Question

Compare the Poisson approximation with the correct binomial probability for the following cases: (a) \(P(X=2\\}\) when \(n=8, p=0.1\). (b) \(P[X=9\\}\) when \(n=10, p=0.95\). (c) \(P[X=0\\}\) when \(n=10, p=0.1\). (d) \(P\\{X=4\\}\) when \(n=9, p=0.2\).

Short answer

Comparing the computed Binomial probabilities and Poisson approximations for each case, we observe the following: (a) Binomial probability: \(P(X=2) \approx 0.2634\); Poisson approximation: \(P(X=2) \approx 0.2874\) (b) Binomial probability: \(P(X=9) \approx 0.6311\); Poisson approximation: \(P(X=9) \approx 0.6668\) (c) Binomial probability: \(P(X=0) \approx 0.3487\); Poisson approximation: \(P(X=0) \approx 0.3679\) (d) Binomial probability: \(P(X=4) \approx 0.0741\); Poisson approximation: \(P(X=4) \approx 0.0983\) The Poisson approximation provides a reasonable estimation of the binomial probability, especially when the success probability is low and the number of trials is high. However, its accuracy decreases as the success probability approaches 1.

Step-by-step solution

Case (a) - Compute the binomial probability

For \(P(X=2)\) when \(n=8, p=0.1\), the binomial probability is given by: \[P(X=2) = {n \choose x} p^x (1-p)^{n-x}\] \[P(X=2) = {8 \choose 2} (0.1)^{2} (0.9)^{6}\]

Case (a) - Compute the Poisson approximation

Now compute the Poisson approximation using the formula: \[P(X=k) \approx \frac{e^{-\lambda} \lambda^k}{k!}\] Where \(\lambda=np\): \[\lambda = 8 \cdot 0.1 = 0.8\] \[P(X=2) \approx \frac{e^{-0.8} 0.8^2}{2!}\]

Case (b) - Compute the binomial probability

For \(P[X=9]\) when \(n=10, p=0.95\), the binomial probability is given by: \[P(X=9) = {10 \choose 9} (0.95)^{9} (0.05)^{1}\]

Case (b) - Compute the Poisson approximation

Now compute the Poisson approximation with \(\lambda=np\): \[\lambda = 10 \cdot 0.95 = 9.5\] \[P(X=9) \approx \frac{e^{-9.5} 9.5^9}{9!}\]

Case (c) - Compute the binomial probability

For \(P[X=0]\) when \(n=10, p=0.1\), the binomial probability is given by: \[P(X=0) = {10 \choose 0} (0.1)^{0} (0.9)^{10}\]

Case (c) - Compute the Poisson approximation

Now compute the Poisson approximation with \(\lambda=np\): \[\lambda = 10 \cdot 0.1 = 1\] \[P(X=0) \approx \frac{e^{-1} 1^0}{0!}\]

Case (d) - Compute the binomial probability

For \(P\\{X=4\\}\) when \(n=9, p=0.2\), the binomial probability is given by: \[P(X=4) = {9 \choose 4} (0.2)^{4} (0.8)^{5}\]

Case (d) - Compute the Poisson approximation

Now compute the Poisson approximation with \(\lambda=np\): \[\lambda = 9 \cdot 0.2 = 1.8\] \[P(X=4) \approx \frac{e^{-1.8} 1.8^4}{4!}\] Now that we have computed the binomial probabilities and Poisson approximations for each case, we can compare the results: (a) Binomial probability: \(P(X=2) \approx 0.2634\); Poisson approximation: \(P(X=2) \approx 0.2874\) (b) Binomial probability: \(P(X=9) \approx 0.6311\); Poisson approximation: \(P(X=9) \approx 0.6668\) (c) Binomial probability: \(P(X=0) \approx 0.3487\); Poisson approximation: \(P(X=0) \approx 0.3679\) (d) Binomial probability: \(P(X=4) \approx 0.0741\); Poisson approximation: \(P(X=4) \approx 0.0983\) The results show that the Poisson approximation provides a reasonable estimation of the binomial probability, especially in cases where the success probability is low, and the number of trials is high. However, as the success probability gets closer to 1, the approximation becomes less accurate.