Question
The number of combines sold each year by a dealership that specialises in agricultural equipment is a Poisson random variable with an average of 4.
What is the probability that the dealership will sell:
3.1 Less than five combines in a given year.
3.2 More than three combines in a period of six months.
3.3 Exactly eight combines in a two year period.
Question 4
If X is a Poisson random variable with a mean of 5, find:
4.1 P( X=3)
4.2 P( X≤3)
4.3 P( X>3)
What is the probability that the dealership will sell:
3.1 Less than five combines in a given year.
3.2 More than three combines in a period of six months.
3.3 Exactly eight combines in a two year period.
Question 4
If X is a Poisson random variable with a mean of 5, find:
4.1 P( X=3)
4.2 P( X≤3)
4.3 P( X>3)
Answers
GPT-4o mini
To solve these problems, we can use the properties of the Poisson distribution. The probability mass function for a Poisson random variable \( X \) with rate parameter \( \lambda \) is given by:
\[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
\]
Where:
- \( \lambda \) is the average or mean number of occurrences,
- \( k \) is the actual number of occurrences,
- \( e \) is approximately equal to 2.71828.
### Question 3
For the first part, we are given that the average number of combines sold per year \( \lambda = 4 \).
#### 3.1: Less than five combines in a given year
We want to calculate \( P(X < 5) \) or \( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \).
Calculating each probability:
\[
P(X = k) = \frac{4^k e^{-4}}{k!}
\]
1. \( P(X=0) = \frac{4^0 e^{-4}}{0!} = e^{-4} \)
2. \( P(X=1) = \frac{4^1 e^{-4}}{1!} = 4 e^{-4} \)
3. \( P(X=2) = \frac{4^2 e^{-4}}{2!} = \frac{16}{2} e^{-4} = 8 e^{-4} \)
4. \( P(X=3) = \frac{4^3 e^{-4}}{3!} = \frac{64}{6} e^{-4} = \frac{32}{3} e^{-4} \)
5. \( P(X=4) = \frac{4^4 e^{-4}}{4!} = \frac{256}{24} e^{-4} = \frac{32}{3} e^{-4} \)
Now summing these up:
\[
P(X < 5) = e^{-4} \left( 1 + 4 + 8 + \frac{32}{3} + \frac{64}{24} \right)
\]
Calculating the numbers gives:
\[
1 + 4 + 8 + \frac{32}{3} = 13 + \frac{32}{3} = \frac{39 + 32}{3} = \frac{71}{3}
\]
Convert back into a normal fraction gives:
\[
P(X < 5) = e^{-4} \left( \frac{71}{3} \right) = \frac{71 e^{-4}}{3}
\]
Using \( e^{-4} \) (approximately 0.0183), you can compute this numerically for an approximate value.
#### 3.2: More than three combines in a period of six months
For a six-month period, the mean number of combines is \( \lambda = 4/2 = 2 \).
We want \( P(X > 3) = 1 - P(X \leq 3) \).
Calculating \( P(X \leq 3) \):
\[
P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
\]
Calculating each:
1. \( P(X=0) = e^{-2} \)
2. \( P(X=1) = 2 e^{-2} \)
3. \( P(X=2) = \frac{4}{2} e^{-2} = 2 e^{-2} \)
4. \( P(X=3) = \frac{8}{6} e^{-2} = \frac{4}{3} e^{-2} \)
Summing:
\[
P(X \leq 3) = e^{-2} \left( 1 + 2 + 2 + \frac{4}{3} \right)
\]
Convert into a common denominator to calculate:
\[
1 + 2 + 2 = 5, \frac{4}{3} = \frac{4}{3} \Rightarrow \text{convert 5 to } \frac{15}{3} = \frac{19}{3}
\]
So,
\[
P(X \leq 3) = e^{-2} \left( \frac{19}{3} \right)
\]
Then
\[
P(X > 3) = 1 - e^{-2} \left( \frac{19}{3} \right)
\]
#### 3.3: Exactly eight combines in a two year period
For two years, the average is \( \lambda = 4 \times 2 = 8 \).
We want \( P(X = 8) \).
\[
P(X=8) = \frac{8^8 e^{-8}}{8!}
\]
Calculating this gives us with \( e^{-8} \) approximately \( 0.000335 \). You can use \( 8! \approx 40320 \).
### Question 4
Given \( \lambda = 5 \) for this Poisson random variable.
#### 4.1: \( P(X = 3) \)
\[
P(X=3) = \frac{5^3 e^{-5}}{3!} = \frac{125 e^{-5}}{6}
\]
Calculate \( e^{-5} \) (approximately \( 0.00674 \)) for numerical results.
#### 4.2: \( P(X \leq 3) \)
Calculate cumulatively:
\[
P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
\]
1. \( P(X=0) = e^{-5} \)
2. \( P(X=1) = 5 e^{-5} \)
3. \( P(X=2) = \frac{25}{2} e^{-5} = 12.5 e^{-5} \)
4. \( P(X=3) = \frac{125}{6} e^{-5} \)
Combine:
\[
P(X \leq 3) = e^{-5} (1 + 5 + 12.5 + \frac{125}{6})
\]
Finding a common denominator again gives us a full total.
