Chapter 9.2, Problem 63E

(a)

To determine

To Explain: the shape, centre and variability of the distribution of the random variable X-Y.

Answer to Problem 63E

About normal with 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.

Explanation of Solution

Given:

  $\begin{array}{c}{\mu }_X=5.3\\ {\sigma }_X=0.01\\ {\mu }_Y=5.26\\ {\sigma }_Y=0.02\end{array}$

Assume

  $\begin{array}{c}X=\text{diameter of randomly selected case}\\ \text{Y=diameter of randomly selected DVD}\end{array}$

The mean of the difference of 2 random variables is

  $\begin{array}{c}{\mu }_{X-Y}={\mu }_X-{\mu }_Y\\ =5.3-5.26\\ =0.04\end{array}$

The variance of the difference of 2 random variables is

  $\begin{array}{c}{\sigma }^2{}_{X-Y}={\sigma }^2{}_X+{\sigma }^2{}_Y\\ ={0.01}^2+{0.02}^2\\ =0.0001+0.0004\\ =0.0005\end{array}$

The standard deviation is

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

X-Y is important to the DVD manufacture, because if 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.

(b)

To determine

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

Answer to Problem 63E

96.33%

Explanation of Solution

Given:

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

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The DVD fits in the case, when the difference X-Y is positive.

The Z-score is

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

Find the corresponding probability using the normal probability table $P\left(z<-1.79\right)$ is given in the row starting with -1.7 and in the column starting with .09 of the standard normal probability table.

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

(c)

To determine

To find: the probability that all DVDs fit in their cases.

Answer to Problem 63E

2.378%

Explanation of Solution

Given:

  $\begin{array}{c}n=100\\ p=0.9633\end{array}$

Formula used:

  $P\left(X=k\right)=C{}_k\cdot p^k\cdot {\left(1-p\right)}^{n-k}$

Calculation:

Binomial probability is

  $P\left(X=k\right)=C{}_k\cdot p^k\cdot {\left(1-p\right)}^{n-k}$

Find the definition of binomial probability at k=100:

  $\begin{array}{c}P\left(X=100\right)=C{}_{100}\cdot {0.9633}^{100}\cdot {\left(1-0.9633\right)}^{100-100}\\ =0.02378\\ =2.378\%\end{array}$

There is a 2.378% possibility that all of the 100 DVDs fit in their cases.