Question 1189824: 7). Forty-five percent of the adult population in a particular large city are women. A court is to randomly select a jury of 12 adults from the population of all adults of this city.
a) Find the probability that none of the 12 jurors is a woman. (1 point)
b) Find the probability that at most 4 of the 12 jurors are women. (1 point)
c) Let x denote the number of women in 12 adults selected for this jury. Obtain the probability distribution of x.
d) Using the probability distribution obtained in part c, find the following probabilities.
i. P(x > 6)
ii. P(x<=3)
iii. P(2<= x<=7)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The probability distribution for women, let them be x, is 12Cx(0.45)^x*(0.55)^(n-x)
a. this is 0.55^12 all men=0.0008
for at most 4 are women--that is 0,1,2,3
we know 0;
1 is 12*0.45*0.55^11=0.0075
2 is 12C2*0.45^2*0.55^10=0.0339
3= 12C3*0.45^3*0.55^9=0.0923
12C4*0,45^4*0.55^8=0.1700
sum is 0.3045, the answer
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p (x>6), can use calculator 1-binomcdf(12,0.45,6)=0.2607
can check, since it is 1-(sum of prob. for 0-6)=1-0.3045-0.2225-0.1489=0.2606 (rounding difference)
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x<=3, can use the above and it is 0.1345
between 2 and 7
less than 7-less than 2 =0.8883-0.0083=0.88
can check
for 5 12C5*0.45^5*0.55^7=0.2225
for 6 it is 0.2124
for 7 it is 0.1489
between 2 and 7 inclusive it is 0.88
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