Continuous Random Variables

1. Learning Goals

Demonstrate understanding of characteristics of normal distributions.
Calculate accurate probabilities of continuous random variables and interpret them in a variety of settings.
Calculate the standardized value (or $z$-score).

2. Distribution of a Continuous Random Variable

The probability distribution of a continuous random variable $X$ is characterized by its probability density function $f(X)$ satisfying that the probability $P(a\leq X\leq b)$ equals the area above the interval $[a, b]$ but under the graph of the density function $f(X)$ which is also called a density curve.

3. Properties of Distribution of a Continuous Variable

The probability density function $f$ is nonnegative, that is $f(X)\ge 0$.
The total area under a density curve is 1.
The cumulative probability $P(X\le b)$ of a random variable $X$ equals the area under the density curve to the left side of $b$.
By the addition rule of probability, we have
- $$P(a\le X\le b)=P(X\le b)-P(X\le a)$$
- $$P(X\ge b)=1-P(X\le b)$$
As a line segment has no area, we have $P(X\le a)=P(X< a)$ as well as $P(X\ge b)=P(X>b)$

4. Example: An Uniform Distribution

Let $X$ be the amount of time that a commuter must wait for a train. Suppose $X$ has a probability density function $$ f(X)= \begin{cases} 0.1, & 0\leq X\leq 10\\ 0, & \text{otherwise} \end{cases} $$

What is the probability that the commuter’s waiting time is less than 4 minutes?

Solution: The probability $P(X\leq 4)$ is the area under the horizontal line $y=0.1$ to the left of $X=4$. Since $f(X)=0$ for $X<0$, the area is the area of the rectangle with width 4 and height 0.1. So the probability is $P(X\leq 4)=0.1\cdot 4=0.4$.

5. Normal Distribution

A normal distribution has a density function $f(x)=\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}},$ where $\mu$ is the mean, $\sigma$ is the standard deviation, $\pi\approx 3.14159$ and $e\approx 2.71828$. The graph of $f$ is called a normal curve.
We write $X\sim N(\mu, \sigma^2)$ for a normal random variable $X$ with the mean $\mu$ and the standard deviation $\sigma$.
A normal distribution has the following properties:
- The mean, median, and mode are equal.
- The normal curve is bell shaped and symmetric with respect to the mean.
- The total area under the curve and above the $x$-axis is $1$.
- The normal curve approaches, but never touches, the $x$-axis as $x$ goes to $\pm\infty$.

6. Normal Curves with Different Means and Standard Deviations

7. The Empirical Rule for Normal Distributions

For any normal distribution, the proportion of data values within 1, 2, and 3 standard deviations away from the mean are approximately 68.3%, 95.4% and 99.7% respectively.

8. Example: Foot length (1 of 3)

Suppose that foot length of a randomly chosen adult male is a normal random variable with the mean $\mu=11$ and the standard deviation $\sigma=1.5$.

How likely is a male’s foot length to be smaller than 9.5 inches
How likely is a male’s foot length to be bigger than 8 inches

Solution: Let’s first sketch the normal curve.

9. Example: Foot length (2 of 3)

Method 1: Using the Empirical Rule

Note that $9.5 = 11 - 1.5 = \mu - \sigma$. By the symmetry of normal curve, we know that the probability $P(X<9.5)$ is the shaded area on the left. Because the probability of getting a foot length within 1 standard deviation away from the mean is 0.683. Then $$P(X<9.5)=\frac12(1-P(9.5<X<12.5))\approx\frac12(1-0.683)=0.1585.$$ Note that $8=11-2\cdot 1.5=\mu-2\sigma$. Because the probability of getting a foot length within 2 standard deviation away from the mean is 0.954. Then $$P(X>8)=(1-P(X<8))=1-\frac12(1- P(8<X<14))=1-\frac12(1-0.954)=0.977.$$

10. Example: Foot length (3 of 3)

Method 2: Using Excel

The probability $P(X<x)$ for a normal random variable $X$ can be calculated using the Excel function NORM.DIST(x, mean, sd, TRUE).

