Answer:
13.92
Explanation:
We have that the critical z-score associated with 85% to the left is 1.04, we know that by table.
So we have to:
m + z * (sd) = 16
where m is the mean, z is the critical z-scor and sd is the standard deviation, if we replace we are left with:
m + 1.04 * (2) = 16
m = 16 - 1.04 * (2)
m = 13.92
Therefore, the average weight if 85% of cucumbers weigh less than 16 ounces is 13.92