(a) The rate at which a hen lays eggs of the time she lays at Poisson distribution, and $$ 99.73 \% $$ of least one egg each week. (i) Calculate an estimate for the average number of eggs the hen lays each week.

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To solve this problem, we need to understand the properties of the Poisson distribution. The Poisson distribution is characterized by a parameter $$\lambda$$, which represents the average rate (mean) of occurrences in a fixed interval of time or space.

Given that $$99.73\%$$ of the time the hen lays at least one egg each week, we can interpret this as follows:

1. The probability of the hen laying at least one egg in a week is given by:
   $$
   P(X \geq 1) = 1 - P(X = 0)
   $$
   where $$X$$ is the number of eggs laid in a week.

2. For a Poisson distribution, the probability of observing $$k$$ events (in this case, eggs) is given by:
   $$
   P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
   $$
   For $$k = 0$$:
   $$
   P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}
   $$

3. Therefore, we can express the probability of laying at least one egg as:
   $$
   P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-\lambda}
   $$

4. We know that $$P(X \geq 1) = 0.9973$$. Thus, we can set up the equation:
   $$
   1 - e^{-\lambda} = 0.9973
   $$

5. Rearranging gives:
   $$
   e^{-\lambda} = 1 - 0.9973 = 0.0027
   $$

6. Taking the natural logarithm of both sides:
   $$
   -\lambda = \ln(0.0027)
   $$
   $$
   \lambda = -\ln(0.0027)
   $$

7. Now we can calculate $$\lambda$$:
   $$
   \lambda \approx -\ln(0.0027) \approx 5.888
   $$

Thus, the estimated average number of eggs the hen lays each week is approximately $$ \lambda \approx 5.89 $$.
The estimated average number of eggs the hen lays each week is approximately 5.89.

(a) The rate at which a hen lays eggs of the time she lays at Poisson distribution, and \( 99.73 \% \) of least one egg each week. (i) Calculate an estimate for the average number of eggs the hen lays each week.

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