A consensus forecast is the average of a large number of individual analysts\' f

Question

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 1.7 percent. A single analyst is randomly selected. Find the probability that his/her forecast is

a. At least 3.2 percent. (Round your answer to 4 decimal places.)

b. At most 9 percent. (Round your answer to 4 decimal places.)

c. Between 3.2 percent and 9 percent. (Round your answer to 4 decimal places.)

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 4
standard Deviation ( sd )= 1.7
a.
P(X < 3.2) = (3.2-4)/1.7
= -0.8/1.7= -0.4706
= P ( Z <-0.4706) From Standard Normal Table
= 0.319
P(X > = 3.2) = (1 - P(X < 3.2)
= 1 - 0.319 = 0.681
b.
P(X > 9) = (9-4)/1.7
= 5/1.7 = 2.9412
= P ( Z >2.9412) From Standard Normal Table
= 0.0016,
P(X < = 9) = (1 - P(X > 9)
= 1 - 0.0016 = 0.9984
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 3.2) = (3.2-4)/1.7
= -0.8/1.7 = -0.4706
= P ( Z <-0.4706) From Standard Normal Table
= 0.319
P(X < 9) = (9-4)/1.7
= 5/1.7 = 2.9412
= P ( Z <2.9412) From Standard Normal Table
= 0.9984
P(3.2 < X < 9) = 0.9984-0.319 = 0.6794

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