Question

Suppose the number of invasive carp in one mile of river is distributed according to the Poisson distribution with a mean of 8 carp per mile. Use the normal approximation to the Poisson to find the probability of fewer than 70 carp being found in a 10 mile stretch of river.

First, without the continuity correction:

Second, with the continuity correction

Answer 1

To find the probability of fewer than 70 carp being found in a 10-mile stretch of river using the normal approximation to the Poisson distribution, we can follow these steps. Let's first calculate it without the continuity correction and then with the continuity correction.

Step 1: Calculate the parameters of the normal approximation:

Given that the mean of carp per mile is 8 (λ = 8), the mean for a 10-mile stretch would be 10 * λ = 10 * 8 = 80.

The standard deviation for the Poisson distribution is the square root of the mean, so for this case, it is √80 ≈ 8.944.

Step 2: Calculate without the continuity correction:

We want to find the probability of having fewer than 70 carp in a 10-mile stretch, which is equivalent to finding the probability of having 69 or fewer carp in that stretch.

Now, we can use the standard normal distribution to approximate the probability:

Z = (X - μ) / σ

Where: X = Number of carp (69 in this case) μ = Mean of the normal distribution (80) σ = Standard deviation of the normal distribution (8.944)

Z = (69 - 80) / 8.944 ≈ -1.232

Now, we find the probability using the standard normal table or calculator:

P(Z < -1.232) ≈ 0.1093

So, without the continuity correction, the probability of finding fewer than 70 carp in a 10-mile stretch is approximately 0.1093 or 10.93%.

Step 3: Calculate with the continuity correction:

The continuity correction involves adjusting the boundaries of the discrete distribution to account for the fact that we are using a continuous distribution (normal distribution) to approximate a discrete one (Poisson distribution).

When dealing with discrete random variables like the Poisson distribution, we can adjust the boundaries by adding or subtracting 0.5 from the value.

For this case, when finding the probability of having fewer than 70 carp, we can adjust it to finding the probability of having fewer than 69.5 carp.

Now, we calculate the Z-score as follows:

Z = (X - μ) / σ

Z = (69.5 - 80) / 8.944 ≈ -1.154

Now, we find the probability using the standard normal table or calculator:

P(Z < -1.154) ≈ 0.1241

So, with the continuity correction, the probability of finding fewer than 70 carp in a 10-mile stretch is approximately 0.1241 or 12.41%.

Answer 2