Suppose the length of time it takes one variety of plant seeds to germinate is normally distributed with a mean of 14.5 days and a standard deviation of 2.5 days.
What proportion of plant seeds will germinate in less than 10 days, given that they germinate within 16.4 days?
Can someone advise if my conditional probability is correct?
Let X be the length of time from the plant seeds to germinate.
$P(X<10 \mid 0 If so, will the numerator just be P(0< X< 10)?
Let $T$ be the time until germination. Then
\begin{align} & \Pr(T<10\mid T<16.4) = \frac{\Pr(T<10\ \&\ T<16.4)}{\Pr(T<16.4)} = \frac{\Pr(T<10)}{\Pr(T<16.4)} \\[10pt] = {} & \frac{\Pr\left( \frac{T-14.5}{2.5} < \frac{10-14.5}{2.5} \right)}{\Pr\left( \frac{T-14.5}{2.5} < \frac{16.4-14.5}{2.5} \right)} = \frac{\Pr\left( Z < \frac{10-14.5}{2.5} \right)}{\Pr\left( Z < \frac{16.4-14.5}{2.5} \right)} = \frac{\Phi\left( \frac{10-14.5}{2.5} \right)}{\Phi\left(\frac{16.4-14.5}{2.5}\right)}. \end{align}
It is extraordinarily improbable that $T<0$ if $T\sim N(14.5, 2.5^2)$ and that possibility is neglected above. It is impossible that $T<0$ if $T$ is time until germination. It is also impossible for the distribution of the time to be exactly normal, since then the probability of its being less than $0$ would not be exactly $0$.
Yes, that is okay.
Note also that $\{X<10 \cap 0