But then they add this bit:
If a random variable T has probability density function given by
,
and zero elsewhere, then it is known that
and = \frac{2}{a^{2}})
d) Use the above results to find:
i. the average time for a seed to germinate
ii. the standard deviation of the time for a seed to germinate
e) Find the probability that the time taken for a seed to germinate is within two standard deviations of the mean.
f) Calculate the exact value of the median time for a seed to germinate.
From the earlier parts of this question, we had this (

substituted in)
 = 0.5e^{-0.5t})
All we need to do now is recognise that in the second part, setting

gives us the distribution in the earlier parts of the question.
d)
i. the average time for a seed to germinateUsing
 = \frac{1}{a})
, and by recognition, using

(to yield
 = 0.5e^{-0.5t})
), then we get:
ii. the standard deviation of the time for a seed to germinateApplying same principle as before (

):
 = \mbox{E}(T^2) - [\mbox{E}(T)]^2 = \frac{2}{(0.5)^2} - \left(\frac{1}{0.5}\right)^2 = 4)

e)
2 standard deviations from the mean: (the mean is 2, and the standard deviation is also 2)
Lower bound = 2 - 2(2) = -2
Upper bound = 2 + 2(2) = 6
However, the domain of the function is for

. This means we have to integrate
 = 0.5e^{-0.5t})
from 0 to 6, and then -2 to 0 for
 = 0)
(which will just give us zero).
My answer:

(it's a very simple integral, if you need help getting here, just ask)
f)
Integrate
 = 0.5e^{-0.5t})
from 0 to m, and let that integral be equal to 0.5. Find the value of m.
My answer:
)
(arriving here just uses the same integral above with a bit of log laws)