Answer: Second Option
[tex]P(180<X <185)=13.5\%[/tex]
We know that the mean is:
[tex]\mu=175[/tex]
and the standard deviation is:
[tex]\sigma=5[/tex]
We are looking at the percentage of students between 180 centimeters and 185 centimeters in height.
This is:
[tex]P(180<X <185)[/tex]
We calculate the Z-score using the formula:
[tex]Z=\frac{X-\mu}{\sigma}[/tex]
For [tex]X=180[/tex]
[tex]Z_{180}=\frac{180-175}{5}[/tex]
[tex]Z_{180}=1[/tex]
For [tex]X=185[/tex]
[tex]Z_{185}=\frac{185-175}{5}[/tex]
[tex]Z_{185}=2[/tex]
Then we look at the normal table
[tex]P(1<Z<2)[/tex]
[tex]P(1<Z<2)=P(Z<2)-P(Z<1)[/tex]
[tex]P(1<Z<2)=0.9772-0.8413[/tex]
[tex]P(1<Z<2)=0.135[/tex]
[tex]P(180<X <185)=13.5\%[/tex]
Note: You can get the same conclusion using the empirical rule
Look at the attached image for [tex]\mu+ 1\sigma <\mu <\mu + 2\sigma[/tex]
