Answer to Question #220759 in Statistics and Probability for mmmm

Question

Suppose the number of off springs a female eagle raises during its lifetime has a distribution with mean 15 and variance 4.6. In an area, 45 rescued and cured female eagles are set free. If these eagles are assumed to be independent (no lack of male partners and food resources), the total number of off-springs produced by these eagles can be assumed to have a Normal distribution with mean  

Blank 1. Calculate the answer by read surrounding text.

 and variance  

Blank 2. Calculate the answer by read surrounding text.

Answer 1

Let X be the number of offsprings a female eagle raises during its lifetime, that has µ=15 and σ²=4.6.

Let S_n be the total number of offsprings produced by n=45 female eagles. It is denoted as:

$S_n = \sum^{45}_{i=1} X_i$

Here, S_n follows normal distribution with the following mean and variance:

$Mean = n \mu \\ = 45 \times 15 \\ = 675 \\ Variance = n \sigma^2 \\ = 45 \times 4.6 \\ = 207$

Therefore, the value of mean is 675 and variance is 207