Answer: (7.27%, 7.55%)
Step-by-step explanation:
As per given , we have
Sample size : n= 392
Sample mean : 

Critical two-tailed z-value for 95% confidence = 
Required confidence interval would be :

Hence, the required 95% confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies : (7.27%, 7.55%)
The volume of a sphere is:
V=(4πr^3)/3, and you are told that r=5in so
V=(4π125)/3
V=500π/3 in^3
V≈523.60 in^3