Continuous Random Variables

Fei Ye

January 2024

1. Learning Goals


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


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


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 \(\sigma=1.5\).

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


12. Example: Standardization Revisited

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

  1. Find the \(Z\)-score for the value \(X=13\).
  2. 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.

  1. Find \(P(Z<1.21)\).
  2. Find \(P(Z\geq 1.21)\).
  3. 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).

  1. In this case, \(P(Z<1.21)\) =NORM.S.DIST(1.21, TRUE) \(\approx 0.8869\).
  2. Since the total area under the normal curve is 1, $$P(Z\geq 1.21) = 1 - P(Z < 1.21) \approx 1-0.8869=0.1131.$$
  3. 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


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


19. Lab Practice: Normal Variable

  1. Let \(Z\) be a standard normal random variable. Find the probabilities: $$\text{1.}\,\, P(Z<1.58)\quad \text{2.}\,\, P(-0.6<Z<1.67)\quad \text{3.}\,\, P(Z>0.19).$$

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