document.write( "Question 822677: Five out of 50 people called to give money to a charity actually agrees to give money. Based on this percentage, what is the probability that out of the next 25 people called:
\n" ); document.write( "a. At least two agree to give money?
\n" ); document.write( "b. At most 5 agree to give money?
\n" ); document.write( "c. More than 3 agree to give money?
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Algebra.Com's Answer #495017 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
\n" ); document.write( "Hi,
\n" ); document.write( "Five out of 50 people called to give money to a charity actually agrees to give money. p(success) = 5/50 = .1
\n" ); document.write( "BINOMIAL Distribution P(x) = nCx* \"p%5Ex%2Aq%5E%28n-x%29\"
\n" ); document.write( "where p and q are the probabilities of success and failure respectively.
\n" ); document.write( "In this case p = .10 & q = .90 & n = 25 nCx = \"n%21%2F%28x%21%28n-x%29%21%29\"
\n" ); document.write( "For ex: P(x = 25) =\"+%28.10%29%5E25\"
\n" ); document.write( "Recommend using Your TI calculator for cumulative probabilities.
\n" ); document.write( "and stattrek.com as an excellent reference.
\n" ); document.write( "what is the probability that out of the next 25 people called:
\n" ); document.write( "a. At least two agree to give money? P(x≥2) = 1 - binomcdf(25, .10, 1)
\n" ); document.write( "b. At most 5 agree to give money? P(x≤5) = binomcdf(25, .10, 5)
\n" ); document.write( "c. More than 3 agree to give money? P(x>3) = 1 - binomcdf(25, .10, 3)
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