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Researchers studying photoperiodism use the wingspanlebur as an experimental plant. The variable of interest, X, the number of hours of uninterrupted darkness required to produce flowering is normally distributed with a mean of 14.5 hours and standard deviation 1.0 hours. c. What is the probability that out of a random sample of 20 wingspanleburs less than two require between 12 and 15 hours of uninterrupted darkness to produce flowering? I got this far: Let Y=# wingspanleburs requiring between 12 and 15 hours of uninterrupted darkness (out of 20) Y~BIN(n=20, p=0.6853) P(Y<2) = ? But I'm unsure about how to use P(Y<2) to get the answer 4.0 x 10^ -9

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The probability that the hours of darkness required to produce flowering lies between 12 and 15 hours is 0.6853. The probability that only one out of a random sample of 20 requires between 12 and 15 hours of darkness to produce flowering is found from the binomial distribution as follows: \[P(1from20)=\left(\begin{matrix}20 \\ 1\end{matrix}\right)\times 0.6853(1-0.6853)^{19}=3.95\times 10^{-9}\] If you now use the binomial distribution to find the probability that zero out of a random sample of 20 requires between 12 and 15 hours of darkness to produce flowering and add this probability to the one calculated above you will have the solution.
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