Question

Quality control $(2.2,5.3,6.3)$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 measurements from these random samples will vary Normally around the target mean $\mu$, with a standard deviation of $\sigma$. 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 $2\sigma$from the target mean $\mu$? Show your work.

(b) What's the probability that the first sample mean that is greater than $\mu +2\sigma$is the one from the fourth sample taken?

(c) 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$(\mu -\sigma ,\mu +\sigma )$, they will conclude that the process isn't working. Is this a reasonable criterion? Justify your answer with an appropriate probability.

Short answer

a). The probability that at least one of the next two sample means $P(X\ge 1)=0.0975$.

b). $P(1\text{st larger than}\mu +2\sigma \text{on 4th sample})=0.0232$.

c). A reasonable criterion.

Step-by-step solution

Part (a) Step 1: Given Information

Random sample varies normal around the target mean $\mu$ with standard deviation $\sigma$.

Part (a) Step 2: Explanation

Rule of Multiplication:

$P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right)$

Rule of Complement

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

The $68-95-99.7$-rule informs us that $95\%$ of the mean of the sample is within $2\sigma$ of the mean and therefore $5\%$ of the mean of the samples is greater than $2\sigma$ of the mean.

$P(within2\sigma fromthemean)=95\%=0.95$

$\mathrm{P}\left(\text{greater than}2\sigma \text{from the mean}\right)=5\%=0.05$

Part (a) Step 3: Explanation

Rule of Multiplication:

$P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right)$

Suppose that $\mathrm{X}$is the number of samples, which implies that the mean is more than $P(X=0)=P(\text{within}2\sigma \text{from the mean}{)}^2$

$=0.{95}^2$

$=0.9025$

Rule of Complement

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

It could then determine the chances of having more than $2\sigma$from the mean of minimum one of the two samples.

$P(X\ge 1)=1-P(X=0)$

$=1-0.9025$

$=0.0975$

Part (b) Step 1: Given Information

Random sample varies normal around the target mean $\mu$with standard deviation $\sigma$.

Part (b) Step 2: Explanation

Multiplication rule:

$P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right)$

The $68-95-99.7$-rule informs us that $95\%$of the mean of the sample is within $2\sigma$of the mean and therefore $5\%$of the mean of the samples is greater than $2\sigma$of the mean.

$P(\text{larger than}\mu +2\sigma )=2.5\%=0.025$$P(\text{less than}\mu +2\sigma )=97.5\%=0.975$

Multiplication rule:

$P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right)$

If the $4^{\text{th}}$sample is the $1^{\text{st}}$sample with a mean larger than $\mu +2\sigma$,

Then the $3$previous samples have a mean less than $\mu +2\sigma$.

localid="1650392399658" $P(1\text{st larger than}\mu +2\sigma \text{on 4th sample})=P(\text{less than}\mu +2\sigma {)}^3\times P(\text{larger than}\mu +2\sigma )$

localid="1650392417061" $$\begin{gathered} =0.{975}^3\times 0.025 \\ =0.0232 \end{gathered}$$

Part (c) Step 1: Given Information

Random sample varies normal around the target mean $\mu$ with standard deviation $\sigma$.

Part (c) Step 2: Explanation

Multiplication rule:

$P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right)$

The $68-95-99.7$-rule informs us that $95\%$of the mean of the sample is within $2\sigma$of the mean and therefore $5\%$of the mean of the samples is greater than $2\sigma$of the mean.

Sample mean in $(\mu -\sigma ,\mu +\sigma )$.

Probability $=68\%=0.68$.

Sample mean not in $(\mu -\sigma ,\mu +\sigma )$.

Probability $=32\%=0.32$.

Part (c) Step 3: Explanation

Multiplication rule:

$P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right)$

Suppose $\mathrm{X}$is the number of samples means in $(\mu -\sigma ,\mu +\sigma )$from the$5$-sample means.

Sample mean not in $(\mu -\sigma ,\mu +\sigma )$

$P(X=0)=P\left(\text{Sample mean not in}\right(\mu -\sigma ,\mu +\sigma ){)}^5=0.{32}^5=0.003355$

Sample mean not in $(\mu -\sigma ,\mu +\sigma )$.

$P(X=0)=5\times P\left(\text{Sample mean not in}\right(\mu -\sigma ,\mu +\sigma ){)}^4$

$\times P\left(\text{Sample mean not in}\right(\mu -\sigma ,\mu +\sigma \left)\right)$

$=5\times 0.{32}^4\times 0.68=0.035652$

Add the associating probability

$P(X\le 1)=0.003355+0.035652$

$=0.039007$

$=3.9007\%$

Since the probability is below $5$percent, it is unlikely that this occurrence will happen by chance and this is thus a reasonable criterion.