- #1
denian
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the lifespan of a species of plant is a random variable T ( tens of days ). the probability density function is given by
f(t) = { (1/8)e^(-t/8) , t > 0
{ 0, otherwise
(i) find the cumulative distribution function of T and sketch its graph
i get this answer
F(t) = { 0, t < or = 0
{ 1 - e^(-t/8) , t > 0
(ii) find the probability, to three decimal places, that a plant of that species randomly chosen has a lifespan of more than 20 days
i do it this way..
F(2) = ...
= 0.221
P(lifespan > 20 days) = 1 - 0.221 = 0.779
(iii) calculate the expected lifespan of that species of plant
i get E(t) = 8
therefore, expected lifespan of that species is 80 days.
please tell me if i am correct in the questions above. sorry, i tried to use Latex but my attemp failed.
f(t) = { (1/8)e^(-t/8) , t > 0
{ 0, otherwise
(i) find the cumulative distribution function of T and sketch its graph
i get this answer
F(t) = { 0, t < or = 0
{ 1 - e^(-t/8) , t > 0
(ii) find the probability, to three decimal places, that a plant of that species randomly chosen has a lifespan of more than 20 days
i do it this way..
F(2) = ...
= 0.221
P(lifespan > 20 days) = 1 - 0.221 = 0.779
(iii) calculate the expected lifespan of that species of plant
i get E(t) = 8
therefore, expected lifespan of that species is 80 days.
please tell me if i am correct in the questions above. sorry, i tried to use Latex but my attemp failed.