Question #164791

Consider all samples of size 5 from this population. 

2,5,6,8,10,12,13

1. Compute the mean and the standard deviation of the population. 

2. List all samples of size 5 and compute the mean for each sample. 

3. Construct the sampling distribution of the sample means.

4. Calculate the mean of the sampling distribution of the sample means. 

Compare this to mean of the population. 

5. Calculate the standard deviation of the sampling distribution of the 

sample means . Compare this to the standard deviation of the 

population. 


Expert's answer

1. Compute the mean and the standard deviation of the population. 

E[X]=(2+5+6+8+10+12+13)/7=8E[X]=(2+5+6+8+10+12+13)/7=8

E[X2]=(22+52+62+82+102+122+132)/7=542/7=7737E[X^2 ]=(2^2+5^2+6^2+8^2+10^2+12^2+13^2 )/7=542/7=77\frac{3}{7}

Var[X]=E[X2]E[X]2=773764=1337Var[X]=E[X^2 ]-E[X]^2=77 \frac{3}{7}-64=13\frac{3}{7}

σ(X)=Var[X]=94/7=3.6645\sigma(X)=\sqrt{Var[X] }=\sqrt{94/7}=3.6645


2. List all samples of size 5 and compute the mean for each sample. 



3. Construct the sampling distribution of the sample means.

P(Xˉ5=6.2)=P(Xˉ5=6.6)=P(Xˉ5=6.8)=P(Xˉ5=7.0)=P(Xˉ5=7.2)=P(Xˉ5=7.4)=P(Xˉ5=7.8)=P(Xˉ5=8.0)=P(Xˉ5=8.6)=P(Xˉ5=8.8)=P(Xˉ5=9.0)=P(Xˉ5=9.2)=P(Xˉ5=9.6)=P(Xˉ5=9.8)=1/21P(\bar X_5=6.2)= P(\bar X_5=6.6)=P(\bar X_5=6.8)=P(\bar X_5=7.0)=P(\bar X_5=7.2)=P(\bar X_5=7.4)=P(\bar X_5=7.8)=P(\bar X_5=8.0)=P(\bar X_5=8.6)=P(\bar X_5=8.8)=P(\bar X_5=9.0)=P(\bar X_5=9.2)=P(\bar X_5=9.6)=P(\bar X_5=9.8)=1/21

P(Xˉ5=8.2)=P(Xˉ5=8.4)=2/21P(\bar X_5=8.2)=P(\bar X_5=8.4)=2/21

P(Xˉ5=7.6)=3/21P(\bar X_5=7.6)=3/21


4. Calculate the mean of the sampling distribution of the sample means. Compare this to mean of the population. 

E[Xˉ5]=(6.2+6.6+6.8+7.0+7.2+7.4+7.8+8.0+8.6+8.8+9.0+9.2+9.6+9.8+28.2+28.4+37.6)/21=8.0=E[X]E[\bar X_5 ]=(6.2+6.6+6.8+7.0+7.2+7.4+7.8+8.0+8.6+8.8+9.0+9.2+9.6+9.8+2\cdot 8.2+2\cdot 8.4+3\cdot 7.6)/21=8.0=E[X]

5. Calculate the standard deviation of the sampling distribution of the sample means . Compare this to the standard deviation of the population.

E[Xˉ52]=(6.22+6.62+6.82+7.02+7.22+7.42+7.82+8.02+8.62+8.82+9.02+9.22+9.62+9.82+28.22+28.42+37.62)/21=1362.8/21=6494105E[\bar X_5^2] =(6.2^2+6.6^2+6.8^2+ 7.0^2+7.2^2+7.4^2+7.8^2+8.0^2+8.6^2+ 8.8^2+9.0^2+9.2^2+9.6^2 +9.8^2+2⋅8.2^2+2⋅8.4^2 +3⋅7.6^2)/21=1362.8/21=64\frac{94}{105}

Var[Xˉ5]=E[Xˉ52]E[Xˉ5]2=649410564=94105Var[\bar X_5 ]=E[\bar X_5^2 ]-E[\bar X_5 ]^2=64\frac{94}{105}-64=\frac{94}{105}

σ(Xˉ5)=Var[Xˉ5]=94/105=σ(X)/15=0.946\sigma(\bar X_5 )=\sqrt{Var[\bar X_5 ]}=\sqrt{94/105}=\sigma(X)/\sqrt{15}=0.946


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