Question

These data can be approximated quite well by a N(3.4, 3.1) model. Economists become alarmed when productivity decreases. According to the normal model what is the probability that the percent change in

asked Mar 9, 2021 69.5k views

1 Answer

Ask a Question

Gelunox · Answer 1 · 2021-03-14T00:49:08+0000

Answer:

First part

P(X< 3.4-0.6*3.1) = P(X<1.54)

And for this case we can use the z score formula given by:

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

And using this formula we got:

P(X<1.54) = P(Z<(1.54 -3.4)/(3.1))= P(Z<-0.6)

And we can use the normal standard table or excel and we got:

P(Z<-0.6) = 0.274

Second part

For the other part of the question we want to find the following probability:

P(-1.715 <X< 7.12)

And using the score we got:

P(-1.715 <X< 7.12)=P((-1.715-3.4)/(3.1) < Z< (7.15-3.4)/(3.1)) = P(-1.65< Z< 1.210)

And we can find this probability with this difference:

P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837

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".

Solution to the problem

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

$X \sim N(3.4,3.1)$

Where
$\mu=3.4$ and
$\sigma=3.1$