\(\displaystyle z=\frac{(x-{\mu})\sqrt{n}}{\sigma}\)

\(\displaystyle z=\frac{(2.5-2.9)\sqrt{50}}{.7}=-4.04\)

You are correct.

This means the probability is very low that it will be less than 2.5, considering the standard deviation is only .7

Looking beyond the scope of the table, the probability would be .0000267

or about .00267%

In other words, very low.

Do you have access to a calcualtor or computer that will do integration?.

Here is the formula used to find those z table values:

\(\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z}e^{\frac{-t^{2}}{2}}dt\)

I used z=-4.04 and plugged it into my TI.