Question 862717
a) standard error of the mean = 38000 / square root(56) = 5078
b) normal
c) P(x > or = 124000) = 1 - P(x<124000)
we calculate the z score for P(x<124000)
124000 - 120000 / 5078 = 0.78771169751870815282 = 0.79
consult z table for probability assocated with z score = .79
P(x > or = 124000) = 1 - .79 = 0.21
d) P(x > or = 112000) = 1 - P(x<112000)
we calculate the z score for P(x<112000)
112000 - 120000 / 5078 = -1.58
consult -z table for probability assocated with z score = -1.58
P(x > 112000) = 1 - P(x<112000) = 1 - 0.06 = 0.94
e) P(x > 112000 and x < 124000) = P(x<124000) - P(x<112000) = 0.79 - 0.06 = 0.73