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soccergal12 Group Title

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

  • 2 years ago
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  1. kropot72 Group Title
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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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  2. soccergal12 Group Title
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    thank you

    • 2 years ago
  3. kropot72 Group Title
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    You're welcome :)

    • 2 years ago
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