The normal distribution curve for the problem is shown below
We need to standardise the value X=405.5 by using the formula
[tex]z-score= \frac{X-Mean}{Standard Deviation} [/tex]
[tex]z-score= \frac{405.5-402.7}{8.8} =0.32[/tex]
We now need to find the probability of z=0.32 by reading the z-table
Note that z-table would give the reading to the left of z-score, so if your aim is to work out the area to the right of a z-score, then you'd need to do:
[tex]P(Z\ \textgreater \ z)=1 - P(Z\ \textless \ z)[/tex]
from the z-table, the reading [tex]P(Z\ \textless \ 0.32)[/tex] gives 0.6255
hence,
[tex]P(Z\ \textgreater \ 0.32)=1-0.6255=0.3475[/tex]
The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%
