Question

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 ?.

Answer 1

The sampling distribution of the difference in sample means x⁻₁ - x⁻₂ is; 3.5 - 3.5 = 0

Difference in sample means

We are told that;

• A fair six-sided die, with sides numbered 1 through 6, will be rolled a total of 15 times.

• x⁻₁ represents the average of the first ten rolls.

• x⁻₂ represents the average of the remaining five rolls.

Now, the average of the largest and lowest numbers of the six sided die is;

E(x) = (6 + 1)/2 = 3.5

Thus, the average mean of the first ten rows is expressed as;

E(x⁻₁) = (n * E(x))/n

E(x⁻₁) = (10 * 3.5)/10

E(x⁻₁) = 3.5

The average mean of the last five rolls will be;

E(x⁻₂) = (n * E(x))/n

E(x⁻₂) = (5 * 3.5)/5

E(x⁻₂) = 3.5

Thus;

x⁻₁ - x⁻₂ = 3.5 - 3.5 = 0

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