Answer to Question #312726 in Statistics and Probability for kyu

Question

1. Consider the set of even single-digit numbers {0, 2, 4, 6, 8}.

a. Make a list of possible sample size of 2 that can be taken from this sets of numbers.

b. Construct the sampling distribution of the sample means for the size of 3 and the standard variation.

2. In the numbers {1, 3, 5, 7, 9} construct the following:

a. List of possible sample size of 3 that can be taken from this sets of numbers.

b. Sampling distribution of the sample means for the size of 3 and the standard variation.

3. A sample size of 36 is to be selected from a population that has a mean of µ = 45 and standard deviation s of 10.

a. Find the mean of the sampling distribution.

b. Find the standard variation of the sampling distribution.

c. What is the probability that this sample mean will be between 40 and 50?

Answer 1

1. (a) possible samples

$N=5~~~n=2$

Sample no

$=C_5^2= \binom{5}{2}=10$ samples

$(0,2)(0,4)(0,6)(0,8)(2,4)(2,6)(2,8)(4,6)(4,8)(6,8)$

(b) Sampling distribution of the sample means for the size of 3 and the standard variation.

$N=5~~~n=3$

sample no

$=C_5^3= \binom{5}{3}=10$ samples

$\def\arraystretch{1.5}\begin{array}{c:c:c}no.&samples&means\\\hline1.&0,2,4&2.00\\\hdashline2.&0,2,6&2.67\\\hdashline3.&0,2,8&3.33\\\hdashline4.&0,4,6&3.33\\\hdashline5.&0,4,8&4.00\\\hdashline6.&0,6,8&4.67\\\hdashline7.&2,4,6&4.00\\\hdashline8.&2,4,8&4.67\\\hdashline9.&2,6,8&5.33\\\hdashline10.&4,6,8&6.00\\\hline\end{array}$

$\def\arraystretch{1.5}\begin{array}{c:c:c:c:c}\bar X&f&f(\bar X)&\bar X f(\bar X)&(\bar X)^2f(\bar X)\\\hline2.00&1&1/10&0.200&0.400\\\hdashline2.67&1&1/10&0.267&0.711\\\hdashline3.33&2&1/5&0.67&2.222\\\hdashline4.00&2&1/5&0.800&3.200\\\hdashline4.67&2&1/5&0.933&4.356\\\hdashline5.33&1&1/10&0.533&2.844\\\hdashline6.00&1&1/10&0.60&3.600\\\hline\sum &10&1&4.00&17.33\\\hline\end{array}$

Standard variation

$\sigma=\sqrt{\sum (\bar X)^2f(\bar X)-(\sum \bar X f(\bar X)^2}$

$\sigma = \sqrt {17.33-4^2}$

$\sigma =1.153$

2. (a) Possible samples

$N=5~~~n=3$

Sample no

$C_5^3= \binom{5}{3}=10$ samples

$(1,3,5)(1,3,7)(1,3,9)(1,5,7)(1,5,9)(1,7,9)(3,5,7)(3,5,9)(3,7,9)(5,7,9)$

(b) Sampling distribution of the sample means for the size of 3 and the standard variation.

$\def\arraystretch{1.5}\begin{array}{c:c:c}no.&samples&means\\\hline1&1,3,5&3.00\\\hdashline2.&1,3,7&3.67\\\hdashline3.&1,3,9&4.33\\\hdashline4.&1,5,7&4.33\\\hdashline5.&1,5,9&5.00\\\hdashline6.&1,7,9&5.67\\\hdashline7.&3,5,7&5.00\\\hdashline8.&3,5,9&5.67\\\hdashline9.&3,7,9&6.33\\\hdashline10.&5,7,9&7.00\\\hline\end{array}$

$\def\arraystretch{1.5}\begin{array}{c:c:c:c:c}\bar X&f&f(\bar X)&\bar X f(\bar X)&(\bar X)^2f(\bar X)\\\hline3.00&1&1/10&0.300&0.900\\\hdashline3.67&1&1/10&0.367&1.346\\\hdashline4.33&2&1/5&0.867&3.750\\\hdashline5.00&2&1/5&1.000&5.000\\\hdashline5.67&2&1/5&1.134&6.430\\\hdashline6.33&1&1/10&0.633&4.00\\\hdashline7.00&1&1/10&0.700&4.900\\\hline\sum &10&1&5.00&26.327\\\hline\end{array}$

Standard variation

$\sigma=\sqrt{\sum (\bar X)^2f(\bar X)-(\sum \bar X f(\bar X)^2}$

$\sigma= \sqrt {26.327-5^2}$

$\sigma=1.152$

3. $N=5~~\mu=45~~\sigma=10$

(a) sample mean is equal to the population mean

$\overline{X}=\mu=45$

(b) Standard deviation

$\sigma_{\bar x}=\dfrac{\sigma}{\sqrt n}=\dfrac{10}{\sqrt{36}}$

$=1.667$