Thread: Central Limit Theorem

Just wanted to state that I appreciate all the help I can get here - you guys were more than helpful in answering one of my questions earlier today.

A large truckload of apples arrives at a packing plant. 175 apples are randomly sampled from the truck and tested for quality. Suppose it is known that 7% of all apples on the truck have discoloration, bruises, or other defects. The plant will only accept the truckload of apples if less than 4% of the sampled apples are bruised, blemished, or have some other defect. What is the probability that the load will be accepted by the plant?

There were other parts of the question I think I was able to answer but I am completely clueless how to approach and set this one up. Any help would certainly be appreciated, thank you so much!

Originally Posted by cechmanek32

Just wanted to state that I appreciate all the help I can get here - you guys were more than helpful in answering one of my questions earlier today.

A large truckload of apples arrives at a packing plant. 175 apples are randomly sampled from the truck and tested for quality. Suppose it is known that 7% of all apples on the truck have discoloration, bruises, or other defects. The plant will only accept the truckload of apples if less than 4% of the sampled apples are bruised, blemished, or have some other defect. What is the probability that the load will be accepted by the plant?

There were other parts of the question I think I was able to answer but I am completely clueless how to approach and set this one up. Any help would certainly be appreciated, thank you so much!

Let X be the random variable number of bruised apples.

X ~ Binomial(n = 175, p = 0.04)

Calculate

(since 4% of 175 is 7).

The normal approximation to the binomial distribution can be used (why?).

So this would then go:

P(x<7)= P(x=0) + P(x=1) + ... P(x=6)

And the numbers I would plug in would be like this:?

P(.04^0*.96^4) this would be what it is for p(x=0), correct? Do I have to do something that involves like 4 choose 0? I'm confused about that part.

Originally Posted by cechmanek32

So this would then go:

P(x<7)= P(x=0) + P(x=1) + ... P(x=6)

And the numbers I would plug in would be like this:?

P(.04^0*.96^4) this would be what it is for p(x=0), correct? Do I have to do something that involves like 4 choose 0? I'm confused about that part.

Originally Posted by mr fantastic

[snip]
The normal approximation to the binomial distribution can be used (why?).

Have you been taught the normal approximation to the binomial distribution? Where are you stuck in applying it?

Yes, it has been covered in class. However, I always have trouble taking the numbers from the problem (or even after I equate other steps) and plugging them into the equation. I've been having an awfully hard time trying to figure this type of stuff out.