Within a computer program, the number of bugs (i.e., coding errors) per lines of code has a Poisson distribution with an average of fifteen

Question

Within a computer program, the number of bugs (i.e., coding errors) per lines of code has a Poisson distribution with an average of fifteen bugs per 1,000 lines. a. Find the probability that there will be exactly eight bugs in 1,000 lines of code. b. Find the probability that there will be at least eight bugs in 1,000 lines of code. c. Find the probability that there will be at least one bug in 1,000 lines of code. d. Find the probability that there will be no more than

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MichaelMet 3 years 2021-07-23T20:03:02+00:00 1 Answers 4 views 0

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  1. thienthi
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    2021-07-23T20:04:57+00:00
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    Answer:

    P(X=8) = 0.0194

    P(X \ge 8)= 0.9820

    P(X \ge 1)= 1

    P(X\le 1) = 0.0000049

    Step-by-step explanation:

    Given

    \lambda = 15

    Poisson distribution is given by:

    P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}

    Solving (a): 8 bugs

    This implies that:

    x = 8

    So, we have:

    P(X=8) = \frac{15^8 * e^{-15}}{8!}

    P(X=8) = \frac{783.99418938}{40320}

    P(X=8) = 0.0194

    Solving (b): At least 8 bugs

    This is represented as: P(X \ge 8)

    Using complement rule:

    P(X \ge 8)= 1 - P(X<8)

    Where

    P(X<8) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)

    P(X<8) = \frac{15^1 * e^{-15}}{1!} + \frac{15^2 * e^{-15}}{2!} + \frac{15^3 * e^{-15}}{3!} + \frac{15^4 * e^{-15}}{4!} + \frac{15^5 * e^{-15}}{5!} + \frac{15^6 * e^{-15}}{6!} + \frac{15^7 * e^{-15}}{7!}

    P(X<8) = (\frac{15^1}{1!} + \frac{15^2}{2!} + \frac{15^3 }{3!} + \frac{15^4}{4!} + \frac{15^5}{5!} + \frac{15^6}{6!} + \frac{15^7}{7!}) e^{-15}

    P(X<8) = (15 + 112.5 + 562.5 + 2109.375 + 6328.125 + 15820.3125 + 33900.6696429) *e^{-15}

    P(X<8) = 58848.4821429 *e^{-15}

    P(X<8) = 0.0180

    So:

    P(X \ge 8)= 1 - P(X<8)

    P(X \ge 8)= 1 - 0.0180

    P(X \ge 8)= 0.9820

    Solving (c): At least 1

    This is represented as: P(X \ge 1)

    Using complement rule:

    P(X \ge 1)= 1 - P(X<1)

    P(X<1) = P(X = 0)

    P(X<1) = \frac{15^0 e^{-15}}{0!}

    P(X<1) = \frac{e^{-15}}{1}

    P(X<1) = e^{-15}

    So:

    P(X \ge 1)= 1 - P(X<1)

    P(X \ge 1)= 1 - e^{-15

    P(X \ge 1)= 0.99999969409

    P(X \ge 1)= 1

    Solving (d): Not more than 1

    This implies at most 1.

    It is represented as:

    P(X\le 1)

    It is calculated using:

    P(X\le 1) = P(X = 0) + P(X =1)

    P(X = 0) = e^{-15}

    P(X=1) = \frac{15^1 * e^{-15}}{1!}

    P(X=1) = 15 * e^{-15}

    So:

    P(X\le 1) = e^{-15} + 15 * e^{-15}

    P(X\le 1) = 0.00000489443

    P(X\le 1) = 0.0000049

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