Posted by Cecile on .
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(1p)]^(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.
Thank you

Statistics  Central Limit Theorem 
drwls,
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!)
= 0.001437
P(2) = (0.7)^23*(0.3)^2* [24*25/(1*2)]
= 0.007390
P(5) = (0.7)^20*(0.3)^5*[25!/(20!*5!)]
= 0.10302
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