In this case, $P(X<9.5)=$ =NORM.DIST(9.5, 11, 1.5, TRUE) $\approx 0.1587.$

Since $P(X>8) = 1 - P(X < 8)$ and $P(X < 8)=$ NORM.DIST(8, 11, 1.5, TRUE), it follows that $$P(X>8) \approx 1 - 0.02275 = 0.97725.$$

11. Standard Normal Distribution

A normal distribution is called a standard normal distribution if the mean is $\mu=0$ and the standard deviation is $\sigma=1$.
A random normal variable can be standardized by the following formula $z=\frac{x-\mu}{\sigma}.$ We call the value $z$ the $Z$-score of $x$. In Excel, the $Z$-score of $x$ can be calculated using the function STANDARDIZE().
Standardization preserves probability: $$\textstyle P(a<X<b)=P\left(\frac{a-\mu}{\sigma}< Z < \frac{b-\mu}{\sigma}\right).$$
The probability $P(Z< z)$ of a standard normal random variable $Z$ can be found using the Excel function NORM.S.DIST(z, TRUE) or the standard normal distribution table.
Recall that the probability $P(X< x)$ of a normal random variable $X$ can be calculated using the Excel function NORM.DIST(x, mean, sd, TRUE).

12. Example: Standardization Revisited

Let $X$ be a normal random variable with the mean $\mu = 8$ and the standard deviation $\sigma=2$.

Find the $Z$-score for the value $X=13$.
Find the $X$-value for the $Z$-score $z=-0.6$.

Solution: The $z$-score for the value $X=13$ is $$z=\dfrac{x-\mu}{\sigma}=\dfrac{13-8}{2}=\dfrac{5}{2}=2.5.$$

The $X$-value for the the $Z$-score $z=-0.6$ is $$x=z\cdot\sigma+\mu=-0.6\cdot 2+8=-1.2+8=6.8.$$

13. Example: Probabilities of Standard Normal Variables

Let $Z$ be a standard normal random variable.

Find $P(Z<1.21)$.
Find $P(Z\geq 1.21)$.
Find $P(0<Z\leq 1.21)$.

Solution: To find the probability of getting a value less than $z$, we may use the Excel function NORM.S.DIST(z,TRUE).

In this case, $P(Z<1.21)$ =NORM.S.DIST(1.21, TRUE) $\approx 0.8869$.
Since the total area under the normal curve is 1, $$P(Z\geq 1.21) = 1 - P(Z < 1.21) \approx 1-0.8869=0.1131.$$
By the symmetry, $P(Z<0)=0.5$ and $$P(0<Z<1.21) = P(Z < 1.21) - P(Z < 0) \approx 0.8869-0.5=0.3869.$$

14. Example: Heights of 25-year-old women

The heights of 25-year-old women in a certain region are approximately normally distributed with mean 62 inches and standard deviation 4 inches. Find the probability that a randomly selected 25-year-old woman is more than 67 inches tall.

Solution: Let $X$ be the height of a randomly selected 25-year-old. From the question, $X$ is approximately normally distributed with mean 62 inches and standard deviation 4 inches. Since $P(X>67)=1-P(X\leq 67)$, to find $P(X>67)$, we find $P(X\leq 67)$ first.

Method 1: Using NORM.DIST(x, mean, SD, TRUE)

$P(X\le 67)$ NORM.DIST(67, 62, 4, TRUE) $\approx 0.8944$.

Method 2: Using NORM.S.DIST(z, TRUE)

The $Z$-score is $z=\frac{67-62}{4}=1.25$ and $P(X\le 67)=P(Z<1.25)=$ NORM.S.DIST(1.25,TRUE) $\approx 0.8944$.

