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3 years ago

Find the mean, median, and mode for each set of data: {8,2,6,8,4,2,3,4,8}

Mathematics

2 answers:

Akimi4 [234]3 years ago

8 0

Mode: 8

2, 2, 3, 4, 4, 6, 8, 8

Median: 4

Mean: 4.625 (37/8)

Tcecarenko [31]3 years ago

3 0

In order- 2,2,3,4,4,6,8,8,8

answer-
mean = 5. add all numbers divide by how many numbers there are

median = 4. put in order smallest - largest, cross one off each side until there is one left

mode = 8. whatever number appears the most

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A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

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

6 0

3 years ago

Someone be as kind as to offer some help?

RideAnS [48]

Answer:

x = 4 and y = 0

Step-by-step explanation:

Using the elimination method:

4(-x + 5y=-4) --> -4x + 20y = -16.

Now we add -4x + 20y = 016 and 4x + 3y = 16. This gives us 23y = 0 or just y =0.

Now we can find x by plugging in y.

-4x + 0 = -16.

Divide by -4 on both sides -> x = 4.

If you want to double check you can plug your values into -4x + 20y = -16.

-4(4) +20(0) = -16.

4 0

3 years ago

Please really need help on this please

Vika [28.1K]

0,1 0,7 I think I use to do these

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3 years ago

Guys please help with this question thanks

erastova [34]

= 16 + 49 - 3(11) - 4(10)
= 16 + 49 - 33 - 40
= -8

3 0

3 years ago

The answer to this question HELP ME PLS

Papessa [141]

The answer is a because 552.50 x .12 is 66.30. 66.30 minus 552.50 is 486.50 and that divided by 7 is 69.45

8 0

3 years ago