# Thread: Normal Approximation: Mike scores a birdie on about 20.9% of all the holes he plays.

1. ## Normal Approximation: Mike scores a birdie on about 20.9% of all the holes he plays.

I'm unsure on how to start this problem. I tried to make a tree diagram but to no avail did it help out.

Question:
On average, Mike Weir scores a birdie on about 20.9% of all the holes he plays. Mike is in contention to win a PGA golf tournament but he must birdie at least 4 holes of the last 6 holes he plays.Find the probability, as a percent correct to one decimal place, that Mike will win.

2. Originally Posted by Math1059
On average, Mike Weir scores a birdie on about 20.9% of all the holes he plays. Mike is in contention to win a PGA golf tournament but he must birdie at least 4 holes of the last 6 holes he plays.Find the probability, as a percent correct to one decimal place, that Mike will win.

Presumably this question is in a section about the normal approximation to the binomial distribution. The question requires application of the binomial distribution; you could do that directly (because the number of trials is small), but your mention of the normal approximation suggests that you must have been told to use it.

3. I was able to identify that the question required normal approximation to the binomial distribution based on the probability of success being the same for each birdie. I'm assuming the equation needed would be P(x)= (nCx)p^x(1-P)^n-x. The problem I have is identifying the values to sub in.

4. Originally Posted by Math1059
I was able to identify that the question required normal approximation to the binomial distribution based on the probability of success being the same for each birdie. I'm assuming the equation needed would be P(x)= (nCx)p^x(1-P)^n-x. The problem I have is identifying the values to sub in.
Well, that's the binomial distribution itself, which as I said you could just as well use directly rather than the approximation. Are you saying that the problem itself didn't mention that you must use the normal approximation, so that you had to "identify" this fact for yourself (with my help); or that it does say this? Please state the entire problem word for word, as we ask you to in the Read before Posting announcement. This saves a lot of wasted time.

But let's suppose that you want to use the binomial distribution. If you do recognize this problem as binomial, you should see from the problem what n and p are. What are they?

Then x is the remaining detail. Do you see that "at least 4 holes of the last 6 holes he plays" means "either 4, or 5, or 6"? You would take each of those as x, and add the probabilities you get.

The normal approximation is intended to avoid having to repeatedly apply the binomial formula this way. It is not really a good approximation with numbers as small as these, but if you were told to use it (perhaps in the introduction to a set of problems), do so. What were you told about it?

5. The problem it self did not mention to use normal approximation. If binomial distribution was to be used (correct me if i'm wrong), N= 4, P=0.209.

6. Originally Posted by Math1059
The problem it self did not mention to use normal approximation. If binomial distribution was to be used (correct me if i'm wrong), N= 4, P=0.209.
Okay; then why did you use the term Normal Approximation in the title of your post? I'll guess that is the title of the section in your text, and this problem is just an introduction to the need for it.

Anyway, yes, p = 0.209; but n is not 4. That is one of the values to use for x. What is the definition of n in the binomial distribution?

7. N is equal to number of trials.

8. Originally Posted by Math1059
N is equal to number of trials.
Right. And how many trials are there here?

9. 6 trials?

10. Originally Posted by Math1059
6 trials?
Right. So you have n=6, p=0.209, and x=4,5,6. Put those in the formula and add them up, and you have the answer.

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