Question

A consensus forecastis the average of a large number of individual analysts’ forecasts. Suppose the individual forecasts fora particular interest rate are normally distributed with a mean of 5.0 percent and a standard deviation of 1.2 percent Asingle analyst is randomly selected. Find the probability that his/ her forecast isa At least 3.5 percentb At most 6 percentc Between 3.5 percent and 6 percent

Answer 1

Answer:

a) $P(X\geq 3.5)=P(\frac{X-\mu}{\sigma}\geq \frac{3.5-\mu}{\sigma})=P(Z\geq \frac{3.5-5.0}{1.2})=P(Z\geq -1.25)$

And we can find this probability using the z table or excel

$P(Z \geq <-1.25)=1-P(Z<-1.25)=1-0.10565=0.894$

b) $P(X\leq 6.0)=P(\frac{X-\mu}{\sigma}\leq \frac{6.0-\mu}{\sigma})=P(Z\leq \frac{6.0-5.0}{1.2})=P(Z\leq 0.833)$

And we can find this probability using the z table or excel:

$P(Z \leq <0.833)=0.798$

c) $P(3.5<X<6)=P(\frac{3.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{3.5-5.0}{1.2}<Z<\frac{6-5.0}{1.2})=P(-1.25<z<0.833)$

And we can find this probability on this way:

$P(-1.25<Z<0.833)=P(Z<0.833)-P(Z<-1.25)$

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

$P(-1.25<z<0.833)=P(Z<0.833)-P(Z<-1.25)=0.798-0.106=0.692$

Step-by-step 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 interest rate of a population, and for this case we know the distribution for X is given by:

$X \sim N(5.0,1.2)$  

Where $\mu=5.0$ and $\sigma=1.2$

We are interested on this probability

$P(X\geq 3.5)$

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

$z=\frac{x-\mu}{\sigma}$

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

$P(X\geq 3.5)=P(\frac{X-\mu}{\sigma}\geq \frac{3.5-\mu}{\sigma})=P(Z\geq \frac{3.5-5.0}{1.2})=P(Z\geq -1.25)$

And we can find this probability using the z table or excel

$P(Z \geq <-1.25)=1-P(Z<-1.25)=1-0.10565=0.894$

Part b

We are interested on this probability

$P(X\leq 6.0)$

$P(X\leq 6.0)=P(\frac{X-\mu}{\sigma}\leq \frac{6.0-\mu}{\sigma})=P(Z\leq \frac{6.0-5.0}{1.2})=P(Z\leq 0.833)$

And we can find this probability using the z table or excel:

$P(Z \leq <0.833)=0.798$

Part c

$P(3.5<X<6)=P(\frac{3.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{3.5-5.0}{1.2}<Z<\frac{6-5.0}{1.2})=P(-1.25<z<0.833)$

And we can find this probability on this way:

$P(-1.25<Z<0.833)=P(Z<0.833)-P(Z<-1.25)$

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

$P(-1.25<z<0.833)=P(Z<0.833)-P(Z<-1.25)=0.798-0.106=0.692$