8.2: Applications of Normal Random Variables

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When working with random variables we adopt the following vocabulary.

Definition:

We say that a certain population has or follows a normal distribution with parameters $\mu$ and $\sigma$ when the following is true:

If we do the census, find the mean $\mu$ and standard deviation $\sigma$, and plot the histogram, then the shape of the distribution is a normal probability density curve with parameters $\mu$ and $\sigma$.
If we let $X$ be a randomly selected observation from such population then $X \sim N(\mu,\sigma)$.

Therefore, $P(a<X<b)$, the probability that $X$ is between $a$ and $b$ can be interpreted in the following two ways.

First, we can literally interpret it as the probability that a randomly selected value is between $a$ and $b$.
Alternatively, the same quantity represents the proportion of the population that has a value between $a$ and $b$.

Similarly, $P(X<c)$, the probability that $X$ is less than $c$ can be interpreted in the following two ways.

First, we can literally interpret it as the probability that a randomly selected value is less than $c$.
Alternatively, the same quantity represents the proportion of the population that has a value less than $c$.

Finally, $P(X>c)$, the probability that $X$ is greater than $c$ can be interpreted in the following two ways.

First, we can literally interpret it as the probability that a randomly selected value is greater than $c$.
Alternatively, the same quantity represents the proportion of the population that has a value greater than $c$.

Example $\PageIndex{1}$

Consider the heights of the males in the U.S. We may assume that the population of heights has a normal distribution with $\mu=175$ cm and $\sigma=15$ cm.

Find the proportion of the population that is shorter than $178$ cm.
Find the probability that a randomly selected US male has the height between $170$ cm and $180$ cm.
Find the $75$-th percentile of the population.

Solution

Let's introduce a random variable $X$ to represent the height of a randomly selected male, then

$X \sim N(175,15)$

\(P(X<178)=P(\frac{X-\mu}{\sigma}<\frac{178-175}{15})=P(Z<0.20)=0.579

$clipboard_e6a38171875a1548b12b2d527c12377c0.png$

$P(170<X<180)=P(\frac{170-175}{15}<\frac{X-\mu}{\sigma}<\frac{180-175}{15})=P(-0.33<Z<0.20)=P(Z<0.20)-P(Z<-0.33)=0.6293-0.3707=0.2586$

$clipboard_ef6d41dda0dfffae4e10490202cbe9b27.png$

$x_{0.25}=\mu+z_{0.25}\sigma=175+0.67\cdot15=185.05$

$clipboard_e9a69d3c1836841f9915e5c54f8e9e350.png$

Example $\PageIndex{2}$

As reported in Runner’s World magazine, the times of the finishers in the New York City 10-km run are normally distributed with mean $61$ minutes and standard deviation $9$ minutes.

Find the probability that a randomly selected runner finished in more than $55$ minutes?
What proportion of the runners finished in between $60$ and $70$ minutes?
What is the $10$-th percentile of the finishing times?

Solution

Let's introduce a random variable $X$ to represent the time of a randomly selected marathon finisher, then

$X \sim N(61,9)$

\(P(X>55)=P(\frac{X-\mu}{\sigma}>\frac{55-61}{9})=P(Z>-0.67)=0.7486

$clipboard_e5aae904cc3909ef928a7482bb433ee08.png$

$P(60<X<70)=P(\frac{60-61}{9}<\frac{X-\mu}{\sigma}<\frac{70-61}{9})=P(-0.11<Z<1)=P(Z<1)-P(Z<-0.11)=0.8413-0.4562=0.3851$

$clipboard_eea948f5a4379128bcca864502d0ebe9d.png$

$x_{0.90}=\mu+z_{0.90}\sigma=61+1.28\cdot9=72.52$

$clipboard_e36914ff1dcd8f6110472edd4ae366c06.png$

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Solution

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Definition:

Example \(\PageIndex{1}\)

Solution

Example \(\PageIndex{2}\)

Solution