Question

Packaging DVDs $(6.2,5.3)$ A manufacturer of digital video discs (DVDs) wants to be sure that the DVDs will fit inside the plastic cases used as packaging. Both the cases and the DVDs are circular. According to the supplier, the diameters of the plastic cases vary Normally with mean $\mu =5.3$inches and standard deviation $\sigma =0.01$inch. The DVD manufacturer produces DVDs with mean diameter$\mu =5.26$inches. Their diameters follow a Normal distribution with $\sigma =0.02$inch.

a. Let X = the diameter of a randomly selected case and Y = the diameter of a randomly selected DVD. Describe the shape, center, and variability of the distribution of the random variable X−Y. What is the importance of this random variable to the DVD manufacturer?

b. Calculate the probability that a randomly selected DVD will fit inside a randomly selected case.

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

Short answer

The required answers:

Part a) About normal with a mean $0.04$and standard deviation $0.02236$

$X-Y$is positive then the DVD will fit in the case but if the difference

$X-Y$is negative, and then the DVD will not fit in the case.

Part b) $96.33\%$

Part c) There is a $2.378\%$ chance that all 100 DVDs will fit in their cases.

Step-by-step solution

Part a) Step 1: Given information

$$\begin{gathered} \mu x=5.3 \\ \sigma x=0.01 \\ \mu y=5.26 \\ \sigma y=0.02 \end{gathered}$$

Let,

$X=$diameter of the randomly chosen case

$Y=$the diameter of a randomly chosen DVD

Part a) Step 2: The objective is to explain the shape, center, and variability of the distribution of the random variable X-Y

The average of the difference between two random variables is:

$$\begin{gathered} \mu X-Y=\mu X-\mu Y \\ =5.3-5.26 \\ =0.04 \end{gathered}$$

The variance of the difference of $2$random variables is as follows:

$$\begin{gathered} {\sigma }^{2}X-Y={\sigma }^{2}X+{\sigma }^{2}Y \\ ={(0.01)}^2+{(0.02)}^2=0.0001+0.0004 \\ =0.0005 \end{gathered}$$

The standard deviation is as follows:

$$\begin{gathered} \sigma X-Y=\sqrt{{\sigma }^{2}X-Y} \\ ==\sqrt{0.0005} \\ =0.02236 \end{gathered}$$

$X-Y$is important to the DVD manufacturers, because of the difference.

$X-Y$is positive then the DVD will fit in the case but if the difference

$X-Y$ is negative, and then the DVD will not fit in the case.

Part b) Step 1: Given information

$$\begin{gathered} \mu =0.04 \\ \sigma =0.0005 \\ x=0 \end{gathered}$$

Part b) Step 2: The objective is to Calculate the probability that a randomly selected DVD will fit inside a randomly selected case. 

Formula used:

$z=\frac{x-\mu }{\sigma }$

When the difference $X-Y$is positive, the DVD fits in the case.

The $Z$-score is :

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0-0.04}{0.02236} \\ =-1.7 \end{gathered}$$

Using the normal probability table, determine the corresponding probability. $P(z<-1.79)$is given in the standard normal probability table in the row beginning with $-1.7$and the column beginning with $0.09$

$$\begin{gathered} P(X-Y>0)=P(Z>-1.79) \\ =1-P(Z<-1.79) \\ =1-0.0367 \\ =0.9633 \\ =96.33\% \end{gathered}$$

Part c) Step 1: Given information

$$\begin{gathered} n=100 \\ p=0.9633 \end{gathered}$$

Part c) Step 2: The objective is to find the probability that all DVDs fit in their cases.

Formula used:

$P(X=k)={}^{n}C_k·p^k·(1-p{)}^{n-k}$

Binomial probability is defined as:

$P(X=k)={}^{n}C_k·p^k·(1-p{)}^{n-k}$

At $k=100$find the definition of binomial probability.

role="math" localid="1654440514831" $$\begin{gathered} P(X=100)={}_{100}C^{100}·0.{9633}^{100}·{(1-0.9633)}^{100-100} \\ =0.02378 \\ =2.378\% \end{gathered}$$

There is a $2.378\%$ chance that all 100 DVDs will fit in their cases.