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)
![= P((108 - 123)/(11) < z < (153-123)/(11))](https://img.qammunity.org/2020/formulas/computers-and-technology/college/v01ptpqdsij1xp41xk3wrnm7re6b85qzxr.png)
![= P(-1.36 < z < 2.72)](https://img.qammunity.org/2020/formulas/computers-and-technology/college/3vc2luxrqd39dfy3o5ndvur783bc7bl5lq.png)
![= P(z < 2.72) - P(z < -1.36)](https://img.qammunity.org/2020/formulas/computers-and-technology/college/3r9t7gz2iwndtl5rftijhe8x135614n6l0.png)
[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%