Chapter 9.2, Problem 61E

a)

To determine

The shape, center and spread of the distribution of the random variable X-Y and the importance of random variable to CD manufacturer.

Explanation of Solution

Given:Let

X = The diameter of randomly selected CD

Y = the diameter of randomly selected case.

The plastic cases vary normally with mean diameter of 4.2 inches and standard deviation of 0.05 inches. The CD’s vary normally with mean diameter of 4 inches and standard deviation of 0.1 inches.

Concept used:The addition or subtraction of two normal distributions (X,Y) will also be a normal distribution.

For X − Y the mean and standard deviation are calculated using the below shown formula

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

Calculation:The Mean of X − Y is calculated as shown below

  $\begin{array}{l}{\mu }_{X-Y}={\mu }_X-{\mu }_Y\\ {\mu }_{X-Y}=4-4.2=-0.2\end{array}$

The standard deviation of X − Y is calculated as shown below

  $\begin{array}{l}{\sigma }_{X-Y}=\sqrt{{\sigma }_X^2+{\sigma }_Y^2}\\ =\sqrt{{0.1}^2+{0.05}^2}=0.112\end{array}$

Conclusion:

The shape of the distribution X − Y is also normal distribution with its center (mean) -0.2 and spread (standard deviation) of 0.112 inches. The random variable X − Y is important to CD manufacturer because CD has to fit into the case. The CD fit into the case only if the random variable has negative values.

b)

To determine

The probability that a randomly selected CD will fit inside a randomly selected case.

Explanation of Solution

Given:Let

X = The diameter of randomly selected CD

Y = the diameter of randomly selected case.

The distribution of X − Y is also normal distribution with mean -0.2 and standard deviation 0.112.

Concept used:The Z score is the distance of any data point from its mean in terms of standard deviation and for a random normal variable X with mean µ and standard deviation s is calculated as shown below

  $Z=\frac{X-\mu }{\sigma }$

Calculation:The CD fit into the case only if the random variable has negative values.

The probability that randomly selected CD will fit inside randomly selected case is

P(X − Y = 0)

The Z score for X − Y = 0 is calculated as shown below

  $\begin{array}{l}Z=\frac{\left(X-Y\right)-{\mu }_{X-Y}}{{\sigma }_{X-Y}}\\ Z=\frac{0-(-0.2)}{0.112}=1.79\end{array}$

  $P\left(X–Y\le 0\right)=P(Z\le 1.79)$

using standard normal probabilities P(Z = 1.79) = 0.9633

Conclusion:

The probability that a randomly selected CD will fit into a randomly selected case is 0.9633

c)

To determine

The probability that a randomly selected 100 CD will fit inside a randomly selected 100 cases.

Explanation of Solution

Given:The probability that a randomly selected CD will fit into a randomly selected case is 0.9633

Concept used:If the probability of an event A to occur be p and event A occur independently then probability that event A occur n times is pⁿ.

Calculation:The probability that one randomly selected CD will fit into one randomly selected case is 0.9633.

The probability that 100 randomly selected CD will fit into 100 randomly selected cases is 0.9633¹⁰⁰ = 0.0238

Conclusion:

The probability that randomly selected 100 CD’s will fit into a randomly selected 100 cases is 0.0238