Please help with statistics work. SOS. I might cry

You're looking for

[tex]\mathbb P(175\le X\le190)[/tex]

where [tex]X[/tex] is the random variable representing the amount of lumber harvested. Transform this to the standard normal distribution, using

[tex]Z=\dfrac{X-\mu_X}{\sigma_X}\implies X=\mu_X+\sigma_XZ[/tex]

where [tex]\mu_X=172[/tex] is the mean and [tex]\sigma_X=12.4[/tex] are the mean and standard deviation of [tex]X[/tex], respectively. So the probability in terms of [tex]Z[/tex] is

[tex]\mathbb P(175\le172+12.4Z\le190)=\mathbb P\left(\dfrac3{12.4}\le Z\le\dfrac{18}{12.4}\right)[/tex]
[tex]\approx\mathbb P(0.2419\le Z\le1.4516)[/tex]

[tex]Z[/tex] is a continuous random variable, so the above can be split up as

[tex]\mathbb P(Z\le1.4516)-\mathbb P(Z\le0.2419)[/tex]

and consulting a table, you end up with a probability (proportion) of about 0.3311.