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 worker output per hour from the previous quarter is more than 0.6 standard deviations below the mean? .0228 Incorrect: Your answer is incorrect. Question 3. What is the probability that the percent change in worker output from the previous quarter is between -1.715 and 7.12? Use the normal model mentioned at the beginning of question 2.

Answer 1

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:

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

And using this formula we got:

$P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)$

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:

$P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)$

And we can find this probability with this difference:

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

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".  

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 $\sigma=3.1$

First part

And for this case we want this probability:

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

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

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

And using this formula we got:

$P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)$

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:

$P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)$

And we can find this probability with this difference:

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