Chapter 10.2, Problem 36E

Solution

a)

The shape, center and spread of the distribution M – W.

Given:M be height of a randomly selected young man which follow a normal distribution with mean 69.3 inches and standard deviation 2.8 inches.

B be height of a randomly selected young woman which follow a normal distribution with mean 64.5 inches and standard deviation 2.5 inches.

Concept used:If X, Y are normally distributed variables with mean µ_X , µ_Y and standard deviation s_X, s_Y respectively, then variable X − Y is also normally distributed with

  $\begin{array}{l}{\mu }_{X-Y}={\mu }_X-{\mu }_Y\\ {\sigma }_{X-Y}=\sqrt{{\sigma }_X^2+{\sigma }_Y^2}\end{array}$

Calculation:

As given M, W are normally distributed, the distribution of M − W is also normally distributed and its mean and standard deviation are calculated as shown below

  $\begin{array}{l}{\mu }_{M-W}={\mu }_M-{\mu }_W=69.3-64.5=4.8\text{ inches}\\ {\sigma }_{M-W}=\sqrt{{\sigma }_M^2+{\sigma }_W^2}=\sqrt{{2.8}^2+{2.5}^2}=3.75\text{ inches}\end{array}$

Conclusion:The distribution of M − W is also normal distribution (symmetric shape) with its center at (mean) 4.8 inches and standard deviation (spread) 3.75 inches.

b)

The probability that a randomly selected young man is at least 2 inches taller than a randomly selected young woman.

Given:The distribution of M − W is normal with mean 4.8 inches and standard deviation 3.75 inches.

Concept used:The Z score is calculated as shown below

  $Z=\frac{X-\mu }{\sigma }$

Calculation:

The probability that a randomly selected young man is at least 2 inches taller than a randomly selected young woman is calculated as shown below

  $\begin{array}{l}P(M-W\ge 2)=P\left(\frac{(M-W)-{\mu }_{M-W}}{{\sigma }_{M-W}}\ge \frac{2-4.8}{3.75}\right)\\ =P(Z\ge -0.75)=0.7734\end{array}$

With reference to Z table P(Z = -0.75) is 0.7734.

Conclusion:The probability that a randomly selected young man is at least 2 inches taller than a randomly selected young woman is 0.7734.