# Looking for different methods to solve these...

#### Mozz

##### New Member
1. Suppose that 10% of a population carries antibodies to a virus. For a research project, you need 38 people with antibodies to the virus. How many people should you test so that the expected number of people with the antibodies is 38?
Round up to a whole number.

2. Suppose that 14% of a population carries antibodies to a virus. For a research project, you need 10 people with antibodies to the virus. How many people must you test to be 95% certain of getting at least 10 people with the antibodies?
The problem will need some trial and error..you are trying to find n so that P(X>=10)=0.95

This is for a stats class I am taking. We are in chapter 4, Discrete Random Variables. There are no other questions like it, in the chapter.

I addition to steps to solve, by hand and/or with calc, I would like the name of this type of question so I can learn about them for the exam.

Thank you in advance for any help

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#### katxt

##### Active Member
It looks like the Binomial distribution. Is that in chapter 4?
The first question is about the expected value. Find the equation in your book which involves expected value, p and n.
The second question is about the cumulative distribution. There is no simple formula for this, but if you have Excel there is a binomial function which could help to find P(x>=10) for p = 0.14 to find n.

#### Mozz

##### New Member
All the sample questions have a sample (n). I am lost on what to do with these questions. Expected value Binomial Distribution has an n in the examples I find.

#### Mozz

##### New Member
That works. Thank you!

For the first one I was doing a proportion, and getting 380. I got 380 when it did E=np, as 38=n(.10)

Still unsure about the second question. The x>= part and the .95. Is there a table for this, or a formula? This textbook is awful.

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#### katxt

##### Active Member
There is no formula or table for x>=, but there is a way to get x<= so you need to convert P(x>=10) = 0.95 into something like P(everything else) = 1-0.95
I don't know what your calculator will do, but in Excel
=BINOMDIST(4,20,0.14,1) = 0.8625 means P(x<=4 from n=20 total when each has 14% chance) = 0.8625. The 1 at the end gives the <= part. Experiment with that for a bit.

#### Mozz

##### New Member
Is the answer to question 2, 71.428?

#### katxt

##### Active Member
P(X>=10)=0.95 is the same as P(X<=9)=1- 0.95 = 0.05
Try =BINOMDIST(9,72,0.14,1) and see if it 0.05 (near enough). If it's not, change the 72 until it is.