Question

Suppose 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 inches. 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 is calculated.

(a) Describe the shape, center, and spread 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 means the diameter is less than 3.99 inches or greater than 4.01 inches since this indicates that some lids will be too small or too large for the cups. If the sample mean is less than 3.99 or greater than 4.01, all the lids from

that hour is thrown away since 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? Show your work.

Additionally, to look for any trends, each hour the company records the value of the sample mean on a chart, like the one below.

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 means increased in 3 consecutive samples, this would suggest that something might be wrong with the machine.

If this trend can be noticed before the sample means gets larger than 4.01, 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? Show your work.

(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? Show your work.

(e) Which of the following results gives more convincing evidence that the machine needs to be shut down? Explain.

1. Getting a single sample mean below 3.99 or above 4.01 OR

2. Taking 5 consecutive samples and having at least 4 of the sample means be between 4.00 and 4.01.

(f) Suggest a different rule (other than 1 and 2 stated in part (e)) for stopping the machine before it starts producing lids that have to be thrown away. Assuming that the machine is working properly, calculate the probability that the machine will be shut down when using your rule.

Short answer

a. Shape: Unskewed

Center: Mean

Variance ($1.6\times {10}^{-5}$)

b. 1

c. 0.5

d. $\frac{3}{16}$

e. Getting a single sample mean below 3.99 or above 4.01

f. Use the 6-sigma limits, the probability is equal to 0.9973.

Step-by-step solution

Part (a) Step 1. Given Information

Let X denotes the diameter of iid's.

$X~N\left(4,0.02\right)$

Also, the sample size is equal to 25.

Part (a) Step 2. Shapes , Center and Spread

Since the graph is normal,

Shape: Unskewed and is equal to 0.

Center: Mean and is equal to 4.

Spread: Variance

$\frac{0.0004}{25}=1.6\times {10}^{-5}$

We accept a lot of particular hrs. if

$3.99<\overline{X}<4.01$

Else we reject it.

Part (b) Step 1. Probability of getting a mean diameter less than 3.99 inches. 

$P\left(3.99<\overline{X}<4.01\right)=P\left(\frac{3.99-4}{0.001}<Z<\frac{4.01-4}{0.001}\right)$

where

$\mu =4$, $\frac{\sigma }{\sqrt{n}}=0.001$and $\frac{\overline{x}-\mu }{{\displaystyle \frac{\sigma }{\sqrt{n}}}}$

$\begin{array}{rcl}P\left(3.99<\overline{X}<4.01\right)& =& P\left(-10<Z<10\right)\\ & =& 1\end{array}$

Part (c) Compute Probability

$\begin{array}{rcl}P\left(4<\overline{X}<4.01\right)& =& P\left(0<Z<10\right)\\ & =& 0.5\end{array}$

Part (d) Compute the probability

Since this is the case of binomial distribution with$n=5$and

Part (e) Step 1. Compute the probability

$\begin{array}{rcl}P\left(\overline{X}<3.99or\overline{X}>4.01\right)& \approx & 0\\ P\left(Y=4or5\right)& =& \frac{3}{16}\end{array}$

Part (e) Step 2. Explanation

Since the event one, that is we get a sample mean diameter of iid's less than 3.99 or greater than 4.01, so we can say that this will be greater evidence in favor of the machines need to shut down.

Part (f) Step 1. Explanation

A much better rule will be to calculate the 6 sigma limits.

$P\left(\left|X-\mu \right|<3\sigma \right)\approx 0.9973$