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Engineering Mathematics Test 11

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Engineering Mathematics Test 11
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  • Question 1
    1 / -0
    In a certain factory of turning razor blades, there is a small chance (1/500) for any blade to be defective. The blades are supplied in packets of 10. Use Poisson distribution to calculate the approximate number of packets containing at least one defective blade in a consignment of 10,000 packets
    Solution

    Explanation:

    Let X = Number of defective Blades in a packet

    Given:

    n=10,p=1500

    np = λ = 0.02

    P(X ≥ 1) = 1 – P(X < 1)

    P(X ≥ 1) = 1 – (P(X = 0))

    ∴ P(X ≥ 1) = 1 – 0.9802

    P(X ≥ 1) = 0.0198

    ∴ Expected number of packets with at least one defective one = 10,000 × 0.0198 = 198
  • Question 2
    1 / -0
    Let X be a binomial random variable with mean 1 and variance 34. The probability that X takes the value of 3 is
    Solution

    Concept:

    Binomial distribution

    P(X=r)=ncrprqnr

    Mean = np

    Variance = npq

    Standard deviation =npq

    Calculation:

    Mean = np = 1

    Variance = npq = 3/4

    p=14,q=34,n=4

    P(X=3)=4c3(14)3(34)43

    P(X=3)=4×164×34=364

  • Question 3
    1 / -0

    Probability density function of a random variable X is given below

    f(x)={0.25if1x50otherwise

    P (X ≤ 4) is

    Solution

    Concept:

    The probability density function of a random variable is,

    P(x4)=4f(x)dx

    =1(0) dx+14(0.25)dx+4(0)dx

    =14(x)14=14(41)

    P(x4)=34

  • Question 4
    1 / -0
    Out of 800 families with 4 children each, how many families would be expected to have 2 boys and 2 girls.
    Solution

    Explanation:

    P(2 boys and 2 girls) =P(X=2)=4C2(12)2(12)42=6×(12)4

    ∴ P(2 boys and 2 girls) =P(X=2)=38

    Now,

    Number of families having 2 boys and 2 girls = N ⋅ P(X = 2)

    (where N is the total number of families considered)

    ∴ Number of families having 2 boys and 2 girls =800×38

    ∴ Number of families having 2 boys and 2 girls = 300 
  • Question 5
    1 / -0
    If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 3)2] equals ______.
    Solution

    Concept:

    In case of Poisson distribution, mean and variance are same.

    Calculation:

    Given, mean = 5

    E[(X + 3)2] = E [X2 + 6X + 9] = E[X2] + E[6X] + E[9]

    Variance = E[X2] - (E[X])2

    As, mean = variance = 5

    Mean = E[X]

    5 = E[X2] -  (5)2

    5 = E[X2] - 25

    E[X2] = 30

    So, E[(X+3)2] = 30 + 6 × 5 + 9 = 69
  • Question 6
    1 / -0
    At a busy traffic intersection, the probability p of an individual car having an accident is p = 0.0001. If 1000 cars pass through the intersection between 5 P.M and 6 P.M. every day, then the probability that two or more car accidents occur during that period is ______. (use Poisson distribution)
    Solution

    Explanation:

    Let X = number of accidents between 5 p.m and 6 p.m.

    For Poisson distribution,

    λ = np = (1000)(0.0001) = 0.1

    P(X=x)=eλλxx!(x=0,1,2,.)

    Now,

    Required Probability = P(X ≥ 2)

    Required Probability = 1 – P(X < 2)

    Required Probability = 1 – {P(X = 0) + P(X = 1)}

    Required Probability = 1 – e-0.1 (1 + 0.1)

    Required Probability = 0.0045

  • Question 7
    1 / -0

    A continuous random variable  has a probability density function as

    f(x) = 3x2, 0 ≤ x ≤ 1, Find a and b such that

    (i) P(X ≤ a) = P(X > a) and

    (ii) P(X > b) = 0.05

    Solution

    Explanation:

    (i) P(X ≤ a) = P(X > a)

    0a3x2dx=a13x2dx

    a3 = 1 – a3

    a3=12

    a = 0.7937

    (ii) P(X > b) = 0.05

    b13x2dx=0.05

    b3 = 0.95

    ∴ b = 0.9830

  • Question 8
    1 / -0
    A die is tossed 180 times. Using normal distribution find the probability that the face 4 will turn up atleast 35 times (Area under the normal curve between Z = 0 and Z = 1 is 0.3413)
    Solution

    Explanation:

    Mean=μ=np=

    Mean=(180)16

    Mean = 30

    Standarddeviation=σ=npq

    Standard deviation = 5

    Now,

    Let x = number of times the face 4 will turn up

    z=xμσ

    z=x305

    When x = 35, z = 1

    Required probability = P(z ≥ 1)

    Required probability = 1 – P(z < 1)

    Required probability = 1 – (0.5 + P(0 < z < 1))

    Required probability = 1 – (0.5 + 0.3413)

    Required probability = 0.1587

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