Suppose that a certain type of metalic tube contains on the average five defects per
900 feet. Let X be the number of defects per 900 feet
1. What distribution does X follows ?
2. What is the probability that a tube 900 feet long contains no defects ?
3. What is the probability that a tube 1200 feet long contains more than 3
defects ?
4. Suppose in this part that X represents the number of defective of a tube that
is 900 feet long. Find P(µ − 1.2σ < X < µ + 1.2σ), and compare it with
Chebeyshev’s theorem.
1. The distribution that X follows is a Poisson distribution.
2. The average number of defects per 900 feet is given as 5. In a Poisson distribution, the probability of zero defects is given by the formula P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average number of defects.
Substituting λ = 5 and k = 0 into the formula, we get P(X = 0) = (e^(-5) * 5^0) / 0! = (e^(-5) * 1) / 1 = e^(-5) ≈ 0.0067.
Therefore, the probability that a tube 900 feet long contains no defects is approximately 0.0067.
3. We need to find the probability that a tube 1200 feet long contains more than 3 defects. As the average number of defects per 900 feet is 5, the average number of defects per 1200 feet would be (5/900) * 1200 = 6.67.
Using the Poisson distribution formula, we can calculate P(X > 3) = 1 - P(X ≤ 3) = 1 - [(e^(-6.67) * 6.67^0) / 0! + (e^(-6.67) * 6.67^1) / 1! + (e^(-6.67) * 6.67^2) / 2! + (e^(-6.67) * 6.67^3) / 3!]
Calculate this expression to find the probability.
4. Assuming X represents the number of defects in a tube that is 900 feet long, we can find P(µ - 1.2σ < X < µ + 1.2σ).
First, we need to find the mean (µ) and standard deviation (σ) of the Poisson distribution. The mean (µ) is given as 5, and the standard deviation (σ) of a Poisson distribution is equal to the square root of the mean.
So σ = √5 ≈ 2.236.
Now, we can calculate the probability using the Chebyshev's theorem, which states that for any distribution (not just Poisson), the probability of a value falling within k standard deviations of the mean is at least 1 - 1/k^2.
Here, k = 1.2, so the probability P(µ - 1.2σ < X < µ + 1.2σ) is at least 1 - 1/(1.2)^2.
Evaluate this expression to compare it with the probability obtained using the Poisson distribution.
1. The distribution that X follows is a Poisson distribution. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space when the events occur randomly and independently with a known average rate.
2. To find the probability that a tube 900 feet long contains no defects, we can use the Poisson distribution formula. The mean (average) number of defects per 900 feet is given as 5. The formula for the probability of having exactly k events in a Poisson distribution is:
P(X = k) = (e^(-λ) * λ^k) / k!
Where λ is the mean (average) number of events, and e is the base of the natural logarithm.
In this case, we want to find the probability of having zero defects, so k = 0. Plugging in the values into the formula:
P(X = 0) = (e^(-5) * 5^0) / 0!
Simplifying further:
P(X = 0) = (e^(-5) * 1) / 1
P(X = 0) = e^(-5)
Using a calculator or a table of values, you can evaluate the exponential function e^(-5) to get the probability.
3. To find the probability that a tube 1200 feet long contains more than 3 defects, we can also use the Poisson distribution formula. However, we need to adjust the mean (average) accordingly because the length of the tube is different.
Given that the tube is 900 feet long and has an average of 5 defects, the mean can be adjusted as follows:
Mean for a 1200 feet tube = (Mean for a 900 feet tube) * (Length of 1200 feet tube / Length of 900 feet tube)
Mean for a 1200 feet tube = 5 * (1200 / 900)
Once you have the adjusted mean, you can use the Poisson distribution formula to find the probability of having more than 3 defects.
4. In this part, we are given that X represents the number of defects in a tube that is 900 feet long. The notation µ represents the mean of the distribution, and σ represents the standard deviation.
To find P(µ - 1.2σ < X < µ + 1.2σ), we need to calculate the values of µ and σ. In a Poisson distribution, the mean (µ) and the standard deviation (σ) are both equal to the square root of the average number of defects.
First, calculate the mean and standard deviation:
µ = sqrt(5)
σ = sqrt(5)
Next, calculate the lower and upper bounds for the range:
Lower bound = µ - 1.2σ
Upper bound = µ + 1.2σ
Finally, use the Poisson distribution formula to calculate the probability:
P(µ - 1.2σ < X < µ + 1.2σ) = P(Lower bound < X < Upper bound)
Compare the result with Chebyshev's theorem, which states that for any distribution with a mean µ and standard deviation σ, the probability of being within k standard deviations of the mean is at least (1 - 1/k^2).