#### 4.3: \( P(X > 3) \)
This is simply the complement:
\[
P(X > 3) = 1 - P(X \leq 3)
\]
Now you can compute the results using the values of \( e^{-5} \).
This concludes the necessary calculations!
\[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
\]
Where:
- \( \lambda \) is the average or mean number of occurrences,
- \( k \) is the actual number of occurrences,
- \( e \) is approximately equal to 2.71828.
### Question 3
For the first part, we are given that the average number of combines sold per year \( \lambda = 4 \).
#### 3.1: Less than five combines in a given year
We want to calculate \( P(X < 5) \) or \( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \).
Calculating each probability:
\[
P(X = k) = \frac{4^k e^{-4}}{k!}
\]
1. \( P(X=0) = \frac{4^0 e^{-4}}{0!} = e^{-4} \)
2. \( P(X=1) = \frac{4^1 e^{-4}}{1!} = 4 e^{-4} \)
3. \( P(X=2) = \frac{4^2 e^{-4}}{2!} = \frac{16}{2} e^{-4} = 8 e^{-4} \)
4. \( P(X=3) = \frac{4^3 e^{-4}}{3!} = \frac{64}{6} e^{-4} = \frac{32}{3} e^{-4} \)
5. \( P(X=4) = \frac{4^4 e^{-4}}{4!} = \frac{256}{24} e^{-4} = \frac{32}{3} e^{-4} \)
Now summing these up:
\[
P(X < 5) = e^{-4} \left( 1 + 4 + 8 + \frac{32}{3} + \frac{64}{24} \right)
\]
Calculating the numbers gives:
\[
1 + 4 + 8 + \frac{32}{3} = 13 + \frac{32}{3} = \frac{39 + 32}{3} = \frac{71}{3}
\]
Convert back into a normal fraction gives:
\[
P(X < 5) = e^{-4} \left( \frac{71}{3} \right) = \frac{71 e^{-4}}{3}
\]
Using \( e^{-4} \) (approximately 0.0183), you can compute this numerically for an approximate value.
#### 3.2: More than three combines in a period of six months
For a six-month period, the mean number of combines is \( \lambda = 4/2 = 2 \).
We want \( P(X > 3) = 1 - P(X \leq 3) \).
Calculating \( P(X \leq 3) \):
\[
P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
\]
Calculating each:
1. \( P(X=0) = e^{-2} \)
2. \( P(X=1) = 2 e^{-2} \)
3. \( P(X=2) = \frac{4}{2} e^{-2} = 2 e^{-2} \)
4. \( P(X=3) = \frac{8}{6} e^{-2} = \frac{4}{3} e^{-2} \)
Summing:
\[
P(X \leq 3) = e^{-2} \left( 1 + 2 + 2 + \frac{4}{3} \right)
\]
Convert into a common denominator to calculate:
\[
1 + 2 + 2 = 5, \frac{4}{3} = \frac{4}{3} \Rightarrow \text{convert 5 to } \frac{15}{3} = \frac{19}{3}
\]
So,
\[
P(X \leq 3) = e^{-2} \left( \frac{19}{3} \right)
\]
Then
\[
P(X > 3) = 1 - e^{-2} \left( \frac{19}{3} \right)
\]
#### 3.3: Exactly eight combines in a two year period
For two years, the average is \( \lambda = 4 \times 2 = 8 \).
We want \( P(X = 8) \).
\[
P(X=8) = \frac{8^8 e^{-8}}{8!}
\]
Calculating this gives us with \( e^{-8} \) approximately \( 0.000335 \). You can use \( 8! \approx 40320 \).
### Question 4
Given \( \lambda = 5 \) for this Poisson random variable.
#### 4.1: \( P(X = 3) \)
\[
P(X=3) = \frac{5^3 e^{-5}}{3!} = \frac{125 e^{-5}}{6}
\]
Calculate \( e^{-5} \) (approximately \( 0.00674 \)) for numerical results.
#### 4.2: \( P(X \leq 3) \)
Calculate cumulatively:
\[
P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
\]
1. \( P(X=0) = e^{-5} \)
2. \( P(X=1) = 5 e^{-5} \)
3. \( P(X=2) = \frac{25}{2} e^{-5} = 12.5 e^{-5} \)
4. \( P(X=3) = \frac{125}{6} e^{-5} \)
Combine:
\[
P(X \leq 3) = e^{-5} (1 + 5 + 12.5 + \frac{125}{6})
\]
Finding a common denominator again gives us a full total.
#### 4.3: \( P(X > 3) \)
This is simply the complement:
\[
P(X > 3) = 1 - P(X \leq 3)
\]
Now you can compute the results using the values of \( e^{-5} \).
This concludes the necessary calculations!