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elena-s [515]
2 years ago
9

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
Mathematics
1 answer:
Alex Ar [27]2 years ago
8 0

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

Where

P(X

P(X

P(X

P(X

P(X

P(X

So:

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

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

P(X

P(X

P(X

P(X

So:

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

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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