Answer :
Answer:
[tex]P(X>60)=P(Z>1)=1-P(Z<1)=1-0.841=0.159[/tex]
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:
[tex]X \sim N(45,15)[/tex]
Where [tex]\mu=45[/tex] and [tex]\sigma=15[/tex]
We are interested on this probability
[tex]P(X>60)[/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>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-45}{15})=P(Z>1)[/tex]
And we can find this probability on this way:
[tex]P(Z>1)=1-P(Z<1)=1-0.841=0.159[/tex]