Question

Suppose a random sample of n measurements is selected from a population with $u=100$mean and variance role="math" localid="1657967387987" ${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\mathbf{100}$. For each of the following values of n, give the mean and standard deviation of the sampling distribution of the sample mean.

1. role="math" localid="1657967260825" $\mathit{n}\mathbf{=}\mathbf{4}$
   
2. $\mathit{n}\mathbf{=}\mathbf{25}$
   
3. $\mathit{n}\mathbf{=}\mathbf{10}\mathbf{0}$
   
4. $\mathit{n}\mathbf{=}\mathbf{50}$
   
5. $\mathit{n}\mathbf{=}\mathbf{500}$
   
6. $\mathit{n}\mathbf{=}\mathbf{1000}$
   

Short answer

$$\begin{gathered} a.\mu =100,\&=5 \\ b.\mu =100,\&=2 \\ c.\mu =100,\&=1 \\ d.\mu =100,\&=1.41 \\ e.\mu =100,\&=0.447 \\ f.\mu =100,\&=0.316 \end{gathered}$$

Step-by-step solution

Given information

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

Calculating the mean and the standard deviation of the sampling distribution of the sample mean  x forn=4

1)

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

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

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

Therefore, mean of sampling distribution of the sample mean is

$$\begin{gathered} {\mu }_x=\mu \\ =100 \end{gathered}$$

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{10}{\sqrt{4}} \\ =5 \end{gathered}$$

Hence,localid="1657969487969" $\mu =100,\&=5$

Calculating the mean and the standard deviation of the sampling distribution of the sample mean x¯ for n=25

2)

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

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

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

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

$$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =100 \end{gathered}$$

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{10}{\sqrt{25}} \\ =2 \end{gathered}$$

Hence,$\mu =100\&=2$$\mu =100,\&=2$

Calculating the mean and the standard deviation of the sampling distribution of the sample mean x¯ for n=100

3)

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

localid="1661429991977" ${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$and localid="1661429999797" ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

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

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

$$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =100 \end{gathered}$$

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{10}{\sqrt{100}} \\ =1 \end{gathered}$$

Hence, $\mu =100,\&=1$

Calculating the mean and the standard deviation of the sampling distribution of the sample mean x¯ for n=50

4)

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

localid="1661430016184" ${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$and localid="1661430023389" ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

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

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

$$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =100 \end{gathered}$$

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{10}{\sqrt{50}} \\ =\frac{10}{7.07} \\ =1.41 \end{gathered}$$

Hence, $\mu =100,\&=1.41$

Calculating the mean and the standard deviation of the sampling distribution of the sample mean x¯ for n=500

5)

According to properties of the Sampling distribution of $\overline{\mathbf{x}}$localid="1661430049695" ${\mathit{\mu }}_{\overline{\mathbf{x}}}\mathbf{=}\mathit{\mu }$

and localid="1661430042149" ${\mathit{\sigma }}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

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

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

localid="1657972123948" $$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =100 \end{gathered}$$

Standard deviation of sampling distribution of the sample mean is

localid="1657972184826" $$\begin{gathered} {\sigma }_{\overline{x}}=\frac{10}{\sqrt{500}} \\ =\frac{10}{22.36} \\ =0.447 \end{gathered}$$

Hence, $\mu =100,\&=0.447$

Calculating the mean and the standard deviation of the sampling distribution of the sample mean  x¯ for n=1000 

6)

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

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

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

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

$$\begin{gathered} {}_{\overline{x}}=\mu \\ =100 \end{gathered}$$

Standard deviation of sampling distribution of the sample mean is

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{10}{\sqrt{500}} \\ =\frac{10}{31.62} \\ =0.316 \end{gathered}$$

Hence,$\mu =100,\&=0.316$