Question 1189824
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