Chapter 10.2, Problem 73E

(a)

To determine

To find out what is the probability that at least one of the next two sample means will fall more than $2{\sigma }_x$ from the target mean ${\mu }_x$ .

Answer to Problem 73E

  $P(X\ge 1)=0.0975$ .

Explanation of Solution

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

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,

  $P(A\text{ and }B)=P(A)\times P(B)$

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

Thus,

  $\begin{array}{l}$ P(X=0)=P{(\text{within 2 standard deviation from the mean)}}^2$={0.95}^2\\ =0.9025\end{array}$

Now, by the complement rule we have,

  $P(\text{ not }A)=1-P(A)$

We can then determine the probability of having at least one of the two sample means more than $2$ standard deviation of the mean. This is,

  $\begin{array}{l}P(X\ge 1)=1-P(X=0)\\ =1-0.9025\\ =0.0975\end{array}$

(b)

To determine

To calculate the probability that the first sample mean that is greater than $2{\sigma }_x+{\mu }_x$ is the one from the fourth sample taken.

Answer to Problem 73E

  $P(\text{ First greater than }2\sigma +\mu \text{ on fourth sample)}=0.0232$

Explanation of Solution

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

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.

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,

  $P(A\text{ and }B)=P(A)\times P(B)$

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{array}{l}P(\text{ First greater than }2\sigma +\mu \text{ on fourth sample)}\\ $ =P{(\text{less than }2\sigma +\mu )}^3\times P(\text{greater than }2\sigma +\mu )$={0.975}^3\times 0.025\\ =0.0232\end{array}$

  $\begin{array}{l}P(\text{ First greater than }2\sigma +\mu \text{ on fourth sample)}\\ $ =P{(\text{less than }2\sigma +\mu )}^3\times P(\text{greater than }2\sigma +\mu )$={0.975}^3\times 0.025\\ =0.0232\end{array}$

(c)

To determine

To find the probability that at least $4\text{ of the 5}$ most recent sample means falls outside the interval assuming the process is working properly and explain is this a reasonable criteria.

Answer to Problem 73E

It is a reasonable criteria.

  $P(X\le 1)=0.039007$ .

Explanation of Solution

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

The $68-95-99.7$ tells us that $68\%$ of the sample mean is within one standard deviation of the mean and thus $32\%$ of the sample mean is more than one standard deviations of the mean.

By multiplication rule we know that,

  $P(A\text{ and }B)=P(A)\times P(B)$

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

  $\begin{array}{l}$ P(X=0)=P{(\text{Sample mean not in }(\mu +\sigma ,\mu -\sigma ))}^5$={0.32}^5\\ =0.003355\end{array}$

  $\begin{array}{l}$ P(X=1)=5\times P{(\text{Sample mean not in }(\mu +\sigma ,\mu -\sigma ))}^4\times P(\text{Sample mean in }(\mu +\sigma ,\mu -\sigma ))$=5\times {0.32}^4\times 0.68\\ =0.035652\end{array}$

Thus, we have,

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

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