Suppose in the population, the Anger-Out score for men is two points higher than it is for women. The population variances for men and women are both 20. Assume the Anger-Out scores for both genders are normally distributed. Given this information about the population parameters: (a) What is the mean of the sampling distribution of the difference between means?
(b) What is the standard error of the difference between means?
(c) What is the probability that you would have gotten this mean difference or less in your sample?

a)= 2

b) 6.324

c) P= 0.1217

Step-by-step explanation:

a) The mean of the sampling distribution of X`1- X`2 denoted by ux`-x` = u1-u2 is equal to the difference between population means i.e = 2 ( given in the question)

b) The standard deviation of the sampling distribution of X`1- X`2 ( standard error of X`1- X`2) denoted by σ_X`1- X`2 is given by

σ_X`1- X`2 = √σ²/n1 +σ²/n2

Var ( X`1- X`2) = Var X`1 + Var X`2 =  σ²/n1 +σ²/n2

so

σ_X`1- X`2 =√20 +20 = 6.324

if the populations are normal the sampling distribution X`1- X`2 , regardless of sample sizes , will be normal with mean u1-u2 and variance σ²/n1 +σ²/n2.

Where as Z is normally distributed with mean zero and unit variance.

If we take X`1- X`2= 0 and u1-u2= 2  and standard deviation of the sampling distribution = 6.324 then

Z= 0-2/ 6.342= -0.31625

P(-0.31625<z<0)= 0.1217

The probability would be 0.1217

welll its a very hard question but ill

step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this