Originally Answered: Normal Distributions Statistics?
In R, the name of the normal cumulative distribution function is "pnorm" (the cumulative distribution is what you want to calculate the probabilities in your question). By default, pnorm uses the standard normal distribution (with mean = 0 and sd = 1), but this function allows you to optionally give other values for the mean and standard deviation if you ever want to work with the normal distribution in other forms. For your questions, pnorm(x) will give you the probability that Z is less than or equal to x, that is, P[Z ≤ x].
P[Z < 1] = P[Z ≤ 1] (since the normal distribution is continuous), so
the calculation in R of P[Z < 1] is simply
which evaluates to 0.8413447 (to seven decimal places).
You can calculate P[Z > 0.25] two ways. The first way is to recognize that
P[Z > 0.25] = 1 - P[Z ≤ 0.25], so in R you would write
1 - pnorm(0.25)
which evaluates to 0.4012937 (to seven decimal places).
However, there is another optional parameter that you can use with the pnorm function that allows you to change the evaluation from the default of the lower tail (or left tail) to the upper tail (or right tail). In this situation, pnorm(x, lower.tail = FALSE) will give you the probability that Z is greater than x, that is, P[Z > x].
Thus, by using lower.tail = FALSE in your evaluation of pnorm, you get the same result, but don't need to subtract the function evaluation from 1.
pnorm(0.25, lower.tail = FALSE)
evaluates to 0.4012937 (to seven decimal places).
For P[-1 < Z < 1] (= P[-1 < Z ≤ 1], since the normal distribution is continuous), it helps to look at a graph if you need more clarity on its meaning. You should easily recognize that
P[-1 < Z ≤ 1] = P[Z ≤ 1] - P[Z ≤ -1], so in R you merely use
pnorm(1) - pnorm(-1)
which evaluates to 0.6826895 (to seven decimal places).
I hope that my explanations have given you what you need to know, but if you want further help, please let me know.
Best wishes in your studies!