $$=12$$
    Hence, option $$A$$ is correct.
  • Question 9
    1 / -0
    Calculate the standard deviation of the first $$13$$ natural numbers.
    Solution
    $$\Rightarrow$$  $$1st$$ $$13$$ natural numbers are $$1,2,3,4,5,6,7,8,9,10,11,12,13.$$
    $$\sum x=\dfrac{n(n+1)}{2}$$
             $$=\dfrac{13(13+1)}{2}$$
             $$=\dfrac{13\times 14}{2}$$
    $$\therefore$$  $$\sum x=91$$
    Now, $$\sum x^2=\dfrac{n(n+1)(2n+1)}{6}$$
                        $$=\dfrac{13\times 14\times 27}{6}$$
    $$\therefore$$  $$\sum x^2=819$$
    $$\Rightarrow$$  $$SD=\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$$
                  $$=\sqrt{\dfrac{819}{13}-\left(\dfrac{91}{13}\right)^2}$$
                  $$=\sqrt{63-49}$$
                  $$=\sqrt{14}$$ 
                  $$=3.74$$
  • Question 10
    1 / -0
    If the variance of a data is $$12.25$$, then the standard deviation is
    Solution
    $$S.D=\sqrt{variance}$$
    $$\Rightarrow S.D=\sqrt{12.25}$$
    $$\Rightarrow S.D=3.5$$
    Option A is correct.
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