SOLUTION: Please help..I am not understanding this..I am taking an online course and I am lost with this 5. Find the probabilities. a.From National Weather Service records, the a

Question 321910: Please help..I am not understanding this..I am taking an online course and I am lost with this
5. Find the probabilities.

a.From National Weather Service records, the annual snowfall in the TopKick Mountains has a mean of 92 inches and a standard deviation  of 12 inches. If the snowfall from 36 randomly selected years are chosen, what it the probability that the snowfall would be less than 95 inches?

b.The loan officer rates applicants for credit. Ratings are normally distributed. The mean is 240 and the standard deviation is 60. If 49 applicants are randomly chosen, what is the probability that they will have a rating between 230 and 260? Round z scores to two decimal places.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a.From National Weather Service records, the annual snowfall in the TopKick Mountains has a mean of 92 inches and a standard deviation  of 12 inches. If the snowfall from 36 randomly selected years are chosen, what is the probability that the snowfall would be less than 95 inches?
----------------
t(95) = (95-92)/[12/sqrt(36)] = 3/4
---
P(x-bar < 95) = P(t < 0.75 when df = 35) = 0.7709
=========================================================
b.The loan officer rates applicants for credit. Ratings are normally distributed. The mean is 240 and the standard deviation is 60. If 49 applicants are randomly chosen, what is the probability that they will have a rating between 230 and 260? Round z scores to two decimal places.
----------------
t(230) = (230-240)/[60/sqrt(49)] = -1.1667
----
t(260) = (260-240)/[60/sqrt(49)] = 2.3333
---------------
P(230<= x-bar <= 260) = P(-1.1667 <= t <= 2.3333 with df=48)
= 0.8635
==================
Cheers,
Stan H.

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