8. over 6 feet. (72 inches) 9. under 65 inches. 10. between 62 inches and 65 inches.

8.

The random variable X is defined as the height of UNLV female student which follows normal distribution with mean 63.7 inches and a standard deviation of 2.3 inches.

The probability that a UNLV female student with over 6 feet(72 inches) is,

$\begin{array}{rcl}P\left(X>72\right)& =& 1-P\left(X<72\right)\\ & =& 1-P\left(\frac{X-63.7}{2.3}<\frac{72-63.7}{2.3}\right)\\ & =& 1-P\left(z<3.61\right)\end{array}$

The probability of z less than 3.61 can be obtained using the excel formula “=NORM.S.DIST(3.61,TRUE)”.

The probability value is 0.9998.

The required probability is,

$\begin{array}{rcl}P\left(X>72\right)& =& 1-P\left(z<3.61\right)\\ & =& 1-0.9998\\ & =& 0.0002\end{array}$

Thus, the probability that a UNLV female student with over 6 feet(72 inches) is 0.0002.

9.

The probability that a UNLV female student with under 65 inches is,

$\begin{array}{rcl}P\left(X<65\right)& =& P\left(\frac{X-63.7}{2.3}<\frac{65-63.7}{2.3}\right)\\ & =& P\left(z<0.565\right)\end{array}$

The probability of z less than 0.565 can be obtained using the excel formula “=NORM.S.DIST(0.565,TRUE)”.

The probability value is 0.7140.

The required probability is,

$\begin{array}{rcl}P\left(X<65\right)& =& P\left(z<0.565\right)\\ & =& 0.7140\end{array}$

Thus, the probability that a UNLV female student with under 65 inches is 0.7140.