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 5.0 percent and a standard deviation of 1.2 percent. A single analyst is randomly selected. Find the probability that his/her forecast is(a) At least 3.5 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)(b) At most 6 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)(c) Between 3.5 percent and 6 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)

Answer 1

a)
`P(X>3.5)=P((X-\mu)/(\sigma)>(3.5-\mu)/(\sigma))=P(Z>(3.5-5)/(1.2))=P(z>-1.25)`

And we can find this probability using the complement rule:

`P(z>-1.25)=1-P(z<-1.25)= 1-0.1057=0.8944`

b)
`P(X<6)=P((X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P(Z<(6-5)/(1.2))=P(z<0.83)`

And we can find this probability uing the normal standard table:

`P(z<0.83)=0.7967`

c)
`P(3.5<X<6)=P((3.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P((3.5-5)/(1.2)<Z<(6-5)/(1.2))=P(-1.25<z<0.83)`

And we can find this probability with this difference:

`P(-1.25<z<0.83)=P(z<0.83)-P(z<-1.25)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(-1.25<z<0.83)=P(z<0.83)-P(z<-1.25)=0.7967-0.1057= 0.6910`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

`X \sim N(5,1.2)`

Where
`\mu=5` and
`\sigma=1.2`

We are interested on this probability

`P(X>3.5)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X>3.5)=P((X-\mu)/(\sigma)>(3.5-\mu)/(\sigma))=P(Z>(3.5-5)/(1.2))=P(z>-1.25)`

And we can find this probability using the complement rule:

`P(z>-1.25)=1-P(z<-1.25)= 1-0.1057=0.8944`

Part b

We are interested on this probability

`P(X<6)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<6)=P((X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P(Z<(6-5)/(1.2))=P(z<0.83)`

And we can find this probability uing the normal standard table:

`P(z<0.83)=0.7967`

Part c

`P(3.5<X<6)=P((3.5-\mu)/(\sigma)<(X-\mu)/(\sigma)<(6-\mu)/(\sigma))=P((3.5-5)/(1.2)<Z<(6-5)/(1.2))=P(-1.25<z<0.83)`

And we can find this probability with this difference:

`P(-1.25<z<0.83)=P(z<0.83)-P(z<-1.25)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(-1.25<z<0.83)=P(z<0.83)-P(z<-1.25)=0.7967-0.1057= 0.6910`