91Ó°ÊÓ

$$\begin{gathered} \mathrm{μ}=100 \\ \mathrm{σ}=15 \end{gathered}$$

Formula used:

$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}$

The z-score is

$$\begin{gathered} =\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{105-100}{15} \\ =0.33 \end{gathered}$$

The associatively probability using table

$$\begin{gathered} P(Xâ‰\yen 105)=P(z>0.33) \\ =P(z<-0.33) \\ =0.3707 \end{gathered}$$

$$\begin{gathered} \backslash mu\&=100\backslash \backslash \\ \backslash sigma\&=15\backslash \backslash \\ n\&=60 \end{gathered}$$

Formula used:

${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

The mean of the sampling distribution of the sample mean $\stackrel{¯}{x}$is

${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}=\frac{15}{\sqrt{60}}=1.9365$

$$\begin{gathered} \mathrm{μ}=100 \\ \mathrm{σ}=15 \\ n=60 \end{gathered}$$

Formula used:

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

The mean of the sampling distribution of the sample mean $\stackrel{¯}{x}$is same to the population standard deviation divided by the square root of the sample size

${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}=100$

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

${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}=\frac{15}{\sqrt{60}}=1.9365$

The z-score is

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{105-100}{1.9365} \\ =2.58 \end{gathered}$$

The corresponding probability using table

$$\begin{gathered} P(\stackrel{¯}{x}â‰\yen 105)=P(z>2.58) \\ =P(z<-2.58) \\ =0.0049 \end{gathered}$$

The answer in (a) could be very different, because it is supposed that the population distribution was normal to find the probability.

The answer in (b) and (c) will not be very different, the reason is that the sample size is 30 or more and, by the central limit theorem, know that the sampling distribution is about normal even if the population distribution is not normal.