Answer:
87.29%
Step-by-step explanation:
Given: Mean= 68.8 inches
Standard deviation= 2.8 inches
Now, finding the percent of the men in the sample were greater than 72 inches.
We know, z-score= [tex]\frac{x-mean}{standard\ deviation}[/tex]
z-score= [tex]\frac{72-68.8}{2.8}[/tex]
⇒ z-score= [tex]\frac{3.2}{2.8} = 1.14[/tex]
∴ z-score= 1.14
Next, using normal distribution table to find percentage.
Coverting 0.8729 into percentage= [tex]0.8729\times 100[/tex]
We get the percentage as 87.29%
Hence, 87.29% of the men in the sample were greater than 72 inches.
