Chapter 7.3, Problem 67E

(a)

To determine

To Explain: the event which is more likely

Answer to Problem 67E

Arbitrary chosen 1 square yard of material and finding 2 or more flaws

Explanation of Solution

The standard deviation of the sampling distribution of the sample mean is equal to the population standard deviation divide by the square root of sample size.

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

This then means that the standard deviation decreases as the sample size increases and therefore the data values are closer to the expected value when using a larger sample size.

This then means that you are less likely to find a mean of 2 or more flaws when the sample size is larger and therefore event with the smaller sample size is more likely to occur.

(b)

To determine

To Calculate: the probability of the first event in part (a)

Answer to Problem 67E

Shape of the population distribution is unknown and the sample is small

Explanation of Solution

  $n=1$

Centre limit theorem: if the sample size is equal to 30 more than that then the sampling distribution of the sample mean $\overline{x}$ is about normal.

Since the sample size of 1 is less than 30, it cannot be used the central limit theorem. In this case, the sampling distribution of the sample mean has the same shape as the population distribution.

Although, the population distribution is unknown and therefore and not knowing the shape of the sampling distribution of the sample mean either, which means that it cannot be find the probability.

(c)

To determine

To Calculate: the probability of the second event in part (a)

Answer to Problem 67E

0.81%

Explanation of Solution

Given:

  $\begin{array}{l}\mu =16\\ \sigma =1.2\\ n=50\\ \overline{x}=2\end{array}$

Formula used:

Calculation:

Since the sample is large because the sample size is more than 30 and the central limit theorem says that the sampling distribution of the sample mean is about normal.

The Z-score is

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{2-1.6}{1.2/\sqrt{50}}=2.36$

The associating probability using the normal probability table

  $P\left(Z<2.36\right)$ is given in the row starting with $.01$ and in the column starting with $.06$ of the standard normal probability table

  $\begin{array}{c}P\left(\overline{X}\ge 2\right)=P\left(Z>2.36\right)\\ =1-P\left(Z<2.36\right)\\ =1-0.9209\\ =0.0081\\ =0.81\%\end{array}$