Question

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice were rolled, and the average of the 10 numbers appearing faceup was recorded. The process was repeated 20 times. Which of the following best describes the distribution being simulate?
(A) A sampling distribution of a sample mean with n = 10, mu_x = 3.5, and sigma_x = 0.54
(B) A sampling distribution of a sample mean with n = 10, mu_x = 3.5, and sigma-x = 1.71
(C) A sampling distribution of a sample mean with n = 20, mu_x = 3 5, and sigma_x = 0.38
(D) A sampling distribution of a sample proportion with n = 10, mu_p = 1/6, and sigma_p = 0.118
(E) A sampling distribution of a sample proportion with n = 20, mu_p = 1/6, and sigma_p = 0.083

Answer 1

Answer:

C) a sample distribution of a sample mean with n = 10  

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up  on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

                            = 3.5       ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

                                = $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

                                = $\frac {91}{6}$

                                $\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

                                    $\simeq 15.166667 - 3.5^{2}$

                                    = 2.91667

   and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

 $Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

      So, we get that $E({\overline}{Y}) = E(Y_{j})$

                                                                 = $E(X_{ij}$  for any i = 1(1)10

                                                                 = 3.5

and,

       $\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10  

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$