Continous Distributions

Dr. Mine Dogucu

Exponential Distribution

pdf

\(X \sim \text{Exp}(\lambda)\)

Expected Value and Variance

\(E(X) = \frac{1}{\lambda}\)

\(Var(X) = \frac{1}{\lambda^2}\)

Example

Suppose your waiting time (in minutes) for a bus is \(X \sim Exp(0.1)\).

\(f(x) =0.1 e^{-0.1 x} \text{ for } \ x \ \epsilon \ [0, \infty)\)

\(f(10) =0.1 e^{-0.1\cdot10}\)

0.1*exp(-0.1*10)
[1] 0.03678794
dexp(x = 10, rate = 0.1)
[1] 0.03678794

Calculating Probability

You show up at the bus stop. What is the probability that you will wait for 10 minutes for the bus?

\(P(X = 10) = 0\)

\(\int_{10}^{10}\lambda e^{-\lambda x}dx\)

Calculating Probability

What is the probability that you will wait for the bus less than 10 minutes?

Cumulative probability!

\(\int_{0}^{10}\lambda e^{-\lambda x}dx\)

\(\int_{0}^{10}0.1 e^{-0.1 x}dx\)

pexp(q = 10, rate = 0.1)
[1] 0.6321206

Calculating Probability

What is the probability that you will wait between 10 to 20 minutes for the bus?

\(P(10<X<20) = ?\)

\(\int_{10}^{20}0.1 e^{-0.1 x}dx\)

\(P(X<20) - P(X<10)\)

pexp(q = 20, rate = 0.1) -
  pexp(q = 10, rate = 0.1)
[1] 0.2325442

Expected Value and Variance

What is the expected value of the waiting time for a bus?

\(E(X) = \frac{1}{\lambda}\)

\(E(X) = \frac{1}{0.1} = 10\) minutes

What is the variance of the waiting time for a bus?

\(Var(X) = \frac{1}{\lambda^2}\)

\(Var(X) = \frac{1}{0.1^2} = 100 \text{ minutes}^2\)

Poisson

Discrete

Number of events (bus)

Every ten minutes we expect 1 bus

Exponential

Continuous

Time between two events

We expect to wait 10 minutes for a bus

Normal Distribution

pdf

\(X \sim \mathcal{N}(\mu, \sigma^2)\)

\(f(x) = \frac{1}{\sigma \sqrt{2\pi}}{e^{-(x-\mu)^2/2\sigma^2}}\)

\(E(X) = \mu\)

\(Var(X) = \sigma^2\)

Characteristics

Mean = Median = Mode

\(P(X \leq \mu) = 0.5\)

Parameters

68.26895% of the data are within 1 standard deviation of the mean.

95.44997% of the data are within 2 standard deviations of the mean.

99.73002% of the data are within 3 standard deviations of the mean.

Example

Midterm exam scores are normally distributed with a mean of 80 and standard deviation of 5 points.

\[X \sim N(80, 5^2)\]

\(f(78)\) =

dnorm(x = 78, mean = 80, sd = 5)
[1] 0.07365403

\(f(92)\) =

dnorm(x = 92, mean = 80, sd = 5)
[1] 0.004478906

Calculating Probability

What is the probability that a student scores below 73 points?

pnorm(q = 73, mean = 80, sd = 5)
[1] 0.08075666


If a student scores 73 points, the student’s percentile rank is 8.0756659%

Calculating Probability

What is the probability that a student scores above 78 points?

1 - pnorm(q = 78, mean = 80, sd = 5)
[1] 0.6554217
pnorm(q = 78, mean = 80, sd = 5, 
      lower.tail = FALSE)
[1] 0.6554217

How many standard deviations away from the mean?

Krishna scored 85 points on the midterm. How many standard deviations is Krishna’s score away from the mean?

\(\frac{85-80}{5} = 1\)

Hira scored 77.5 points on the midterm. How many standard deviations is Hira’s score away from the mean?

\(\frac{77.5-80}{5} = -0.5\)

Z-score

\[Z = \frac{X-\mu}{\sigma} \]

Calculating z-scores

We can calculate Z-scores regardless of the distribution of the data. In other words, data do not need to be normally distributed. Z-scores only show how many standard deviations a value is away from the mean.

Interpreting z-scores

If we know that the data follow a normal distribution then we can use z-scores for probabilistic interpretations.

