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#1 |
Dec 2003
Hopefully Near M48
6DE16 Posts |
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I'm preparing for a Math Exam next Monday and this is a problem on a Past Paper. Could someone help me out (but don't provide the answer, just a hint please)?
A coin is biased so that when it is tossed the probability of obtaining heads is 2/3. The coin is tossed 1800 times. Let X be the number of heads obtained. Find: a) The mean value of X b) The standard deviation of X Of course, a) is easy to find. The mean of X = (2/3)*1800 = 1200. |
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#2 |
"William"
May 2003
New Haven
23·5·59 Posts |
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There is a simple and well known formula for the variance of binomial distribution - you should probably look it up and memorize it for the exam. If you must derive it, then find the variance of the outcome for flipping the coin one time, then note that 1800 flips is the sum of 1800 iid random variable, so their variances add.
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#3 |
Sep 2002
2·331 Posts |
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Hope this helps.
http://psych.rice.edu/online_stat/ch.../binomial.html http://cnx.rice.edu/content/m11024/latest/ Last fiddled with by dsouza123 on 2003-12-13 at 02:47 |
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#4 |
Cranksta Rap Ayatollah
Jul 2003
10100000012 Posts |
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X is a random variable with binomial distribution
The probabilty of getting n heads out of N trials with a coin that has probabilty p of coming up heads (and probabilty q=1-p of getting tails) is (N,n)*pnqN-n = N!/n!(N-n)! * pn(1-p)N-n where (N,n) is a binomial coefficient the standard deviation is Sqrt[N*p*(1-p)] Sorry if I gave too much away |
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#5 |
Dec 2003
Hopefully Near M48
2×3×293 Posts |
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A little too much. Oh well, I have the answer now. The standard deviation is 20. Thanks
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