Chapter 12, Problem 46PT4

a)

To determine

To explain shape, center and spread of the sampling distribution of sample mean.

Answer to Problem 46PT4

The sampling distribution of sample mean is approximately Normal distribution with mean 4 and standard deviation 0.04

Explanation of Solution

Given:

Population mean = $\mu$ = 4

Population standard deviation = $\sigma$ = 0.02

Sample size = n = 25

When the population has approximately Normal distribution then the sampling distribution of sample mean also has approximately Normal distribution.

The mean will be as original.

  ${\mu }_{\overline{x}}=4$

The standard deviation of the sampling distribution of sample mean is,

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.02}{\sqrt{25}}=0.004$

Hence, the sampling distribution of sample mean is approximately Normal distribution with mean 4 and standard deviation 0.04.

b)

To determine

To find $P(\overline{X}<3.99or\overline{X}>4.01)$ .

Answer to Problem 46PT4

  $P(\overline{X}<3.99or\overline{X}>4.01)=0.0124$

Explanation of Solution

Given:

The mean will be as original.

  ${\mu }_{\overline{x}}=4$

The standard deviation of the sampling distribution of sample mean is,

  ${\sigma }_{\overline{x}}=0.004$

Formula:

Z-score is,

  $Z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$

Assume that ${\overline{x}}_1=3.99$

Using z-score,

  $z=\frac{3.99-4}{0.004}=-2.50$

Assume that ${\overline{x}}_2=4.01$

Using z-score,

  $z=\frac{4.01-4}{0.004}=2.50$

Therefore,

  $\begin{array}{l}P(\overline{X}<3.99or\overline{X}>4.01)=P(Z<-2.50orZ>2.50)\\ P(\overline{X}<3.99or\overline{X}>4.01)=2\times P(Z<-2.50)\\ P(\overline{X}<3.99or\overline{X}>4.01)=2\times 0.0062=0.0124\end{array}$

Using excel formula, =NORMSDIST(-2.5)

Hence, $P(\overline{X}<3.99or\overline{X}>4.01)=0.0124$

c)

To determine

To find $P(4.00<\overline{X}<4.01)$ .

Answer to Problem 46PT4

  $P(4.00<\overline{X}<4.01)=0.4938$

Explanation of Solution

Given:

The mean will be as original.

  ${\mu }_{\overline{x}}=4$

The standard deviation of the sampling distribution of sample mean is,

  ${\sigma }_{\overline{x}}=0.004$

Formula:

Z-score is,

  $Z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$

Assume that ${\overline{x}}_1=4.00$

Using z-score,

  $z=\frac{4.00-4}{0.004}=0$

Assume that ${\overline{x}}_2=4.01$

Using z-score,

  $z=\frac{4.01-4}{0.004}=2.50$

Therefore,

  $\begin{array}{l}P(4.00<\overline{X}<4.01)=P(0<Z<2.50)\\ P(4.00<\overline{X}<4.01)=P(Z<2.50)-P(Z<0)\\ P(4.00<\overline{X}<4.01)=0.9938-0.5000\\ P(4.00<\overline{X}<4.01)=0.4938\end{array}$

Using excel formula, =NORMSDIST(2.5) and =NORMSDIST(0)

Hence, $P(4.00<\overline{X}<4.01)=0.4938$

d)

To determine

To find $P(X\ge 4)$

Answer to Problem 46PT4

The probability that below the upper boundary of 4.01 is 0.1799.

Explanation of Solution

Given:

The number of trials = n = 5

Probability of success = p = 0.4938

Formula:

Binomial probability is,

  $P(X=x)=C{}_x\times p^x\times {(1-p)}^{n-x}$

Using the formula of Binomial distribution,

  $\begin{array}{l}P(X\ge 4)=P(X=4)+P(X=5)\\ P(X\ge 4)=C{}_4\times {0.4938}^4\times {(}^1+C{}_5\times {0.4938}^5\times {(}^1\\ P(X\ge 4)=0.1505+0.0294\\ P(X\ge 4)=0.1799\end{array}$

Hence,

e)

To determine

To explain which results given more convincing evidence that the machine needs to be shut down.

Answer to Problem 46PT4

The probability of getting a single sample mean below 3.99 or above 4.01 is 0.0124

Explanation of Solution

Given:

Result 1: The probability of getting a single sample mean below 3.99 or above 4.01 is 0.0124

Result 2: The probability of having at least 4 of the 5 sample means of 5 consecutive samples between 4.00 and 4.01 is 0.1799.

The machine needs to be shut down if their probability is smaller. Therefore, the probability of getting a single sample mean below 3.99 or above 4.01 is 0.0124 which is small. Hence, this result is more convincing.

f)

To determine

To find the probability that the machine will be shut down when using the rule.

Answer to Problem 46PT4

The probability that the machine will be shut down when using the rule is 0.0145

Explanation of Solution

Given:

  $P(4.00<\overline{X}<4.01)=0.4938$

The possible rule will be “stop the production process if there are 6 consecutive sample means that are between 4.00 and 4.01”.

Therefore, the probability would be,

  ${[P(4.00<\overline{X}<4.01)]}^6={[0.4938]}^6=0.0145$