Respuesta :
Answer:
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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] 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 [tex]\mu = 3n[/tex], 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 [tex]P(X = 0)[/tex] when [tex]\mu = 3*1 = 3[/tex]. So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]
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
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]. Then
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]
[tex]P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.8008[/tex]
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 [tex]n = 0.5[/tex]
[tex]\mu = 3n = 3*0.5 = 1.5[/tex]
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 [tex]\mu = 1.5[/tex]. So
[tex]P(X = 0) = \frac{e^{-1.5}*1.5^{0}}{(0)!} = 0.2231[/tex]
0.2231 = 22.31% probability of no off-the-job accidents during the next six months.
