We have pool with fish inside. The weight of the fishes is normally distributed.
$\mu=900, \sigma=150$ (The wight is at grams).
- We pull out 10 fish. What is the probability that there was $A$ fish that their wight is over 800 when $2\le A\le9$? My answer was this: $P(X>800)=0.7454$ So: $$p=\sum_{i=2}^{9}{10 \choose i}\cdot\left(0.7454\right)^{i}\cdot\left(0.2546\right)^{10-i}$$
- We pull out fish, until the 3rd fish is over 1KG. What is the expectation and what is the variance of the amount of the fish that we pull out until it happens?
I don't how to solve it but what I thinking is about the the expectation and the variance of $NB$ distribution, I'm right?
And I add another Q: It's possible that $P(X>800)=P(X>1000)?$
Thank you!!