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

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 .

The calculation of mean is as follows,

The variance of a single roll of a die is,

The variance of the mean of the random variables is,

Substitute the value.

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

Thus, the correct option is (a).

Learn more about what standard deviation is here:

link

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

and

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,

= The number which comes up  on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

=

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

and,

=

                                =

                                =

                                15.166667

so, =

                                    = 2.91667

   and =

Now we get that,

We get that are iid RV's ∀ j = 1(1)20

Let,

      So, we get that

                                                                 =  for any i = 1(1)10

                                                                 = 3.5

and,

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

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

and

y=x+9

?

step-by-step explanation:

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)