SOLUTION: . In a certain Caribbean country, the average monthly income per family is US$5,600 with a standard deviation of US$3,200. If a random sample of 100 families is selected, calcu

Algebra ->  Probability-and-statistics -> SOLUTION: . In a certain Caribbean country, the average monthly income per family is US$5,600 with a standard deviation of US$3,200. If a random sample of 100 families is selected, calcu      Log On


   

Question 912513: . In a certain Caribbean country, the average monthly income per
family is US$5,600 with a standard deviation of US$3,200. If a
random sample of 100 families is selected, calculate the probability
that the sample mean monthly income is:
(a)between US$4,000 and US$6,000
(b) less than US$5000
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
m = 5600, sd = 3200
z+=blue+%28x+-+mu%29%2Fblue%28sigma%2Fsqrt%28n%29%29
a) P(4000 < x < 6000) = normalcdf(4000,6000,5600,320) Using TI
0r
P(4000 < x < 6000) = P( z < 400/320) - P(z < -1400/320)
b)P(x < 5000) = P(z < 600/320)
compute z, use table etc to find Probability