Question

Suppose the age of people in a certain population are distributed normally with a mean of 37.5 years and standard deviation of 6.2 years. What is the probability of randomly selecting a person who is over 45 years old given that they are older than 40.

Answer 1

Answer:

$P(X>45 | X>40)= \frac{P(X>45 \cap X>40)}{P(X>40)}= \frac{P(X>45)}{P(X>40)}= \frac{0.113}{0.343}= 0.329$

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 age of a population, and for this case we know the distribution for X is given by:

$X \sim N(37.5,6.2)$  

Where $\mu=37.5$ and $\sigma=6.2$

We are interested on this probability:

$P(X>45 | X>40)= \frac{P(X>45 \cap X>40)}{P(X>40)}= \frac{P(X>45)}{P(X>40)}$

We can begin finding $P(X>40)$ using the z score formula given by:

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

Using this formula we have:

$P(X>40)= P(Z>\frac{40-37.5}{6.2}) = P(Z>0.403)$

And using the complement rule and the normal standard table or excel we have this:

$P(Z>0.403)=1-P(Z<0.403)= 1-0.657= 0.343$

Now we can find $P(X>45)$ using the z score formula given by:

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

Using this formula we have:

$P(X>45)= P(Z>\frac{45-37.5}{6.2}) = P(Z>1.210)$

And using the complement rule and the normal standard table or excel we have this:

$P(Z>1.210)=1-P(Z<1.210)= 1-0.887= 0.113$

And replacing into our original probability we got:

$P(X>45 | X>40)= \frac{P(X>45 \cap X>40)}{P(X>40)}= \frac{P(X>45)}{P(X>40)}= \frac{0.113}{0.343}= 0.329$