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1. Researchers studying photoperiodism use the cocklebur as an e of interest, X,

ID: 2936616 • Letter: 1

Question



1. Researchers studying photoperiodism use the cocklebur as an e of interest, X, the number of hours of uninterrupted darkness is normally distributed with a mean of 14.5 hours and standard deviation 1.0 hours xperimental plant. The variable required to produce fowering (a) What is the probability that a random selected cocklebur will require between 12 and 15 hours of uninterrupted darkness to produce flowering? (b) What is the probability that the average number of hours of uninterrupted darkness required to produce flowering for a random sample of 20 cockleburs will be between 12 and 15 hours? (c) What is the probability that out of a random sample of 20 cockleburs less than two reruire hetween 12 and 15 hours of uninterrupted darkness to produce flowering?

Explanation / Answer

Ans:

Given that

mean=14.5

standard dveiation=1

a)

z(12)=(12-14.5)/1=-2.5

z(15)=(15-14.5)/1=0.5

P(-2.5<=z<=0.5)=P(z<=0.5)-P(z<=-2.5)=0.6915-0.0062=0.6853

b)standard error of mean=1/sqrt(20)=0.2236

z(12)=(12-14.5)/0.2236=-11.18

z(15)=(15-14.5)/0.2236=2.236

P(-11.18<=z<=2.236)=P(z<=2.236)-P(z<=-11.18)=0.9873

c)Use binomial distribution with n=20,p=0.6853

P(x<2)=P(x<=1)=BINOMDIST(1,20,0.6853,TRUE)=0