Question

According to a recent current population reports, self-employed individuals in the United States work an average of 45 hours per week, with a standard deviation of 15. If this variable is approximately normally distributed, what proportion averaged more than 60 hours per week?

Answer 1

`P(X>60)=P(Z>1)=1-P(Z<1)=1-0.841=0.159`

Explanation:

1) 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".

2) Solution to the problem

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

`X \sim N(45,15)`

Where
`\mu=45` and
`\sigma=15`

We are interested on this probability

`P(X>60)`

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>60)=P((X-\mu)/(\sigma)>(60-\mu)/(\sigma))=P(Z>(60-45)/(15))=P(Z>1)`

And we can find this probability on this way:

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