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

(a) At least 3.9 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)

(b) At most 7 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)

(c) Between 3.9 percent and 7 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)

Answer 1

Solution :

Given that ,

mean = = 6% = 0.06

standard deviation = = 1.7% = 0.017

( a )

3.9% = 0.039

P(x 0.039 ) = 1 - P(x   0.039)

= 1 - P[(x - ) / ( 0.039 - 0.06 ) / 0.017 ]

= 1 -  P(z -1.24 )   

  Using z table,

= 1 - 0.1075

= 0.8925

Probability = 0.8925

( b )

7% = 0.07

P(x 0.07 )

= P[(x - ) / ( 0.07 - 0.06 ) / 0.017 ]

= P(z 0.59 )

Using z table,

= 0.7224

Probability = 0.7224

( c )

3.9% = 0.039 and 7% = 0.07

P( 0.039 < x < 0.07 )

= P[( 0.039 - 0.06 ) / 0.017 ) < (x - ) /  < ( 0.07 - 0.06) / 0.017 ) ]

= P( -1.24 < z < 0.59 )

= P(z < 0.59 ) - P(z < -1.24 )

Using z table,

= 0.7224 - 0.1075

= 0.6149

Probability = 0.6149