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?

Step-by-step 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:

Where and

We are interested on this probability

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

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

And we can find this probability on this way:

the answer is 54 sq. ft. i just did this question, on odyssey ware, and it was right. make brainliest.

4+4+2x

or

2x+8

step-by-step explanation: