Chapter 10.2, Problem 75E

(a)

To determine

To find: the probability that greater than two standard deviations from the target mean would fall from the at least of the next two sample means.

Answer to Problem 75E

  $P\left(X\ge 1\right)=0.0975$

Explanation of Solution

Given:

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

Formula used:

Rule of Multiplication:

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

Rule of Complement

  $P\left(\text{not }A\right)=1-P\left(A\right)$

Calculation:

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 (within $2\sigma$ from the mean) = 95% = 0.95

P (greater than $2\sigma$ from the mean) = 5% =0.05

Rule of Multiplication:

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

Suppose that X is the number of samples, which implies that the mean is more than $2\sigma$ . $\begin{array}{c}P\left(X=0\right)=P{\left(\text{within }2\sigma \text{ from the mean}\right)}^2\\ ={0.95}^2\\ =0.9025\end{array}$

Rule of Complement

  $P\left(\text{not }A\right)=1-P\left(A\right)$

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

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

(b)

To determine

To find: the probability that the mean of the 1^st sample is greater than $\mu +2\sigma$ is that taken from the 4^th sample.

Answer to Problem 75E

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

Explanation of Solution

Given:

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

Formula used:

Multiplication rule:

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

Calculation:

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.

  $\begin{array}{l}P\left(\text{larger than }\mu \text{+2}\sigma \right)=2.5\%=0.025\\ P\left(\text{less than }\mu \text{+2}\sigma \right)=97.5\%=0.975\end{array}$

Multiplication rule:

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

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

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

  $\begin{array}{c}P\left(\text{1st larger than }\mu \text{+2}\sigma \text{ on 4th sample}\right)=P{\left(\text{less than }\mu \text{+2}\sigma \right)}^3\times P\left(\text{larger than }\mu \text{+2}\sigma \right)\\ ={0.975}^3\times 0.025=0.0232\end{array}$

(c)

To determine

To explain: they will assume that it is not working in the process. That this is a reasonable criterion justifies the response with a sufficient probability.

Answer to Problem 75E

Reasonable criterion

Explanation of Solution

Given:

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

Formula used:

Multiplication rule:

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

Calculation:

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.

  $\begin{array}{l}\text{Sample mean in }\left(\mu -\sigma ,\mu \text{+}\sigma \right)\\ \text{Probability}=68\%=0.68\\ \text{Sample mean not in }\left(\mu -\sigma ,\mu \text{+}\sigma \right)\\ \text{Probability}=32\%=0.32\end{array}$

Multiplication rule:

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

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

  $\begin{array}{c}\text{Sample mean not in }\left(\mu -\sigma ,\mu \text{+}\sigma \right)\\ P\left(X=0\right)=P{\left(\text{Sample mean not in }\left(\mu -\sigma ,\mu \text{+}\sigma \right)\right)}^5={0.32}^5=0.003355\\ \text{Sample mean not in }\left(\mu -\sigma ,\mu \text{+}\sigma \right)\\ P\left(X=0\right)=5\times P{\left(\text{Sample mean not in }\left(\mu -\sigma ,\mu \text{+}\sigma \right)\right)}^4\times P\left(\text{Sample mean not in }\left(\mu -\sigma ,\mu \text{+}\sigma \right)\right)\\ =5\times {0.32}^4\times 0.68=0.035652\end{array}$

Add the associating probability

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

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