Answer:
47.72%
Step-by-step explanation:
Given
Mean (u) = 15
SD = 4
P(15<x<23) = [tex]P(\frac{x-u}{\sigma}<z<\frac{x-u}{\sigma})[/tex]
= [tex]P(\frac{15-15}{4}<z<\frac{23-15}{4})\\[/tex]
=P(0<z< 2)
=P(z<2) - P(z<0)
Use standard normal table to find probability of the z score
= 0.9772 - 0.5
= 0.4772 [tex]\approx\\[/tex] 47.72%
