How do I calculate phi(z) to 4dp when my Normal Distribution tables are to 2dp? I have two Normal Distributions representing weights: X ~ N(30, 25) and Y ~ N(32, 16)
Calculate the probability an item from distribution Y weighs more than X, i.e P(Y - X > 0).
I did everything correct and ended-up with:
B = Y - X
B ~ N(2, 41)
= 1 - P (B < 0)
= 1 - phi((0 - 2)/root(41))
= 1 - phi(-0.3123)
= 1 - (1 - 0.6217)
= 0.6217
However, their answer:
B = Y - X
B ~ N(2, 41)
P(B > 0) = 1 - P(B < 0) = 1 - 0.3774 = 0.6226
The problem is I looked-up phi(0.31) in the Normal Distribution tables, whereas they calculated phi(0.3123) exactly.
How do I calculate phi(0.3123) exactly? This is an International A Level question. We're not supposed to use the phi(z) integral formula. It's either a table look-up, or calculator exercise.
To calculate a cumulative probability on my calculator it requires a lower and upper bound, but here the lower bound is -infinity?
 A: You can get a good approximation with interpolation
Your tables presumably give

*

*$\Phi(0.31)=0.6217$ though $0.6217195$ is closer

*$\Phi(0.32)=0.6255$ though $0.6255158$ is closer

You want to go $23\%$ of the way between these to approximate $\Phi(0.3123)$ perhaps using

*

*$0.6217 +0.23(0.6255-0.6217) = 0.622574$ or $0.6226$ rounded

*or with more digits $0.6217195+0.23(0.6255158-0.6217195)=0.6225926$

*which are close to  $\Phi(0.3123)=0.6225937$
A: Three possible approaches:
(1) Interpolate from a printed table as in @Henry's Answer (+1).
(2) Learn to use a statistical calculator for greatest accuracy.
(3) Use statistical software. Almost any of the software in
common use will do.  Results from R, where pnorm is the standard normal CDF, are shown below:
1 - pnorm(-.3123)
[1] 0.6225937
1 - pnorm(-.31)
[1] 0.6217195

Sometimes a little accuracy is lost doing the arithmetic for standardization.
In R, pnorm with appropriate parameters can be used to avoid
standardizing:
1 - pnorm(0, 2, sqrt(41))
[1] 0.6226118

You say, "We're not supposed to use the phi(z) integral formula. It's either a table look-up, or calculator exercise." I don't doubt that
this is true. Using the PDF $\varphi(z)$ of standard normal
would not be helpful. However, if there is a prohibition on using computer
software, then maybe someone should consider whether that prohibition is
appropriate.
I am old enough to remember days when calculators were absolutely prohibited
in mathematics and statistics courses. Rules do change over time. One of the things I don't regret forgetting is how
to 'extract a square root' by longhand.
In my opinion, yours may be the last generation of
statistics students who routinely use printed tables of
the standard normal CDF to solve such problems.
A: As you implied in the question, one of the ways to compute $\Phi(0.3123)$
is to integrate the part of the standard normal probability density,
$f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2},$ up to $x = 0.3123$:
$$ \Phi(0.3123) =
 \int_{-\infty}^{0.3123} \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,\mathrm dx, $$
except that your calculator does not allow you to enter $-\infty$ as the lower boundary of the integral.
As has been pointed out in comments, you can get a result to a large number of decimal places by using a suitably large negative real number instead of $-\infty$.
But I find it a bit more satisfying to do the integral like this:
\begin{align}
 \Phi(0.3123)
& = \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,\mathrm dx
    + \int_0^{0.3123} \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,\mathrm dx \\
& = \frac12 + \int_0^{0.3123} \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,\mathrm dx. 
\end{align}
Now you don't have to remember how negative the lower boundary has to be in order to have a sufficient number of decimal places of accuracy.
Just integrate from $0$ to $0.3123$ and add $0.5.$
Some tables for calculating the standard normal distribution actually use that form of the integral, that is, they give an integral from $0$ to a non-negative upper boundary instead of from $-\infty$ to some upper boundary.
For $\Phi(x)$ where $x < 0$ you can use the symmetry of the density function,
for example,
\begin{align}
\Phi(-0.4)
& = \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,\mathrm dx
    - \int_{-0.4}^0 \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,\mathrm dx \\
& = \frac12 - \int_0^{0.4} \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,\mathrm dx. 
\end{align}
