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April 25, 2014

April 25, 2014

Posted by **Cecile** on Tuesday, January 17, 2012 at 2:22am.

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.

Thank you

- Statistics - Central Limit Theorem -
**drwls**, Tuesday, January 17, 2012 at 6:22amI 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

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