91Ó°ÊÓ

It is given that random samples of n measurements are selected from a population with mean $\mathrm{μ}=100$and variance${\mathrm{σ}}^2=100$

1)

According to properties of the Sampling distribution of $\overline{\mathbf{x}}$${\mathit{μ}}_{\mathbf{x}}\mathbf{=}\mathit{μ}$

and${\mathit{σ}}_{\overline{\mathbf{x}}\mathbf{=}}\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}$

But, mean is 100 and standard deviation is$\sqrt{100}$i.e.$\mathrm{σ}=10$

Therefore, mean of sampling distribution of the sample mean is

$$\begin{gathered} {\mathrm{μ}}_x=\mathrm{μ} \\ =100 \end{gathered}$$

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{10}{\sqrt{4}} \\ =5 \end{gathered}$$

Hence,localid="1657969487969" $\mathrm{μ}=100,\&=5$

2)

According to properties of the Sampling distribution of $\stackrel{\mathbf{¯}}{\mathbf{x}}{\mathit{μ}}_{\mathbf{x}}\mathbf{=}\mathit{μ}$

and ${\mathit{σ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}$

But, mean is 100 and standard deviation is $\sqrt{100}$i.e.$\mathrm{σ}=10$

Therefore, the mean of sampling distribution of the sample mean is

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

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{10}{\sqrt{25}} \\ =2 \end{gathered}$$

Hence,$\mathrm{μ}=100\&=2$$\mathrm{μ}=100,\&=2$

According to properties of the Sampling distribution of $\stackrel{\mathbf{¯}}{\mathbf{x}}$

localid="1661429991977" ${\mathit{μ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\mathit{μ}$and localid="1661429999797" ${\mathit{σ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}$

But, mean is 100 and standard deviation is $\sqrt{100}$i.e.$\mathrm{σ}=10$

Therefore, the mean of sampling distribution of the sample mean is

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

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{10}{\sqrt{100}} \\ =1 \end{gathered}$$

Hence, $\mathrm{μ}=100,\&=1$

According to properties of the Sampling distribution of $\stackrel{\mathbf{¯}}{\mathbf{x}}$

localid="1661430016184" ${\mathit{μ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\mathit{μ}$and localid="1661430023389" ${\mathit{σ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}$

But, mean is 100 and standard deviation is $\sqrt{100}$i.e.$\mathrm{σ}=10$

Therefore, the mean of sampling distribution of the sample mean is

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

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{10}{\sqrt{50}} \\ =\frac{10}{7.07} \\ =1.41 \end{gathered}$$

Hence, $\mathrm{μ}=100,\&=1.41$

According to properties of the Sampling distribution of $\stackrel{\mathbf{¯}}{\mathbf{x}}$localid="1661430049695" ${\mathit{μ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\mathit{μ}$

and localid="1661430042149" ${\mathit{σ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}$

But, mean is 100 and standard deviation is $\sqrt{100}$i.e.$\mathrm{σ}=10$

Therefore, the mean of sampling distribution of the sample mean is

localid="1657972123948" $$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ} \\ =100 \end{gathered}$$

Standard deviation of sampling distribution of the sample mean is

localid="1657972184826" $$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{10}{\sqrt{500}} \\ =\frac{10}{22.36} \\ =0.447 \end{gathered}$$

Hence, $\mathrm{μ}=100,\&=0.447$

According to properties of the Sampling distribution of $\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{}{\mathit{μ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\mathit{μ}$

and${\mathit{σ}}_{\stackrel{\mathbf{¯}}{\mathbf{x}}}\mathbf{=}\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}$

But, mean is 100 and standard deviation is$\sqrt{100}$i.e.$\mathrm{σ}=10$

Therefore, the mean of sampling distribution of the sample mean is

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

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{10}{\sqrt{500}} \\ =\frac{10}{31.62} \\ =0.316 \end{gathered}$$

Hence,$\mathrm{μ}=100,\&=0.316$