The number of peanuts in bags is normally distributed with a mean of 193.8 peanuts and a standard deviation of 4.1 peanuts. What is the z-score of a bag containing 185 peanuts?
−2.47
−2.15
−1.90
−1.22
The "formula" for a z score is
[ given score - mean score ] / standard deviation = z score
So....
[ 185 - 193.8 ] / 4.1 ≈ -2.15
+155 
