Answer:
[tex]P(X>117000)=P(\frac{X-\mu}{\sigma}>\frac{117000-\mu}{\sigma})=P(Z>\frac{117000-100000}{20000})=P(z>0.85)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel:
[tex]P(z>0.85)=1-P(z<0.85)=1-0.802=0.198[/tex]
The firgure attached illustrate the problem
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 retirement savings of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(100000,20000)[/tex]
Where [tex]\mu=100000[/tex] and [tex]\sigma=20000[/tex]
We are interested on this probability
[tex]P(X>117000)[/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>117000)=P(\frac{X-\mu}{\sigma}>\frac{117000-\mu}{\sigma})=P(Z>\frac{117000-100000}{20000})=P(z>0.85)[/tex]
And we can find this probability using the complement rule and the normal standard table or excel:
[tex]P(z>0.85)=1-P(z<0.85)=1-0.802=0.198[/tex]
The firgure attached illustrate the problem
