Mathematics, 21.01.2020 05:31 gracemartin4535

Asimulation 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

Answers: 1

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Answers

We can describes the distribution being simulate by finding the mean and standard deviation. The standard deviation is the square root of variance.

The correct option is (a).

Given:

The faces were numbered 1 through 6.

The average number appearing is n=10 .

The calculation of mean is as follows,

$\mu_{\overline{x}}=\dfrac {1+2+3+4+5+6}{6}\\\mu_{\overline{x}}=3.5$

The variance of a single roll of a die is,

$\sigma_x^2=\sum_{i=1}^6 \frac16 \left(i-3.5 \right)^2=\dfrac{35}{12}$

The variance of the mean of the random variables is,

$Var(\overline X)=\dfrac{ \Large{\sigma_{x}^2}}n$

Substitute the value.

$Var(\overline X)=\dfrac{\left(\dfrac{35}{12}\right)}{10}\\Var(\overline X)=0.291$

As we know that standard deviation is the square root of variance.

$\sigma_{\bar X}=\sqrt{Var(\overline X)} \\=\sqrt{0.291}\\\sigma_{\bar{x}}\approx 0.54$

Thus, the correct option is (a).

Learn more about what standard deviation is here:

link

Answer from: mhwalters

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

Answer from: Quest

10 glasses of juice could be made with 5 carrots, 1,500 orange juice, 2,000 papaya juice.

Answer from: Quest

345

step-by-step explanation:

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Asimulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. r...