Question

A company manufactures plastic lids for disposable coffee cups. When the manufacturing process is working correctly, the diameters of the lids are approximately Normally distributed with a mean diameter of 4 inches and a standard deviation of 0.02 inch. To make sure the machine is not producing lids that are too big or too small, each hour a random sample of 25 lids is selected and the sample mean x−x¯ is calculated.
(a) Describe the shape, center, and variability of the sampling distribution of the sample
mean diameter, assuming the machine is working properly.
The company decides that it will shut down the machine if the sample mean diameter is less than3.99inches or greater than4.01inches, because this indicates that some lids will be too small or too large for the cups. If the sample mean is less than3.99or greater than4.01, all the lids manufactured that hour are thrown away because the company does not want to sell bad products.
(b) Assuming that the machine is working properly, 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?

One benefit of using this type of chart is that out-of-control production trends can be noticed before it is too late and lids have to be thrown away. For example, if the sample mean is consistently greater than 4 (but less than4.01), this would suggest that something might be wrong with the machine. If such a trend is noticed before the sample mean gets larger than4.01, then the machine can be fixed without having to throw away any lids.
(c) Assuming that the manufacturing process is working correctly, what is the probability that the sample mean diameter will be above the desired mean of 4.00 but below the upper boundary of 4.01 ?
(d) Assuming that the manufacturing process is working correctly, what is the probability that in 5 consecutive samples, 4 or 5 of the sample means will be above the desired mean of 4.00 but below the upper boundary of 4.01 ?
(e) Which of the following results gives more convincing evidence that the machine needs to be shut down? Explain your answer.

Short answer

(a) The probability that 4 or 5 of the sample means will be above the desired mean in five consecutive samples is 4 but less than the upper boundary of 4.01 is 0.1799.

(b) A random sample of 25 lids has a 0.0124 chance of having a mean diameter less than 3.99 inches or greater than 4.01 inches.

(c) The probability that the sample mean diameter will be greater than the desired mean but less than the upper limit of 4.01 is 0.4938.

(d) The probability that 4 or 5 of the sample means will be above the desired mean in five consecutive samples is 4 but less than the upper boundary of 4.01 is 0.1799.

(e) The machine must be shut down if the sample mean is less than 3.99 inches or greater than 4.01 inches.

Step-by-step solution

Part (a) Step 1: Given information

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

Explanation

Given that,

$\mu =4,\sigma =4,n=25$

Because the population has a normal distribution, the sample mean's sampling distribution is also normal. So, we've got

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

As a result, the sample mean sampling distribution is essentially normal, with a mean of $4$and a standard deviation of $0.004$.

Part (b) Step 1: Given information

Each hour, the corporation records the value of the sample mean on the chart to determine the likelihood that a random sample of $25$lids will have a mean diameter less than $3.99$inches or larger than$4.01$inches, as well as to discover any trends.

Explanation

Given that,

$\mu =4,\sigma =4,n=25,\overline{x}=3.99or4.01$

Because the population has a normal distribution, the sampling distribution of the sample mean has a normal distribution as well. As a result,

The z-score has the following value:

$$\begin{gathered} z=\frac{\overline{x}-{\mu }_{\overline{x}}}{\sigma x¯} \\ =\frac{\overline{x}-{\mu }_{\overline{x}}}{{\displaystyle \frac{\sigma }{\sqrt{n}}}} \\ =\frac{3.99-4}{{\displaystyle \frac{0.02}{\sqrt{25}}}} \\ =-2.50 \\ z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}} \\ =\frac{\overline{x}-{\mu }_{\overline{x}}}{{\displaystyle \frac{\sigma }{\sqrt{n}}}} \\ =\frac{4.01-4}{{\displaystyle \frac{0.02}{\sqrt{25}}}} \\ =2.50 \end{gathered}$$
The following is the probability:

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

Thus, a random sample of $25$lids has a $0.0124$chance of having a mean diameter of less than $3.99$inches or larger than$4.01$inches.

Part (c) Step 1: Given information

To determine the likelihood that the sample mean diameter would be greater than the target mean$4$but less than the upper boundary of$4.01$.

Explanation

Given that,

$\mu =4,\sigma =4,n=25,\overline{x}=4or4.01$

Because the population has a normal distribution, the sampling distribution of the sample mean has a normal distribution as well. As a result,

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

The following is the probability:

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

As a result, there's a $0.4938$chance that the sample mean diameter will be higher than the targeted mean $4$but lower than the upper boundary of localid="1654456370050" $4.01$.

Part (d) Step 1: Given information

The probability that or $5$of the sample means will be above the target mean in five consecutive samples is $4$but below the upper boundary of$4.01$.

Explanation

Now, we'll use the binomial probability, and we'll have,

We will calculate the probability at k=4,5 as follows:

$$\begin{gathered} P(X=4)={}_{5}C_4\times {{\left(}_{\right)}}^{00.4938}\times {(}_{-}^1 \\ =\frac{5!}{4!(5-4)!}\times {{\left(}_{\right)}}^{0.4938}\times {{\left(}_{\right)}}^{00.5062} \\ =0.1505 \\ P(X=5)={}_{5}C_5\times {{\left(}_{\right)}}^{0.4938}\times {(}_{-}^1 \\ =\frac{5!}{5!(5-5)!}\times {{\left(}_{\right)}}^{0.4938}\times {{\left(}_{\right)}}^{00.5062}=0.0294 \end{gathered}$$

Using the addition rule, we obtain

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

Thus, the probability that four or five of the sample means will be above the desired mean in five consecutive samples is 4 but less than the upper boundary of 4.01 is 0.1799.

Part (e) Step 1: Given information

Explain which of the following results provides more convincing evidence that the machine should be turned off.

Explanation

Thus, the probability that four or five of the sample means will be above the desired mean in five consecutive samples is $4$but less than the upper boundary of $4.01$is $0.1799$.

Part (b) now gives us: A random sample of $25$lids has a $0.0124$chance of having a mean diameter less than $3.99$inches or greater than $4.01$
inches.

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

As a result, the lower the probability, the less likely the event occurs by chance, and thus the more evidence we have that the machine needs to be programmed.