A new car was put into production. It involved many assembly tasks. Each car was inspected at the end of the assembly line and the number of defects per unit was recorded. For the first 100 cars produced, there were 40 defective cars. Some of the cars had no defects, a few had one defect, and so on. The distribution of defects followed a Poisson distribution. Based on the first 100 cars produced, about how many out of every 1,000 cars assembled should have one or more defects?
Poison Disrribution! ... 330
Is there something I am not getting her? The sample was .40 defective.
so in a population of 1000, that is 400 cars.
To estimate the number of cars out of every 1,000 that should have one or more defects, we need to use the information from the first 100 cars produced, where we know that there were 40 defective cars in total.
Using the Poisson distribution, we can calculate the average number of defects per car, which is equal to the total number of defects divided by the number of cars produced:
Average defects per car = Total defects / Number of cars produced
= 40 / 100
= 0.4 defects per car
Next, we can calculate the probability of a car having no defects using the Poisson distribution. The probability mass function (PMF) of the Poisson distribution is given by:
P(X; λ) = (e^-λ * λ^X) / X!
Where:
X is the number of defects (in this case, 0),
λ is the average number of defects per car.
So, the probability of a car having no defects is:
P(X=0) = (e^-0.4 * 0.4^0) / 0!
= (e^-0.4 * 1) / 1
= e^-0.4
Using this probability, we can calculate the probability of a car having one or more defects:
P(X ≥ 1) = 1 - P(X=0)
= 1 - e^-0.4
To estimate the number of cars out of every 1,000 that should have one or more defects, we can multiply this probability by 1,000:
Number of cars with one or more defects = P(X ≥ 1) * 1,000
= (1 - e^-0.4) * 1,000
Calculating this expression, we can estimate the number of cars out of every 1,000 assembled that should have one or more defects.
To determine the number of cars out of every 1,000 that should have one or more defects, we first need to estimate the rate parameter for the Poisson distribution.
The rate parameter (λ) can be estimated using the sample mean, which is equal to the number of defects divided by the number of cars produced. In this case, the sample mean is equal to 40 defects / 100 cars, which is 0.4 defects per car.
The Poisson distribution gives the probability of observing a certain number of events (defects) in a fixed interval (in this case, each car). The parameter λ represents the average number of events per interval.
To calculate the probability of having no defects in one car, we can use the Poisson probability formula:
P(X = 0) = (e^(-λ) * λ^0) / 0!
Where e is the base of the natural logarithm (approximately 2.71828), λ is the rate parameter, and 0! is the factorial of zero (which is equal to 1).
For λ = 0.4, plugging in the values, we get:
P(X = 0) = (e^(-0.4) * 0.4^0) / 0!
= (e^(-0.4) * 1) / 1
≈ 0.6703
Therefore, the probability of having no defects in one car is approximately 0.6703.
To compute the probability of having one or more defects in one car, we can subtract the probability of having no defects from 1:
P(X ≥ 1) = 1 - P(X = 0)
= 1 - 0.6703
≈ 0.3297
Therefore, the probability of having one or more defects in one car is approximately 0.3297.
Now, to estimate the number of cars out of every 1,000 with one or more defects, we can multiply this probability by 1,000:
Number of cars with one or more defects = 0.3297 * 1,000
= 329.7
Rounding off, about 330 out of every 1,000 cars assembled should have one or more defects based on the data from the first 100 cars produced.