The number of nuts in a can of mixed nuts is found to be normally distributed, with a mean of 500 nuts and a standard deviation of 20 nuts. My can of mixed nuts has only 455 nuts. What is the z-score for this can of nuts?

-2.25

Step-by-step explanation:

We have that the mean (m) is equal to 500, the standard deviation (sd) 20

They ask us for P (x <455)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / sd

We have all these values, replacing we have:

z = (455 - 500) / (20)

z = - 2.25