Question

Please help with statistics work. SOS. I might cry

Answer 1

You're looking for


`\mathbb P(175\le X\le190)`

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


`Z=(X-\mu_X)/(\sigma_X)\implies X=\mu_X+\sigma_XZ`

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


`\mathbb P(175\le172+12.4Z\le190)=\mathbb P\left(\frac3{12.4}\le Z\le(18)/(12.4)\right)`

`\approx\mathbb P(0.2419\le Z\le1.4516)`


`Z` is a continuous random variable, so the above can be split up as


`\mathbb P(Z\le1.4516)-\mathbb P(Z\le0.2419)`

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