Engineering Mathematics Miscellaneous

Engineering Mathematics Miscellaneous

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Engineering Mathematics

Engineering Mathematics Miscellaneous
  1. Let X1, X2 be two independent normal random variables with means μ1, μ2 and standard deviations σ1, σ2 respectively. Consider Y = X1 – X2; μ1 = μ2 = 1, σ1 = 1, σ2 = 2, Then,








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    Y is normally distributed wit h mean 0 and variance 5.

    Correct Option: B


    Y is normally distributed wit h mean 0 and variance 5.

  1. Consider a Poisson distribution for the tossing of a biased coin. The mean for this distribution is μ. The standard deviation for this distribution is given by








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    Standard deviation √Variance = √μ

    Correct Option: A

    Standard deviation √Variance = √μ

  1. A machine produces 0, 1 or 2 defective pieces in a day with associated probability of 1/6, 2/3 and 1/6, respectively. The mean value and the variance of the number of defective pieces produced by the machine in a day, respectively, are








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    Let ‘x’ be no. of defective pieces.

    Mean (μ) = E(x) = ∑ x P(x)

    =0 ×
    1
    		+1 ×
    1
    		+2 ×
    1
    636

    = 0 +
    2
    		+
    1
    		= 1
    33

    E(x2) = ∑ x2 P(x)
    =0 ×
    1
    		+1 ×
    2
    		+4 ×
    1
    636

    = 0 +
    2
    		+
    2
    		=
    4
    333

    Variance, V(x) = E(x2) – {E(x)}2
    =
    4
    		- 1 =
    1
    33

    Correct Option: A

    Let ‘x’ be no. of defective pieces.

    Mean (μ) = E(x) = ∑ x P(x)

    =0 ×
    1
    		+1 ×
    1
    		+2 ×
    1
    636

    = 0 +
    2
    		+
    1
    		= 1
    33

    E(x2) = ∑ x2 P(x)
    =0 ×
    1
    		+1 ×
    2
    		+4 ×
    1
    636

    = 0 +
    2
    		+
    2
    		=
    4
    333

    Variance, V(x) = E(x2) – {E(x)}2
    =
    4
    		- 1 =
    1
    33

  1. In the following table, x is a discrete random variable and p(x) is the probability density. The standard deviation of x is








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    Given

    mean (μ) = exp(x) = 1 × 0.3 + 2 × 0.6 + 3 × 0.1
    = 0.3 + 1.2 + 0.3 = 1.8
    E(x2) = ∑ x2P(x)
    = 1 × 0.3 + 4 × 0.6 + 9 × 0.1
    = 0.3 + 2.4 + 0.9 = 3.6
    Variance v (x) = E (x2) – μ2 = 3.6 – (1.8)2
    S.D(σ) = + √υ(x)
    = + √3.6 - (1.8)2
    = √0.36 = 0.6

    Correct Option: D

    Given

    mean (μ) = exp(x) = 1 × 0.3 + 2 × 0.6 + 3 × 0.1
    = 0.3 + 1.2 + 0.3 = 1.8
    E(x2) = ∑ x2P(x)
    = 1 × 0.3 + 4 × 0.6 + 9 × 0.1
    = 0.3 + 2.4 + 0.9 = 3.6
    Variance v (x) = E (x2) – μ2 = 3.6 – (1.8)2
    S.D(σ) = + √υ(x)
    = + √3.6 - (1.8)2
    = √0.36 = 0.6

  1. The standard deviation of a uniformly distributed random variable between 0 and 1 is








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    Variance,

    V(x) =
    (b - a)2
    		=
    (1 - 0)2
    1212

    ∴ Standard deviation,
    σ =
    1
    √12

    Correct Option: A

    Variance,

    V(x) =
    (b - a)2
    		=
    (1 - 0)2
    1212

    ∴ Standard deviation,
    σ =
    1
    √12