Friday, July 31, 2009

Statistics. Need homework help please.?

It was reported that approximately 45% os all university professors are extroverted. Suppose you have classes with six different professors.





(a) What is the probability that all six are extroverts?





(b) What is the probability that none of your professors is an extrovert?





(c) What is the probability that at least two of your professors are extroverts?





(d) In a group of six professors selected at random, what is the expected number of extroverts?


What is the standard deviation?

Statistics. Need homework help please.?
Let X be the number of extroverted professors. X has the binomial distribution with n = 6 trials and success probability p = 0.45





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[X = x] = 0 for any other value of x.





The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.


Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.





X ~ Binomial( n , p )





the mean of the binomial distribution is n * p = 2.7 ← ANSWER TO D


the variance of the binomial distribution is n * p * (1 - p) = 1.485


the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.218606 ← Answer to D





The Probability Mass Function, PDF,


f(X) = P(X = x) is:





P( X = 0 ) = 0.02768064 ← Answer to B


P( X = 1 ) = 0.1358868


P( X = 2 ) = 0.2779502


P( X = 3 ) = 0.3032184


P( X = 4 ) = 0.1860659


P( X = 5 ) = 0.06089428


P( X = 6 ) = 0.008303766 ← Answer to A








The Cumulative Distribution Function, CDF,


F(X) = P(X ≤ x) is:





x


∑ P(X = t) =


t = 0





P( X ≤ 0 ) = 0.02768064


P( X ≤ 1 ) = 0.1635674


P( X ≤ 2 ) = 0.4415177


P( X ≤ 3 ) = 0.7447361


P( X ≤ 4 ) = 0.930802


P( X ≤ 5 ) = 0.9916962


P( X ≤ 6 ) = 1








1 - F(X) is:





n


∑ P(X = t) =


t = x





P( X ≥ 0 ) = 1


P( X ≥ 1 ) = 0.9723194


P( X ≥ 2 ) = 0.8364326 ← Answer to C


P( X ≥ 3 ) = 0.5584823


P( X ≥ 4 ) = 0.2552639


P( X ≥ 5 ) = 0.06919805


P( X ≥ 6 ) = 0.008303766
Reply:This is a binomial distribution question.





a) .83%





b) 2.77%





c)83.64%





d)2.7


e)1.485





Now, go to a binomial distribution formula and see if you get the same answers.. Trials are six, probability is .45. For some of these you need the cumulative density distribution and for some the probability density distribution.





Do each possible combination one at a time on a table starting with 0 extraverts and going to 6 extraverts....that is your pdf...your cdf is the sum of the pdf values.


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