Question

If outliers are discarded, then the retirement savings by residents of Econistan is normally distributed with a mean of $100,000 and a standard deviation of $20,000. a) What is the probability that one randomly selected resident of Econistan will have retirement savings greater than $117,000? Please include a picture and calculations in your answer.

Answer 1

Answer:

$P(X>117000)=P(\frac{X-\mu}{\sigma}>\frac{117000-\mu}{\sigma})=P(Z>\frac{117000-100000}{20000})=P(z>0.85)$

And we can find this probability using the complement rule and the normal standard table or excel:

$P(z>0.85)=1-P(z$

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:

$X \sim N(100000,20000)$  

Where $\mu=100000$ and $\sigma=20000$

We are interested on this probability

$P(X>117000)$

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

$z=\frac{x-\mu}{\sigma}$

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

$P(X>117000)=P(\frac{X-\mu}{\sigma}>\frac{117000-\mu}{\sigma})=P(Z>\frac{117000-100000}{20000})=P(z>0.85)$

And we can find this probability using the complement rule and the normal standard table or excel:

$P(z>0.85)=1-P(z$

The firgure attached illustrate the problem