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 ?
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

Answer 1

Answer:

option B

Step-by-step explanation:

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

$E (x) = \frac{6+1}{2} =\frac{7}{2} =3.5\\\\E(x)=3.5$

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

$\bar x_1 =\frac{ \sum ^{10}_{i=1}xi}{10}$

$E(\bar x_1)=\frac{ \sum ^{10}_{i=1}E(x_1)}{10} \\\\=\frac{10(3.5)}{10} \\\\=3.5\\\\E(\bar x_1) = 3.5----(1)$

now suppose ,

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

average of last 5 roll is

$E(\bar x_2)= \frac{ \sum ^{5}_{i=1}xi}{5} \\\\= \frac{5 \times(3.5)}{5} \\\\=3.5$

Therefore,$\bar x_1 - \bar x_2$

3.5 - 3.5 = 0