If an apple is normaly distributed with a mean mass $\mu$ and variance $\sigma^2$
$$X_i - N(\mu, \sigma)$$
and a mathematician puts apples to a basket one at a time until the total mass of the apples is equal or greater than M and then stops
What is the probability that there is less or equal to n apples in the basket?
I have tried to find a solution to this problem from literature and from internet. One friend of my proposed that the probability to have n apples in the basket is
$$P(N\leq n)=P(\sum_{i=1}^n X_i \geq M)$$
Which I doubt is not true. I think the answear is
$$P(N\leq n) = \sum_i^n P(n-1 failures | succes) = \sum_i^n \frac{P(succes | n-1 failure)P(n-1 failure)}{\sum_k^\infty P(succes | k-1 failure)P(k-1 failure)}$$
Is there a known solution for this problem? Negative binomial distribution is a solution for a similar problem where goal is to get k successes with tossing a coin. Is there a similar distribution for a normaly distributed case i.e. negative normal distribution?