1. ## probability question

30% of all object X is infected with bacteria. A person has purchased 12 object X, what is the probability the consumer will have more than 6 contaminated object X?

Follow up question, suppose a company buys 1000 object X, find an approximate 95% interval for the number of infected object X.

How do u solve this problem?? I've tried using sample proportion, that didnt work for me...and i dont know stnd deviation so i cant do calculate p value either...

any help much appreciated!

2. Your first question you can solve using the Binomial distribution with X~Bin(12, 0.3). You are calculation the P(X>6) therefore, P(X=7)+P(X=8). . .P(X=12)=P(X>6).

For the second part, you are meant to understand the Binomial can be approximated using the Normal distribution with the mean equal to 1000(0.3) and standard deviation equal to the square root of 1000(0.3)(.07). Also, the p-value isn't needed here.

3. Originally Posted by ANDS!
Your first question you can solve using the Binomial distribution with X~Bin(12, 0.3). You are calculation the P(X>6) therefore, P(X=7)+P(X=8). . .P(X=12)=P(X>6).

For the second part, you are meant to understand the Binomial can be approximated using the Normal distribution with the mean equal to 1000(0.3) and standard deviation equal to the square root of 1000(0.3)(.07). Also, the p-value isn't needed here.
for the second part, i get how you get the mean and standard dev. but how do you calculate the interval?

heres what I did:
$m=z*\frac{\sigma}{\sqrt{n}}=1.96(\frac{14.49}{31.6 })=0.898$

so is the interval 300+0.898 and 300-0.898?

4. Assuming you do your calculations right your Confidence Interval will be the calculated mean, plus or minus the z-score corresponding to an area of 0.025 in one tail (I will leave you to figure our why) of your curve.