Answer:
About 220 of the students scored less than 96
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 156 and a standard deviation of 23.
This means that 
Proportion that scored less than 96:
p-value of Z when X = 96. So



has a p-value of 0.00453.
About how many of the students scored less than 96?
0.00453 out of 48592.
0.00453*48592 = 220.1.
Rounding to the closest integer:
About 220 of the students scored less than 96