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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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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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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