SOLUTION: A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a particular interest rate are normally distributed wi
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Question 1198619: A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a particular interest rate are normally distributed with a mean of 4 percent and a standard deviation of 1.0 percent. A single analyst is randomly selected. Find the probability that his/her forecast is
(a) At least 3.3 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)
(b) At most 9 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)
(c) Between 3.3 percent and 9 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)
z-score formula is z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation.
in your problem:
m = 4
s = 1
when x = 3.3, the formula becomes:
z = (3.3 - 4) / 1 = -.7
area to the left of z = -.7 = .2419635782.
area to the right of z = -.7 = 1 minus .2419635782 = .7580364128.
when x = 9, the formula becomes:
z = (9 - 4) / 1 = 5
area to the left of z = 5 = .9999997129.
area to the right of z = 5 = 1 - .9999997129 = .000000287195.
when 3.3 < x < 9, area between those 2 x-values is area to the left of z-score of 9 minus area to the left of z-score of -.7 = .9999997129 minus .2419635782 = .7580361347.
here's what those values look like on a graph.
if i did this correctly, your solutions are:
(a) At least 3.3 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)
.7580
(b) At most 9 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)
.0000 = 0
(c) Between 3.3 percent and 9 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)
