Finding probability of strictrly less than, when expectance and variance are given

Question

A gardener grows cabagge. The weight of a ripe cabbage, $X$, is assumed to have normal distribution with the expectance $\mu = 2.4$ kg and standard weight $\sigma = 0.7$

If we pick a random cabagge, what is the probability that...

1. ... it weighs less than 1.5 kg?
2. ... weighs between 2 and 2.5 kg?
3. ... the difference in weight between two random cabbages is greater than 1 kg?

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(1)

So we have $X\sim Normal(\mu,\sigma)$, and

\begin{align}\\ & P(X\le a) = \phi\left(\frac{a-\mu}{\sigma}\right) \\ & P(X\ge b) = 1-\phi\left(\frac{b-\mu}{\sigma}\right) \\ \end{align}

But what is the formula for strictly less than, i.e. $P(X<c)?$

Edit: Is it $P(X<c)=1-P(X\ge b)?$

Answer 1

Note that, if you are working with a continuous variable you have the following: \begin{equation*} P(X=a) = 0, \hspace{.2cm} \forall a \end{equation*} And \begin{equation*} P(X\leq a) = P(X <a) + P(X=a) = P(X<a) \end{equation*} So the formula is the same, for continuous variables.