Question

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?

Answer 1

`P(X>4)=P((X-\mu)/(\sigma)>(4-\mu)/(\sigma))=P(Z>(4-4.4)/(0.4))=P(z>-1)`

And we can find this probability with the complement rule:

`P(z>-1)=1-P(z<1) = 1-0.841= 0.159`

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:

`X \sim N(4.4,0.4)`

Where
`\mu=4.4` and
`\sigma=0.4`

We are interested on this probability

`P(X>4)`

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>4)=P((X-\mu)/(\sigma)>(4-\mu)/(\sigma))=P(Z>(4-4.4)/(0.4))=P(z>-1)`

And we can find this probability with the complement rule:

`P(z>-1)=1-P(z<1) = 1-0.841= 0.159`