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 6 percent and a standard deviation of 1.3 percent. A single analyst is randomly selected. Find the probability that his/her forecast is
(a) At least 3.8 percent.

(b) At most 8 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)

(c) Between 3.8 percent and 8 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)

Answer 1

Answer:

a) 0.954

b) 0.937

c) 0.891  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  6 percent

Standard Deviation, σ = 1.3 percent

We are given that the distribution of particular interest rate is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(At least 3.8 percent.)

$P(x \geq 3.8)$

$P( x \geq 3.8) = P( z \geq \displaystyle\frac{3.8 - 6}{1.3}) = P(z > -1.69)$

$= 1 - P(z < -1.69)$

Calculation the value from standard normal z table, we have,  

$P(x \geq 3.8) = 1 - 0.046 = 0.954 = 95.4\%$

b) P(At most 8 percent)

$P(x \leq 8) = P(z \leq \displaystyle\frac{8-6}{1.3}) = P(z \leq 1.53)$

Calculating the value from the standard normal table we have,

$P( x \leq 8) =0.937= 93.7\%$

c) P(Between 3.8 percent and 8 percent. )

$P(3.8 \leq x \leq 8) = P(-1.69 \leq z \leq 1.53)\\\\= P(z \leq 1.53) - P(z < -1.69)\\= 0.937 - 0.046 = 0.891 = 89.1\%$

$P(3.8 \leq x \leq 8) = 89.1\%$