Mathematics, 14.07.2020 01:01 babygirl2984

The National Safety Council (NSC) estimates that off-the-job accidents cost U. S. businesses almost $200 billion annually in lost productivity (National Safety Council, March 2006). Based on NSC estimates, companies with 50 employees are expected to average three employee off-the-job accidents per year. Answer the following questions for companies with 50 employees. a. What is the probability of no off-the-job accidents during a one-year period (to 4 decimals)?b. What is the probability of at least two off-the-job accidents during a one-year period (to 4 decimals)?c. What is the expected number of off-the-job accidents during six months (to 1 decimal)?d. What is the probability of no off-the-job accidents during the next six months (to 4 decimals)?

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Answer from: aprilpendergrass

a. 0.0498 = 4.98% probability of no off-the-job accidents during a one-year period

b. 0.8008 = 80.08% probability of at least two off-the-job accidents during a one-year period.

c. The expected number of off-the-job accidents during six months is 1.5.

d. 0.2231 = 22.31% probability of no off-the-job accidents during the next six months.

Step-by-step explanation:

We have the mean during a period, so we use the Poisson distribution.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Companies with 50 employees are expected to average three employee off-the-job accidents per year.

This means that $\mu = 3n$ , in which n is the number of years.

a. What is the probability of no off-the-job accidents during a one-year period (to 4 decimals)?

This is P(X = 0) when $\mu = 3*1 = 3$ . So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

0.0498 = 4.98% probability of no off-the-job accidents during a one-year period.

b. What is the probability of at least two off-the-job accidents during a one-year period (to 4 decimals)?

Either there are less than two accidents, or there are at least two. The sum of the probabilities of these events is 1. So

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$ . Then

$P(X \geq 2) = 1 - P(X < 2)$

In which

P(X < 2) = P(X = 0) + P(X = 1)

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

P(X < 2) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.8008$

0.8008 = 80.08% probability of at least two off-the-job accidents during a one-year period.

c. What is the expected number of off-the-job accidents during six months (to 1 decimal)?

6 months is half a year, so n = 0.5

$\mu = 3n = 3*0.5 = 1.5$

The expected number of off-the-job accidents during six months is 1.5.

d. What is the probability of no off-the-job accidents during the next six months (to 4 decimals)?

This is P(X = 0) when $\mu = 1.5$ . So

$P(X = 0) = \frac{e^{-1.5}*1.5^{0}}{(0)!} = 0.2231$

0.2231 = 22.31% probability of no off-the-job accidents during the next six months.

Answer from: Quest

2500

step-by-step explanation:

Answer from: Quest

11123355543321112345566444

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