91Ó°ÊÓ

Given,

$$\begin{gathered} \mathrm{μ}=0.5 \\ \mathrm{σ}=0.7 \\ n=50 \end{gathered}$$

The central limit theorems say that the sampling distribution of the sample means is about normal, when the sample is sufficiently large. When a sample size of 10 or more is used, it is considered to be appropriately large.

By the central limit theorem, the sampling distribution of the sample means $\stackrel{¯}{x}$ is nearly normal in this example because the sample size is 50, which is greater than 30.

Given,

Formula used:

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \end{gathered}$$

The mean of the sampling distribution of the sample mean $\stackrel{¯}{x}$is equal to the population mean:

${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}=0.5$

The sampling distribution of the sample mean's standard deviation $\stackrel{¯}{x}$is

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{0.7}{\sqrt{50}} \\ =0.0990 \end{gathered}$$

The sample size is 30 or more. We know that the sampling distribution of the sample mean$\stackrel{¯}{x}$is approximately normal because of the central limit theorem.

The z-score is a measure of how well something works.

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0.6-0.5}{0.0990} \\ =1.01 \end{gathered}$$

The associative probability using table

$$\begin{gathered} P(\stackrel{¯}{x}â‰\yen 0.6)=P(z>1.01) \\ =P(Z<-1.01) \\ =0.1562 \end{gathered}$$

Given,

$$\begin{gathered} \mathrm{μ}=0.5 \\ \mathrm{σ}=0.7 \\ n=50 \end{gathered}$$

No

Since this probability is not insignificant, it is possible that the sample mean number of months is thus large just by coincidence.