Question

Packaging CDs (6.2, 5.3) A manufacturer of compact discs (CDs) wants to be sure that their CDs will fit inside the plastic cases they have bought for packaging. Both the CDs and the cases are circular. According to the supplier, the plastic cases vary Normally with mean diameter $\mu =4.2inches$and a standard deviation $\sigma =0.05inches$. The CD manufacturer decides to produce CDs with mean diameter $\mu =4inches$. Their diameters follow a Normal distribution with $\sigma =0.1inches$.

(a) Let X = the diameter of a randomly selected CD and Y = the diameter of a randomly selected case. Describe the shape, center, and spread of the distribution of the random variable X = Y. What is the importance of this random variable to the CD manufacturer?

(b) Compute the probability that a randomly selected CD will fit inside a randomly selected case.

(c) The production process actually runs in batches of 100 CDs. If each of these CDs is paired with a randomly chosen plastic case, find the probability that all the CDs fit in their cases.

Short answer

a. The random variable X - Y has a Normal distribution with mean ${\mu }_{X-Y}=0.2$and standard deviation ${\sigma }_{X-Y}=0.112$. This random variable is important because for a CD to fit in the case, the variable must take on a negative number.

b. $P(X-Y<0)=P(z<1.79)=0.9633$

c.$(0.9633{)}^{100}=0.0238$

Step-by-step solution

Given information

${\mu }_X=4.2inches$

${\mu }_Y=4inches$

${\sigma }_X=0.05inches$

${\sigma }_Y=0.1inches$

Explanation (part a)

The significance of random variable difference X - Y is a misfit of CD's into cases. Some of the cases are too small for the CD's which means the CD manufacturer needs to acquire more than 1 case for each CD.
In real life, the cost for acquiring more cases might be less than the cost to re-engineer the tools for making the variation smaller for either the CD's or the cases. So in real life, the CD manufacturer might be satisfied to not expend excessive costs of re-engineering and makes a business decision to acquire more cases which have a probability of misfit.

Explanation (part b)

Normal distribution with difference in mean $\mu =0.2$and standard deviation${\sigma }_{X-Y}=0.112$

$P(X-Y<0)=P(z<1.79)=0.9633$

Explanation (part c)

The production process actually runs in batches of 100 CDs. If each of these CDs is paired with a randomly chosen plastic case, the probability that all the CDs fit in their cases$(0.9633{)}^{100}=0.0238$