Chapter 12, Problem AP4.46CPT

(a)

To determine

To describe the shape, center, and variability of the sampling distribution of the sample mean diameter, assuming the machine is working properly.

Explanation of Solution

It is given that,

  $\begin{array}{l}\mu =4\\ \sigma =4\\ n=25\end{array}$

Since the population has approximately a normal distribution the sampling distribution of the sample mean also has approximately a normal distribution. Thus, we have,

  $\begin{array}{l}{\mu }_{\overline{x}}=\mu =4\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{0.02}{\sqrt{25}}\\ =0.004\end{array}$

Thus, the sampling distribution of the sample mean is approximately normal with mean $4$ and standard deviation $0.004$ .

(b)

To determine

To find out what is the probability that a random sample of $25$ lids will have a mean diameter less than $3.99$ inches or greater than $4.01$ inches and also to identify any trend, each hour the company records the value of the sample mean on the chart.

Answer to Problem AP4.46CPT

The probability that a random sample of $25$ lids will have a mean diameter less than $3.99$ inches or greater than $4.01$ inches is $0.0124$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}\mu =4\\ \sigma =4\\ n=25\\ \overline{x}=3.99\text{ or 4}\text{.01}\end{array}$

Since the population has approximately a normal distribution the sampling distribution of the sample mean also has approximately a normal distribution. Thus, we have,

The value of z-score is as:

  $\begin{array}{l}z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-{\mu }_{\overline{x}}}{\frac{\sigma }{\sqrt{n}}}\\ =\frac{3.99-4}{\frac{0.02}{\sqrt{25}}}\\ =-2.50\\ z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-{\mu }_{\overline{x}}}{\frac{\sigma }{\sqrt{n}}}\\ =\frac{4.01-4}{\frac{0.02}{\sqrt{25}}}\\ =2.50\end{array}$

The corresponding probability is as:

  $\begin{array}{l}P(\overline{X}\le 3.99\text{ or }\overline{X}\ge 4.01)\\ =P(Z<-2.50\text{ or }Z>2.50)\\ =2P(Z<-2.50)\\ =2(0.0062)\\ =0.0124\\ =1.24\%\end{array}$

Thus, the probability that a random sample of $25$ lids will have a mean diameter less than $3.99$ inches or greater than $4.01$ inchesis $0.0124$ .

(c)

To determine

To find out what is the probability that the sample mean diameter will be above the desired mean $4$ but below the upper boundary of $4.01$ .

Answer to Problem AP4.46CPT

The probability that the sample mean diameter will be above the desired mean $4$ but below the upper boundary of $4.01$ is $0.4938$ .

Explanation of Solution

It is given that:

  $\begin{array}{l}\mu =4\\ \sigma =4\\ n=25\\ \overline{x}=4\text{ or 4}\text{.01}\end{array}$

Since the population has approximately a normal distribution the sampling distribution of the sample mean also has approximately a normal distribution. Thus, we have,

The value of z-score is as:

  $\begin{array}{l}z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-{\mu }_{\overline{x}}}{\frac{\sigma }{\sqrt{n}}}\\ =\frac{4-4}{\frac{0.02}{\sqrt{25}}}\\ =0\\ z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}\\ =\frac{\overline{x}-{\mu }_{\overline{x}}}{\frac{\sigma }{\sqrt{n}}}\\ =\frac{4.01-4}{\frac{0.02}{\sqrt{25}}}\\ =2.50\end{array}$

The corresponding probability is as:

  $\begin{array}{l}P(4<\overline{X}<4.01)\\ =P(0<Z<2.50)\\ =P(Z<2.50)-P(Z<0)\\ =0.9938-0.5\\ =0.4938\\ =49.38\%\end{array}$

Thus, the probability that the sample mean diameter will be above the desired mean $4$ but below the upper boundary of $4.01$ is $0.4938$ .

(d)

To determine

To find out what is the probability that in five consecutive samples, $\text{4 or 5}$ of the sample means will be above the desired mean is $4$ but below the upper boundary of $4.01$ .

Answer to Problem AP4.46CPT

The probability that in five consecutive samples, $\text{4 or 5}$ of the sample means will be above the desired mean is $4$ but below the upper boundary of $4.01$ is $0.1799$ .

Explanation of Solution

Now, we will be using the binomial probability, then we have,

  $\begin{array}{l}n=5\\ p=0.4938\end{array}$

We will evaluate at the probability at $k=4,5$ as:

  $\begin{array}{l}P(X=4)=C{}_4\times {(}^{0.4938}\times {(}^1\\ =\frac{5!}{4!(5-4)!}\times {(}^{0.4938}\times {(}^{0.5062}\\ =0.1505\\ P(X=5)=C{}_5\times {(}^{0.4938}\times {(}^1\\ =\frac{5!}{5!(5-5)!}\times {(}^{0.4938}\times {(}^{0.5062}\\ =0.0294\end{array}$

Now, using the addition rule we get,

  $\begin{array}{l}P(X\ge 4)=P(X=4)+P(X=5)\\ =0.1505+0.0294\\ =0.1799\\ =17.99\%\end{array}$

Thus, the probability that in five consecutive samples, $\text{4 or 5}$ of the sample means will be above the desired mean is $4$ but below the upper boundary of $4.01$ is $0.1799$ .

(e)

To determine

To explain which of the following results give more convincing evidence that the machine needs to be shut down.

Answer to Problem AP4.46CPT

The machine in which the sample mean is below $3.99$ inches or greater than $4.01$ inches needs to be shut down.

Explanation of Solution

Now, from part (b), we get: The probability that a random sample of $25$ lids will have a mean diameter less than $3.99$ inches or greater than $4.01$ inches is $0.0124$ .

And from part (d), we get: The probability that in five consecutive samples, $\text{4 or 5}$ of the sample means will be above the desired mean is $4$ but below the upper boundary of $4.01$ is $0.1799$ .

Thus, the smaller the probability the less likely the event occurs by chance and thus the more evidence we have that the machine needs to be shut down. Therefore, we note that the smaller probability belongs to the event of a single sample mean below $3.99$ inches or greater than $4.01$ inches. Thus, the part (b) machine needs to shut down.