Question 1061750: Generation Y has been defined as those individuals who were born between 1981 and
1991. According to the Project on Student Debt, Generation Y students graduating from
college averaged $23,200 in debt in 2009. Assume the standard deviation for debt is
$7,500 per student. A random sample of 30 service stations were selected.
a. Calculate the standard error of the mean.
b. What is the probability that the sample mean will be less than $24,000.
c. What is the probability that the sample mean will be more than $21,725.
d. What is the probability that the sample mean will be between $21,000 and $26,000?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The SEM is sd/sqrt (30)=$1369.30
The probability the sample mean will be fewer than $24000
If this is population,and given that sd is known or at least assumed, then z=(xbar-mu)/$1369.30
=-800/$1369.30
0.2795
more than $21,725: z>-2275/$1369.30=0.9517
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between $21,000 and 26,000
z between these two values is -3000/$1369.30 and 2000/$1369.30
that is a z between -2.19 and 1.46 and a probability of 0.9136
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