Posted by Cecile on Tuesday, January 17, 2012 at 2:22am.
Actually I'm asking this question because I found my answer for the HW question somewhat weird, but I still can't determine where did I make mistakes.
Given: A basketball player can score on a particular shot with p = 0.3. Let S be the number of successes out of 25 independent shots.
Question: Find approximate probability that S <= 5, then compare to the exact probability
First of all, I think this is a case of Binomial distribution, am I right?
** Here is my work:
P(S <=5)= P(Z <= [5 - n(mean)] / [(standard deviation)* (n^(1/2))]
where I use:
mean = np = 25 * 0.3 = 7.5
sd = [np(1-p)]^(1/2) = 2.29
After doing all the jobs on calculator, I get P(Z <= -1.09) = 1 - P(Z >= 1.09) = 0.1379
** However, when I do P(S = 5), using the probability mass function of Binomial distribution
P(S = 5) = (25C5)(.3^5)(.7^15) = 0.612
which is way off from my estimation P(S <= 5). My bet is that the two numbers have to be somewhat close.
Please help me to check if I did make errors or miss something. Now I'm really confused on trying to figure this out.
- Statistics - Central Limit Theorem - drwls, Tuesday, January 17, 2012 at 6:22am
I believe your first P(Z) number using the normal distribution.
You need to compare that with the sum of the exact probabilities of getting 0, 1, 2, 3, 4 and 5. using the binomial distribution.
P(0) = (0.7)^25 = 0.000134
P(1) = (0.7)^24*(0.3) *25!/(24!*1!)
P(2) = (0.7)^23*(0.3)^2* [24*25/(1*2)]
P(5) = (0.7)^20*(0.3)^5*[25!/(20!*5!)]
You do the others, P(3) and P(4), and add them all. You might be quite close to the normal distribution result.
It looks to me like you did P(5) incorrectly
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