Chapter 7.2, Problem 11E

To determine

(a)

To find:

The probability that a quality control regulator will pull a tube off the assembly line that has a length between 8.6 and 9 cm.

Answer to Problem 11E

Solution:

The probability that a quality control regulator will pull a tube off the assembly line that has a length between 8.6 and 9 cm is 0.2881.

Explanation of Solution

Given:

$\mu =9and\sigma =0.5;$

Description:

Given a population of size $N$ with mean $\left(\mu \right)$ and standard deviation $\left(\sigma \right)$, samples of a fixed discrete number of trials $\left(n\right)$ can be chosen each having a mean called the sample mean $\left(\overline{x}\right)$.

On dealing with the collection of these means for samples chosen from the population, the concept of sampling distribution pops and the statistic dealt with is the sample means, and that the distribution contains all possible samples for the chosen sample size.

The continuity criterion can be extended and here it is known as Central Limit Theorem which states that sampling distribution is approximately normal with:

a) mean ${\mu }_{\overline{x}}=\mu$, and

b) Standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Moreover, the standard score is given by:

$z=\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}$.

Thereafter, the required probability is accordingly obtained by taking into account the specifics in the question.

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{8.6-9}{0.5}\\ & =& -0.8\end{array}$

And,

$\begin{array}{rll}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{9-9}{0.5}\\ & =& 0\end{array}$

Therefore,

$\begin{array}{l}P(8.6<x<9)=P(-0.8<z<0)\\ =P(z<0)-P(z<-0.8)\\ =0.5-0.2119\\ =0.2881\end{array}$

To determine

(b)

To find:

The probability that a random sample of 40 tubes will have a mean of less than 8.8 cm.

Answer to Problem 11E

Solution:

The probability that a random sample of 40 tubes will have a mean of less than 8.8 cm is 0.0057.

Explanation of Solution

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\\ & =& \frac{8.8-9}{\left(\frac{0.5}{\sqrt{40}}\right)}\\ & =& -2.53\end{array}$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\overline{x}<8.8\right)& =& P\left(z<-2.53\right)\\ & =& 0.0057\end{array}$

To determine

(c)

To find:

The probability that a random sample of 35 tubes will have a mean of more than 9.2 cm

Answer to Problem 11E

Solution:

The probability that a random sample of 35 tubes will have a mean of more than 9.2 cm is 0.0089.

Explanation of Solution

Calculation:

The value of the standard score (z-score)

$\begin{array}{rll}z& =& \frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\\ & =& \frac{9.2-9}{\left(\frac{0.5}{\sqrt{35}}\right)}\\ & =& 2.37\end{array}$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\overline{x}>9.2\right)& =& P\left(z>2.37\right)\\ & =& P\left(z<-2.37\right)\\ & \approx & 0.0089\end{array}$

To determine

(d)

To find:

The probability that a random sample of 75 tubes will have a mean that differs from the population mean by more than 0.1 cm.

Answer to Problem 11E

Solution:

The probability that a random sample of 75 tubes will have a mean that differs from the population mean by more than 0.1 cm is 0.0836.

Explanation of Solution

Calculation:

The probability that the sample mean differs from the population mean by more than 0.1cm is calculated as:

$\begin{array}{rll}P\left(\left|\overline{x}-\mu \right|\right)& >& 0.1\\ P\left(\left|\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\right|\right)& >& \frac{0.1}{\frac{0.5}{\sqrt{75}}}\\ P\left(\left|z\right|\right)& >& 1.73\end{array}$

The probability using z table is calculated as:

$\begin{array}{rll}P\left(\left|z\right|>1.73\right)& =& P\left(z<-1.73\right)+P\left(z>1.73\right)\\ & =& 2\times P\left(z<-1.73\right)\\ & =& 2\times 0.0418\\ & =& 0.0836\end{array}$