Normal Approximation Problem Solving

[Math1015]

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.

 

[MarkFL]

I would use:

[box=blue]

Binomial Probability Formula​

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$$P(x)={n \choose x}p^x(1-p)^{n-x}$$[/box]

Now, suppose we want to find the probability of at least $s$ successes out of $n$ trials...so if $X$ is this event, we may write:

\(\displaystyle P(X)=\sum_{k=s}^{n}\left({n \choose k}p^k(1-p)^{n-k}\right)\)

Can you identify $s,\,n,\,p$ for this problem?

 

[Jameson]

The above approach is definitely correct for the exact answer. If you want a normal approximation then we'll use the following formula:

\(\displaystyle Z=\dfrac{Y-np}{\sqrt{np(1-p)}}\stackrel {d}{\longrightarrow} N(0,1)\)

I think from your class you should be able to tell us if you've covered the normal approximation to the binomial distribution yet. Once we know that, we can help you through either approach. :)

https://onlinecourses.science.psu.edu/stat414/node/179

 

[Math1015]

I've for sure seen the binomial probability formula in class and we have covered that formula. I have not seen the other two however. The problem comes in when identifying values for each of the variables.

 

[MarkFL]

I've for sure seen the binomial probability formula in class and we have covered that formula. I have not seen the other two however.

I glossed over the phrase "Normal Approximation" in the thread title, and gave you the method for determining the exact answer. I do recall though, using the normal distribution to approximate the binomial distribution in the one elementary stats class I took long ago. I seem to remember only using it for much larger numbers of trials though, where using a sum such as I posted would be tedious to compute. :)