<PRE><B>Question 1187527</B>
True
False--X must take on whole number values in this instance. Mathematically it does not have to, but with litters, there are no fractional ones.
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e^(-7)*7^1/1!=0.0064, or 0.01
confirm with Poissonpdf(7,1) ENTER
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we know that 1 litter has probability 0.0064 and 0 is e^(-7)=0.0009. They add to 0.0073, and the complement or 0.9927, is the answer, at least two litters. Two decimal places is 0.99
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exactly 3 litters is e^(-7)*7^3/3!=0.0521, or 0.05
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exactly 4 litters is e^(-7)*7^4/24=0.0912, or 0.09
so the answer is that plus 0.0521 and the probability of 2 litters (0.0223)=0.1656, or 0.17</PRE>