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Measures of Dispersion Test - 9

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Measures of Dispersion Test - 9
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  • Question 1
    1 / -0
    Find the Standard Deviation of the following data:
    $$5,9,8,12,6,10,6,8$$
    Solution
    Here $$n=8$$.
    Mean, $$u=\dfrac{\sum x_i}{n}=\dfrac{64}{8}=8$$
    Value of Deviations $$(x_i-u)$$ are
    $$-3,1,0,4,-2,2,-2,0$$
    Variance, $$\sigma ^2=\dfrac{\sum(x_i-u)^2}{n}\\=\dfrac{(-3-8)^2-(1-8)^2-(0-8)^2-(4-8)^2-(-2-8)^2-(2-8)^2-(-2-8)^2-(0-8)^2}{8}\\=\dfrac{38}{8}=4.75$$.
    Standard Deviation, $$S.D=\sqrt{\sigma ^2}$$
    $$\therefore S.D=\sqrt{4.75}=2.17$$
    Ans- Option $$D$$. 
  • Question 2
    1 / -0
    Find the variance of first $$10$$ multiples of $$3$$.
    Solution
    First $$10$$ multiples of $$3$$ are $$3,6,9...30$$.
    This is an A.P.
    $$sum=\dfrac n2 (a+l)= \dfrac {10}{2} \times (3+30)$$
    $$\therefore sum=165$$.
    Mean, $$u=\dfrac {sum}{n}=\dfrac{165}{10}$$.
    Variance, $$\sigma ^2=\dfrac{\sum(x_i ^2)}{n}-u^2$$.
    $$\therefore \sigma ^2=\dfrac{3^2+6^2+...30^2}{10}-{16.5}^2$$.
    $$\therefore \sigma ^2=\dfrac{3\times(1^2+2^2+...10^2)}{10}-{16.5}^2$$.
    $$\therefore \sigma ^2=\dfrac{9\times 10\times (10+1)\times (2\times 10+1)}{6\times 10}-{16.5}^2$$.
    $$\therefore \sigma ^2=346.5-272.25$$
    $$\therefore \sigma ^2=74.25$$
    Ans-Option $$C$$.
  • Question 3
    1 / -0
    Standard Deviation of first n natural numbers.
    Solution
    Mean, $$u=\dfrac{(1+2+3...+n)}{n}$$

    $$\therefore u=\dfrac 12 (n+1)$$

    Variance, $$\sigma ^2=\dfrac{\sum(x_i-u)^2}{n}=\dfrac{\sum x_i^2}{n}-u^2$$

    $$\therefore \sigma ^2= \dfrac{\sum n^2}{n}-{\dfrac 12 (n+1)}^2$$

    $$\therefore \sigma ^2= \dfrac 1n \dfrac{n(n+1)(2n+1)}{n}-({\dfrac 12 (n+1)})^2$$

    $$\therefore \sigma ^2=\dfrac{n^2-1}{12}$$
    Standard Deviation, $$S.D=\sqrt{\sigma ^2}$$
    $$\therefore S.D=$$\sqrt{\dfrac{n^2-1}{12}}$$$$

    Ans- Option $$B$$. 
  • Question 4
    1 / -0
    Variance of first $$n$$ natural numbers.
    Solution
    Mean, $$u=\dfrac{(1+2+3...+n)}{n}$$

    $$\therefore u=\dfrac 12 (n+1)$$

    Variance, $$\sigma ^2=\dfrac{\sum(x_i-u)^2}{n}=\dfrac{\sum x_i^2}{n}-u^2$$

    $$\therefore \sigma ^2= \dfrac{\sum n^2}{n}-{\dfrac 12 (n+1)}^2$$

    $$\therefore \sigma ^2= \dfrac 1n \dfrac{n(n+1)(2n+1)}{n}-({\dfrac 12 (n+1)})^2$$

    $$\therefore \sigma ^2=\dfrac{n^2-1}{12}$$
    Ans-Option $$C$$
  • Question 5
    1 / -0
    Standard Error of Mean is defined as ________.
  • Question 6
    1 / -0

    Directions For Questions

    Factory$$A$$$$B$$
    Number of Workers$$4000$$$$5000$$
    Mean Wages$$3500$$$$3500$$
    Variance in wages$$64$$$$81$$

    ...view full instructions

    Which factory has more variation in wages?
    Solution
    $$(\sigma _A) ^2=64$$ and $$(\sigma _B) ^2=81$$
    $$\because S.D=\sqrt {\sigma ^2}$$
    $$\therefore S.D_A=8$$ and $$S.D_B=9$$
    Since monthly mean wages in two factories are same, and $$ S.D_A < S.D_B$$.
    Thus, the factory $$B$$ has more variation in wages.
    Ans-Option $$B$$.
  • Question 7
    1 / -0
    The Standard error is denoted by__________.
    Solution
    Standard error is a statistical measure that shows the accuracy by which a sample represents the population. Standard error is denoted by both the equations mentioned in option a and b.
  • Question 8
    1 / -0
    A sampling distribution of means is______________.
    Solution
    A sampling distribution is the probability distribution of a given statistics based on random samples, it shows the likely values of the various sample means.
  • Question 9
    1 / -0
    The variance of six observations is given as $$16$$ and mean is $$8$$. If each observation is multiplied by $$3$$.
    Find the new variance:
    Solution
    Mean, $$u=8$$.
    So, $$\dfrac{x_1+x_2...x_6}{6}=8$$.
    $$\therefore x_1+x_2...x_6=48$$.
    Multiplying the above equation by $$3$$ on both the sides we get,
    $$3x_1+3x_2...3x_6=48\times 3$$
    $$\therefore u'=\dfrac{3x_1+3x_2....3x_6}{6}=\dfrac{48 \times 3}{6}$$
    $$\therefore u'=24$$
    Thus, new mean, $$u'=24$$.
    $$\therefore$$ New variance, $$\sigma ^2=\dfrac{(3x_1)^2+...(3x_6)^2}{6}-(24)^2$$
    $$\therefore \sigma ^2=9\times \dfrac{(x_1)^2+....(x_6)^2}{6}-576$$
    $$\therefore \sigma ^2=9\times \dfrac{480}{16}-576$$
    $$\therefore \sigma ^2=720-576$$
    $$\therefore \sigma ^2=144$$
    Thus, New Variance is $$144$$.
    Ans-Option $$D$$
  • Question 10
    1 / -0
    Which of the following is true with respect to range?
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