Answer:
percent of scores in the data set is 90.98%
Step-by-step explanation:
Given data:
Mean = 123
standard deviation 15
Determine the percent of score between 108 to 153
P(108 < X < 153)



[FROM STANDARD Z TABLE obtained the value of z
= 0.9967 - 0.0869 [FROM STANDARD Z TABLE]
= 0.9098
percent of scores in the data set is 90.98%