1. Confusing distribution

The probability function for the number of eggs in a clutch laid by a particular type of bird is:

number of eggs | Probability
4 | 0.15
5 | 0.2
6 | 0.2
7 | 0.4
8 | 0.05

a) What is the probability that the clutch is 6 or more eggs?

b) Find the probability that among eighteen females at most fourteen have clutches of 6 or more; what assumption do you need to make to obtain an answer here?

c) An experimenter wants to identify fecund(fertile) females for use in future experiments. She needs 10 birds that produce clutches of 6 or more. How many females would you advise observing to have a probability of at least 0.90 of identifying (at least) 10 suitable birds?

a) 0.2+0.4+0.05 = 0.65

b+c) Not sure how to start these. They are not typical probability distribution, not sure how to make these kind of calculations?

Be very grateful if someone could give me a hand?

Thanks

2. Originally Posted by craig
The probability function for the number of eggs in a clutch laid by a particular type of bird is:

number of eggs | Probability
4 | 0.15
5 | 0.2
6 | 0.2
7 | 0.4
8 | 0.05

a) What is the probability that the clutch is 6 or more eggs?

b) Find the probability that among eighteen females at most fourteen have clutches of 6 or more; what assumption do you need to make to obtain an answer here?

c) An experimenter wants to identify fecund(fertile) females for use in future experiments. She needs 10 birds that produce clutches of 6 or more. How many females would you advise observing to have a probability of at least 0.90 of identifying (at least) 10 suitable birds?

a) 0.2+0.4+0.05 = 0.65

b+c) Not sure how to start these. They are not typical probability distribution, not sure how to make these kind of calculations?

Be very grateful if someone could give me a hand?

Thanks
b) the number of females that have clutches of size n from a group of 18 females has a binomial distribution so p(n)=b(n;18,0.65)

CB

3. Originally Posted by CaptainBlack
b) the number of females that have clutches of size n from a group of 18 females has a binomial distribution so p(n)=b(n;18,0.65)

CB
Thank you! Just one more little question though

In my stats tables, the probabilities only go up to 0.5, so I can't calculate it the normal way.

Would this be the correct way to do it?

For any $\displaystyle P(X \leq x)$, $\displaystyle F(x,n,p) = F(n-x-1;n,1-p$

Therefore we have, $\displaystyle F(14;18,0.65) = 1-F(3,18,0.35)$, giving us the probability 0.9217?

4. Originally Posted by craig
c) An experimenter wants to identify fecund(fertile) females for use in future experiments. She needs 10 birds that produce clutches of 6 or more. How many females would you advise observing to have a probability of at least 0.90 of identifying (at least) 10 suitable birds?
$\displaystyle P(X \geq 10) \geq 0.9$, this means that $\displaystyle P(X \leq 9) \leq 0.1$

Using the formula above, $\displaystyle F(x,n,p) = F(n-x-1;n,1-p$, this gives me the following:

$\displaystyle F(9,n,0.65) = F(n-9-1;n,0.35) \leq 0.1$

This is where I get stuck?

Thanks again for the help

Craig

5. Any ideas?