Answer:
Step-by-step explanation:
Let X represent the age of men when they marry for the first time in the United States.
Given x bar follows normal with mean = 29 years and sigma = 2.5 years
Sample size =n= 60 men
Std error of sample = [tex]\frac{\sigma}{\sqrt{n} } =\frac{2.5}{\sqrt{60} } \\=0.3227[/tex]
The likelihood that the age when they were first married is less than 29.3. years
[tex]=P(X bar <29.3)=P(Z<\frac{29.3-29}{0.3227} \\=P(Z<0.93)[/tex]
=0.8238