A counselor records the number of disagreements (per session) among couples during group counseling sessions. If the number of disagreements is distributed normally as 4.4 ± 0.4 ( M ± SD) disagreements, then what proportion of couples disagree at least four times during each counseling session?

Respuesta :

Answer:

[tex]P(X>4)=P(\frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma})=P(Z>\frac{4-4.4}{0.4})=P(z>-1)[/tex]

And we can find this probability with the complement rule:

[tex]P(z>-1)=1-P(z<1) = 1-0.841= 0.159[/tex]

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 disagreements of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(4.4,0.4)[/tex]  

Where [tex]\mu=4.4[/tex] and [tex]\sigma=0.4[/tex]

We are interested on this probability

[tex]P(X>4)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X>4)=P(\frac{X-\mu}{\sigma}>\frac{4-\mu}{\sigma})=P(Z>\frac{4-4.4}{0.4})=P(z>-1)[/tex]

And we can find this probability with the complement rule:

[tex]P(z>-1)=1-P(z<1) = 1-0.841= 0.159[/tex]