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A fair six-sided die, with sides numbered 1 through 6, will be rolled a total of 15 times. Let x¯1 represent the average of the first ten rolls, and let x¯2 represent the average of the remaining five rolls. What is the mean μ(x¯1−x¯2) of the sampling distribution of the difference in sample means x¯1−x¯2 ?

A. 3.5/10 - 3.5/5 =-.35
B. 3.5-3.5=0
C. 10-5=5
D. 10(3.5)-5(3.5)=17.5
E. 6(10-5)=30

Respuesta :

shallomisaiah19

Answer:

option B

Step-by-step explanation:

[tex]P(x =x) [{\frac{1}{6} ;x =1,2,3...6][/tex]

[tex]E (x) = \frac{6+1}{2} =\frac{7}{2} =3.5\\\\E(x)=3.5[/tex]

Given random experiment of tossing of 6 sided dice is follow above distribution

Therefore, suppose x₁, x₂ ...x₁₀ are 10 independent and indentical random variable which represent first 10 rolls

Average of first 10 rows equals

[tex]\bar x_1 =\frac{ \sum ^{10}_{i=1}xi}{10}[/tex]

[tex]E(\bar x_1)=\frac{ \sum ^{10}_{i=1}E(x_1)}{10} \\\\=\frac{10(3.5)}{10} \\\\=3.5\\\\E(\bar x_1) = 3.5----(1)[/tex]

now suppose ,

x₁₁,x₁₂, ...x₁₅ are 5 independent and identical random variable which represent last 5 roll

average of last 5 roll is

[tex]E(\bar x_2)= \frac{ \sum ^{5}_{i=1}xi}{5} \\\\= \frac{5 \times(3.5)}{5} \\\\=3.5[/tex]

Therefore,[tex]\bar x_1 - \bar x_2[/tex]

3.5 - 3.5 = 0

The mean [tex]\mu(\bar{x_1}-\bar{x_2})[/tex]  of the sampling distribution of the difference in sample means [tex](\bar{x_1}-\bar{x_2})[/tex] is 0 and this can be determined by using the formula of the average mean.

Given:

  • A fair six-sided die, with sides numbered 1 through 6, will be rolled a total of 15 times.
  • Let [tex]\bar {x_1}[/tex] represent the average of the first ten rolls, and let [tex]\bar {x_2}[/tex] represent the average of the remaining five rolls.

The average mean of the first ten rolls is given by the formula:

[tex]\rm E(\bar{x_1}) = \dfrac{\sum^{10}_{i=1}(x_1)}{10}[/tex]

Now, put the values of known terms in the above equation.

[tex]\rm E(\bar{x_1})=\dfrac{10\times 3.5}{10}[/tex]

[tex]\rm E(\bar{x_1})={3.5}[/tex]

The average mean of the remaining five rolls is given by the formula:

[tex]\rm E(\bar{x_2}) = \dfrac{\sum^{5}_{i=1}(x_i)}{5}[/tex]

Now, put the values of known terms in the above equation.

[tex]\rm E(\bar{x_1})=\dfrac{5\times 3.5}{5}[/tex]

[tex]\rm E(\bar{x_1})={3.5}[/tex]

Now, [tex](\bar{x_1}-\bar{x_2}) = 3.5 - 3.5 = 0[/tex].

The mean [tex]\mu(\bar{x_1}-\bar{x_2})[/tex]  of the sampling distribution of the difference in sample means [tex](\bar{x_1}-\bar{x_2})[/tex] is 0.

For more information, refer to the link given below:

https://brainly.com/question/16217700

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