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The normal distribution curve for the problem is shown below
We need to standardise the value X=405.5 by using the formula
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:
from the z-table, the reading
gives 0.6255
hence,
The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%
We need to standardise the value X=405.5 by using the formula
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:
from the z-table, the reading
hence,
The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%