Respuesta :
Answer:
Therefore the probability of getting exactly 5 defective motors is 0.10082.
Step-by-step explanation:
Poisson distribution:
Poisson distribution is a tool that the number of an event is likely to occur within a time period.
[tex]P(X=x)=\frac{e^{-\mu}\mu^x}{x!}[/tex]
μ= mean number of occurrence in interval.
Given that,
The probability a defective electric motor is 0.01.
Total number of motor is 300.
Mean number= μ=(300×0.01)= 3,
The number of defective motor is 5.
The Poisson distribution
[tex]P(X=5)=\frac{e^{-3}3^5}{5!}[/tex]
=0.10082
Therefore the probability of getting exactly 5 defective motors is 0.10082.
Answer:
Probability that a sample of 300 electric motors will contain exactly 5 defective motors is 0.10082.
Step-by-step explanation:
We are given that a company makes electric motors. The probability an electric motor is defective is 0.01.
And we have to find the probability that a sample of 300 electric motors will contain exactly 5 defective motors.
Firstly, we consider the above situation as of Poisson distribution because here binomial distribution can't be used as sample size is very large.
The probability distribution of Poisson distribution is given as;
[tex]P(X=x) = \frac{e^{-\lambda} \times \lambda^{x} }{x!} ; x = 0,1,2,3,....[/tex]
where, [tex]\lambda[/tex] = chances of electric motor being defective
So, for a sample of 300 electric motors, [tex]\lambda[/tex] = [tex]300 \times 0.01[/tex] = 3
Let X = No. of defective motors
So, X ~ Poisson([tex]\lambda = 3[/tex])
Now, probability that a sample of 300 electric motors will contain exactly 5 defective motors is given by = P(X = 5)
P(X = 5) = [tex]\frac{e^{-3} \times 3^{5} }{5!}[/tex] = [tex]\frac{e^{-3} \times 243 }{120}[/tex] {As 5! = 120}
= [tex]e^{-3} \times 2.025[/tex]
= 0.10082
Therefore, probability that a sample of 300 electric motors will contain exactly 5 defective motors is 0.10082.
