Answer: 0.9958
Step-by-step explanation:
We assume that the heights of men are normally distributed with
Mean : [tex]\mu=69.0\text{ inches}[/tex]
Standard deviation : [tex]\sigma=2.8\text{ inches}[/tex]
Sample size : [tex]n=64[/tex]
The value of z-score is given by:
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
Let x be the height of the randomly selected men.
For x= 68
[tex]z=\dfrac{68-69}{\dfrac{2.8}{\sqrt{64}}}\approx-2.86[/tex]
For x= 70
[tex]z=\dfrac{70-69}{\dfrac{2.8}{\sqrt{64}}}\approx2.86[/tex]
Now, the probability that they have a mean height between 68 and 70 inches is given by :-
[tex]P(68<x<70)=P(-2.86<z<2.86)\\=1-2P(z<-2.86)=1-2(0.0021182)=0.9957636\approx0.9958[/tex]
Hence, the probability that they have a mean height between 68 and 70 inches = 0.9958
