Chapter 10.2, Problem 35E

Solution

a)

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

Given:M be the cholesterol level of a randomly selected 20 − 34 year old man which follow a normal distribution with mean 188 mg/dl and standard deviation 41 mg/dl.

B be the cholesterol level of a randomly selected 14 year old boys which follow a normal distribution with mean 170 mg/dl and standard deviation 30 mg/dl.

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

  $\begin{array}{l}{\mu }_{M-B}={\mu }_M-{\mu }_B=188-170=18mg/dl\\ {\sigma }_{M-B}=\sqrt{{\sigma }_M^2+{\sigma }_B^2}=\sqrt{{41}^2+{30}^2}=50.80mg/dl\end{array}$

Conclusion:The distribution of M − B is also normal distribution (symmetric shape) with its center at (mean) 18 mg/dl and standard deviation (spread) 50.80 mg/dl.

b)

The probability that a randomly selected 14 year old boy has higher cholesterol than a randomly selected man aged 20 to 34.

Given:The distribution of M − B is normal with mean 18 mg/dl and standard deviation 50.8 mg/dl

Concept used:The Z score is calculated as shown below

  $Z=\frac{X-\mu }{\sigma }$

Calculation:

The probability that a randomly selected 14 year old boy has higher cholesterol than a randomly selected man aged 20 to 34 is calculated as shown below

  $\begin{array}{l}P(M-B<0)=P\left(\frac{(M-B)-{\mu }_{M-B}}{{\sigma }_{M-B}}<\frac{0-18}{50.8}\right)\\ =P(Z<-0.35)=0.3632\end{array}$

With reference to Z table P(Z < -0.35) is 0.3632.

Conclusion:The probability that a randomly selected 14 year old boy has higher cholesterol than 20 to 34 year aged man is 0.3632.