Answer to Question #313305 in Statistics and Probability for Wondering2

Question

Student , Height , X²

Hazel Kim , 1.47 , 2.16

Mee Ann Mae , 1.58 , 2.50

Grace Ann , 1.67 , 2.79

Jethro , 1.69 , 2.86

Chelsea Mae , 1.51 , 2.28

TOTAL: 9.92 , 12.59

1. Solve for the mean of the population μ.

2. Solve for the mean of the sampling distribution of the sample means Hr.

3. Compare u and ur.

4. Solve for the variance (a) and the standard deviation (a) of the population.

5. Solve the variance (or) and the standard deviation (ar) of the sampling distribution of the sample means Hr.

6. Compare a and or.

Answer 1

$\mu=1.47+1.58+1.67+1.69+1.51/5=1.58$

2.$\mu(X^2)=2.16+2.5+2.79+2.86+2.28=2.518$

$\sqrt{\mu(X^2)}=1.59$

3. $\mu<\mu(X^2)$

This is due to rounding error.

4.Variance$S^2=\sqrt{\frac{(1.58-1.47)^2+(1.58-1.58)^2+(1.58-1.67)^2+(1.58-1.69)^2+(1.58-1.51)^2}{4}}=\sqrt{\frac{0.0121+0+0.0081+0.0121+0.0049}{4}}=0.096$

Deviation

$\sigma=\sqrt{S^2}=0.31$

5.$S^2(or)=\sqrt{\frac{(2.52-2.16)^2+(2.52-2.5)^2+(2.52-2.79)^2+(2.52-2.86)^2+(2.52-2.28)^2}{4}}=\sqrt{\frac{0.1296+0.0004+0.0729+0.1156+0.0576}{4}}=0.31$

$\sigma(ar)=\sqrt{or}=0.56$

6. ar is a square root of a, as x and x².

$\sqrt{0.31}=0.56$