Answer
Sara's total standard deviation is greater than Reginald's.
Work Step by Step
1st:
Reginald: $z=\frac{y-\mu}{\rho}=\frac{80-80}{4}=0$
Sara: $z=\frac{y-\mu}{\rho}=\frac{88-80}{4}=2.0$
2nd:
Reginald: $z=\frac{y-\mu}{\rho}=\frac{85-70}{15}=1.0$
Sara: $z=\frac{y-\mu}{\rho}=\frac{65-70}{15}=-0.33$
Total:
Reginald: $0+1.0=1.0$
Sara: $2.0-0.33=1.67$
Hence Sara's total standard deviation is greater than Reginald's.
