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Qa - binomial distributions - q2

priya (Student) (64 Points)

02 June 2012

Friends, Not getting any answers now ...Help me with these 2 please...

a) X is a binomial variable such that 2 P(X=2) = P(X=3) and mean of x is known to be 10/3 . What is the probability that x assumes at most the value of 2. [ I got the value of n & p but not after that.]
a) 16/81 b) 17/81 c) 47/243 d) 46/243

b) Assuming that one-third of the population are tea drinkers and each of 1000 enumerators takes a sample of 8 individuals to find out whether they are tea drinkers or not, how many enumerators are expected to report that five or more people are tea drinkers..
a) 100 b) 95 c) 88 d) 90

Thanks in advance..

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6 Replies

Tejaswi Kasturi (student-cpt) (427 Points)
Replied 02 June 2012

1.np = 10/3.2*nc2*p^2*q^(n-2)= nc3*p^3*q^n-3 implies 2*(1-p) = ((n-2)/3)*p implies p =2/3 and n =5 now P(X<=2) = P(
X=0)+P(X=1)+P(X=2) = (1/3)^5 + 5*(2/3)*(1/3)^4 + 10*(2/3)^2*(1/3)^3 = (1+10+40)/243 = 51/243 = 17/81 answer is b

2. assume that being a tea drinker is like taking a flip of coin.(i.e. either the person drinks tea or not) with the probability of heads being 1/3 and tails being 2/3 now P(X>=5) =P(X=5)+P(X=6)+P(X=7)+P(X=8) = (56*8+28*4+8*2+1)/6561 = 577/6561 =0.087943 so for 1000 trials the number of trials which report greater than or equal to 5 is 1000*P(X>=5) = 87.94 = 88 so answer is c

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