Therefore, $P(X>67) \approx 1-0.8944=0.1056.$

15. Cutoff Value for a Given Tail Area

The $k$-th percentile for a random variable $X$ is the value $x_k$ that cuts off a left tail with the area $k/100$, that is $P(X<x_k)=\frac{k}{100}$, where $0\leq k\leq 100$.
Let $c$ be a nonnegative number less than 0.5. The $(100c)$-th percentile for the standard normal distribution is usually denoted as $-z_c$, that is $P(Z<-z_c)=c<0.5$. By symmetry, $z_c$ is the value such that $P(Z> z_c)=c$, that is $P(Z<z_c)=1-c$.
- Note: the cutoff value $z_c$ is positive if $c<0.5$.
- In Excel, if $c<0.5$, the cutoff value $z_c$ such that $P(Z > z_c)=c$ can be calculated using the function NORM.S.INV(1-c).
For a normal random variable $X$ with the mean $\mu$ and standard deviation $\sigma$, the cutoff value $x^*$ with a tail area $p$, can be calculated using the Excel function NORM.INV(p, mean, sd), or the standardization formula, that is, $x^*=z^*\cdot \sigma+\mu,$ where $z^*$ is the cutoff $z$-score with the tail area $p$, that is $z^*=$ NORM.S.INV(p).

16. Example: Cutoff Value

Let $X$ be the normal random variable with mean $6$ and standard deviation $3$. Suppose the value $x^*$ cuts off a left-tail area $0.05$. Find the value $x^*$.

Solution: Since the $x^*$ cuts off a left-tail, to find the value $x^*$:

Method 1: $x^*=$ NORM.INV(0.05, 6, 3) $\approx 1.065.$

Method 2: Let $z^*$ be the cutoff $z$-score with the tail area $0.05$. Then $z^*=$ NORM.S.INV(0.05) $\approx -1.645.$ Then $$x^*=z^*\cdot \sigma+\mu=-1.645\cdot 3+6=1.065.$$

17. Example: Math Course Placement

Scores on a standardized college placement examination are normally distributed with mean 60 and standard deviation 13. Students whose scores are in the top $5\%$ will be placed in a Calculus II course. Find the minimum score needed to be placed in a Calculus II course.

Solution: Let $x^*$ be the minimum score. From the question, we know that $P(X\geq x^*)=5\%$. Equivalently, $P(X<x^*)=1 - 5\%$.

Method 1: $x^*=$ NORM.INV(1-5%, 60, 13) $\approx 81.38.$

Method 2: Let $z^*$ be the cutoff $z$-score with the tail area $1-5%$. Then $z^*=$ NORM.S.INV(1-5%) $\approx 1.64.$ Then $$x^*=z^*\cdot \sigma+\mu=1.64\cdot 13+60=81.32.$$

So the minimum score needed to be placed in a Calculus II course is $82$.

Note: The minimum score is the same as the 95th percentile.

Practice: Dash washing time

Practice: Fruit weight

Practice: Shortest lifespan

Practice: Battery life

Practice: Sum of Normal Random Variables

Let $Z$ be a normal random variable with $\mu=0$ and $\sigma=1$. Let $X$ be a normal random variable with $\mu=4.3$ and $\sigma=1.7$.

Determine the values $P(Z>1) + P(X<6)$ and explain how do you find the value.

Extra Practice

Practice: Area under a normal curve

Practice: Find cutoff $Z$-scores

Practice: Chocolate Chips in Acceptable Cookies

Practice: Blood Pressure

Lab Instructions in Excel

18. Excel Functions for Normal Distributions

Let $Z$ be a standard normal random varaible. In Excel, $P(Z<z)$ is given by NORM.S.DIST(z, TRUE).
Let $X$ be a normal random variable with mean $\mu$ and standard deviation $\sigma$, that is $X\sim N(\mu, \sigma^2)$. In Excel, $P(X<x)$ is given by NORM.DIST(x, mean, sd, TRUE).
When a cumulative probability $p=P(X<x)$ of a normal random variable $X$ is given, we can find $x$ using NORM.INV(p, mean, sd).
When a cumulative probability $p=P(Z<z)$ of a standard normal random variable $Z$ is given, we can find $z$ using NORM.S.INV(p).

19. Lab Practice: Normal Variable

Let $Z$ be a standard normal random variable. Find the probabilities: $$\text{1.}\,\, P(Z<1.58)\quad \text{2.}\,\, P(-0.6<Z<1.67)\quad \text{3.}\,\, P(Z>0.19).$$
Let $X$ be a normal random variable with $\mu=5$ and $\sigma=2$. Find the probabilities: $$\text{1.}\,\, P(-2<X<8)\quad \text{2.}\,\, P(X>-1) \quad \text{3.}\,\, P(X<4).$$