Question

Many manufacturing companies use statistical techniques to ensure that the products they make meet standards. One common way to do this is to take a random sample of products at regular intervals throughout the production shift. Assuming that the process is working properly, the mean measurement $x-$from a random sample varies according to a Normal distribution with mean $\mu \overline{x}$and standard deviation $\sigma \overline{x}$. For each question that follows, assume that the process is working properly.

a. What’s the probability that at least one of the next two sample means will fall more than $2\sigma \overline{x}$from the target mean $\mu \overline{x}$?

b. What’s the probability that the first sample mean that is greater than $\mu \overline{x}+2\sigma \overline{x}$is the one from the fourth sample taken? Plant managers are trying to develop a criterion for determining when the process is not working properly. One idea they have is to look at the $5$most recent sample means. If at least $4$of the $5$fall outside the interval $(\mu \overline{x}-\sigma \overline{x},\mu \overline{x}+\sigma \overline{x})$, they will conclude that the process isn’t working.

c. Find the probability that at least $4$ of the $5$ most recent sample means fall outside the interval, assuming the process is working properly. Is this a reasonable criterion? Explain your reasoning.

Short answer

Part a. $P(X\ge 1)=0.0975$

Part b. $P(Firstgreaterthan2\sigma +\mu onfourthsample)=0.0232$

Part c. It is a reasonable criteria.

$P(X\le 1)=0.039007$

Step-by-step solution

Part a. Step 1. Given information

It is given that random samples vary normally around the target mean with a standard deviation.

Part a. Step 2. Explanation

The$68-95-99.7$tells us that$95\%$of the sample is within$2$standard deviation of the mean and thus$5\%$of the samples mean is more than$2$standard deviation of the mean.

By multiplication rule we know that,

$P(AandB)=P\left(A\right)\times P\left(B\right)$

Let X be the number of sample means that are more than$2$standard deviation of the mean.

Thus,

$P(X=0)=P{(within2s\tan darddeviationfromthemean)}^2=0.{95}^2=0.9025$

Now, by the complement rule we have,

$P(notA)=1-P\left(A\right)$

We can then determine the probability of having at least one of the two sample means more than standard deviation of the mean. This is,

$$\begin{gathered} P(X\ge 1)=1-P(X=0) \\ =1-0.9025 \\ =0.0925 \end{gathered}$$

Part b. Step 1. Explanation

Since the normal distribution is symmetric about its mean, $2.5\%$of the sample mean is greater than $2\sigma +\mu$

By multiplication rule we know that,

$P(AandB)=P\left(A\right)\times P\left(B\right)$

If the fourth sample is the first sample with a mean greater than $2\sigma +\mu$, then the three previous sample had a mean less than $2\sigma +\mu$

$$\begin{gathered} P(firstgreaterthan2\sigma +\mu onfourthsample) \\ =P{(lessthan2\sigma +\mu )}^3\times P(greaterthan2\sigma +\mu ) \\ =0.{975}^3\times 0.025 \\ =0.0232 \end{gathered}$$

Part c. Step 1. Explanation

By multiplication rule we know that,

$P(AandB)=P\left(A\right)\times P\left(B\right)$

Let X be the number of sample means in $(\mu +\sigma ,\mu -\sigma )$out of the five sample means.

$$\begin{gathered} P(X=0)=P(Samplemeannotin{(\mu +\sigma ,\mu -\sigma ))}^5=0.{32}^5=0.003355 \\ P(X=1)=5\times P(Samplemeannotin{(\mu +\sigma ,\mu -\sigma ))}^4\times P(Samplemeanin(\mu +\sigma ,\mu -\sigma \left)\right) \\ =5\times 0.{32}^4\times 0.68 \\ =0.035652 \end{gathered}$$

Thus, we have,

$$\begin{gathered} P(X\le 1)=0.003355+0.035652 \\ =0.039007 \\ =3.9007\% \end{gathered}$$

Since the probability is less than$5\%$ this event is unlikely to happen by chance and thus is a reasonable criteria.