The mass of eggs in a farm is normally distributed with a mean of $575~\text{g}$ and standard deviation of $80~\text{g}$.
(a) An egg can be classified as having a “medium size” if its mass is between
$500$ and $600$ grams. What is the probability that a randomly selected egg from the farm is a medium-sized egg? Use R to find the answer.(b) Let $M$ be the count of medium-sized eggs in a random sample of $100$. What is the distribution of $M$? Using R, find the probability that at least half of the $100$ eggs are medium-sized.
(c) Eggs from this farm are sold by the dozen (12 eggs). Let $D$ be the number of randomly selected eggs we must weigh until we find the 12th medium-sized egg. What is the distribution of $D$ ? Using R, find the probability that we need to weigh at least 20 eggs before we find the 12th medium-sized egg
My attempt using Rstudio for a and b
p <- pnorm(600,575,80)-pnorm(500,575,80)
p
[1] 0.448419
sum(dbinom(50:100, 50, p))
[1] 3.838845e-18
Now for part c do you use the part b and change 50 to 12 to find the distribution and then somehow find the probability using that to find the second part of part c?
$$EDIT$$
Possible answer to part C, by
$$P(X=10) = {20-1 \choose 12-1}0.0448419^{12}(1-0.044819)^{20-12}$$
Thoughts?
Best Answer
The result for (a) is right.
At (b) it seems that you haven´t used the right number for $p=0.448419$ like at (c).
$$P(X\geq 50)=\sum_{k=50}^{100} {{100} \choose k} \cdot 0.448419^k\cdot (1-0.448419)^{100-k}$$
$=0.1744=17.44\%$
At (c) the key word is "at least". You have to sum from $n=20$ to infinity.
$$\sum_{n=20}^{\infty} {{n-1} \choose {12-1}}\cdot 0.448419^{12}\cdot (1- 0.448419)^{n-12}=0.915083\approx 91.51\%$$
Remark
For (c) you can also use the converse proability $P(X\geq 20)=1-P(X\leq 19)$
$$1-\sum_{n=12}^{19} {{n-1} \choose {12-1}}\cdot 0.448419^{12}\cdot (1- 0.448419)^{n-12}$$
$$=1-0.084917=0.915083\approx 91.51\%$$