Normal Distribution vs. Standard Normal (z) Distribution

Normal Distribution

pnorm(q = 90, mean = 80, sd = 5)
[1] 0.9772499

Standard Normal (z) Distribution

pnorm(q = 2, mean = 0, sd = 1)
[1] 0.9772499

SAT scores - z scores

Let X represent the SAT scores and \(X \sim N(1100, 200^2)\)

What is the Z-score of a student who scores 1500?

\(z = \frac{1500-1100}{200} = 2\)

SAT scores - probability

What is the percentile rank of a student who scores 1500?

SAT scores - probability

What is the percentile rank of a student who scores 1500?

pnorm(q = 1500, 
      mean = 1100, 
      sd = 200)
[1] 0.9772499

If you know the order of arguments:

pnorm(1500, 1100, 200)
[1] 0.9772499
pnorm(2, 
      mean = 0, 
      sd = 1)
[1] 0.9772499

The default mean is 0 and sd is 1.

pnorm(2)
[1] 0.9772499

What percentage of test takers score between 800 and 1400?

\(P(800<X<1400) =P(X<1400) - P(X<800)\)

pnorm(q = 1400, mean = 1100, sd = 200) -
  pnorm(q = 800, mean = 1100, sd = 200) 
[1] 0.8663856

Known probability unknown x

If College Board wanted to send a congratulatory email to the top 10% in the SATs, above what SAT score should they consider sending the email to?

qnorm(p = 0.90, 
      mean = 1100, 
      sd = 200) 
[1] 1356.31

Continuous Uniform Distribution

pdf

\(X \sim \mathcal{U}(a,b)\)

Expected Value

\(E(X) = \int_{x \ \epsilon \ S_X} xf(x)dx\)

\(E(X) = \int_a^b x\frac{1}{b-a}dx\)

\(E(X) = \frac{1}{b-a} \big[\frac{x^2}{2}\bigg\rvert_a^b\big]\)

\(E(X) = \frac{1}{b-a} \big[\frac{b^2}{2} - \frac{a^2}{2}\big]\)

\(E(X) = \frac{b^2 - a^2}{2(b-a)}\)

\(E(X) = \frac{b+a}{2}\)

\(E(X^2)\)

\(Var(X) = E(X^2)- [E(X)]^2\)

\(E(X^2) = ?\)

\(E(X^2) = \int_a^b x^2\frac{1}{b-a}dx\)

\(E(X^2) = \frac{1}{b-a} \big[\frac{x^3}{3}\bigg\rvert_a^b\big]\)

\(E(X^2) = \frac{b^3 - a^3}{3(b-a)}\)

Variance

\(Var(X) = E(X^2)- [E(X)]^2\)

\(Var(X) = \frac{b^3 - a^3}{3(b-a)}- (\frac{b+a}{2})^2\)

\(Var(X) = \frac{(b-a) (b^2+ab+a^2)}{3(b-a)}- \frac{(b+a)^2}{4}\)

\(Var(X) = \frac{(b-a)^2}{12}\)

Example

A cellphone has a defective alarm system. Once every day the alarm starts ringing randomly. Let X be the random variable representing the time in the 24 hour system when the alarm goes on.

\(f(x) = \frac{1}{b-a} \text{ for } \ a \leq x \leq b\)

\(f(x) = \frac{1}{24-0} \text{ for } x \ \epsilon [0,24)\)

pdf

dunif(x = 3, min = 0, max = 24)
[1] 0.04166667
dunif(x = 15, min = 0, max = 24)
[1] 0.04166667
dunif(x = 21.25, min = 0, max = 24)
[1] 0.04166667

Calculating Probability

What is the probability that the phone alarm will ring at 3 am?

\(P(X = 3) = 0\)

\(\int_3^{3}\frac{1}{b-a}dx = 0\)

Calculating Probability

What is the probability that the phone alarm will ring at some time between midnight and 8 am?

Cumulative probability!

\(\int_0^{8}\frac{1}{b-a}dx\)

\(\frac{x}{b-a}\bigg\rvert_0^8 = \frac{8}{24} = \frac{1}{3}\)

punif(q = 8, min = 0, max = 24)
[1] 0.3333333

Calculating Probability

What is the probability that the phone alarm will ring at some time between 2 am to 8 am?

\(P(2<X<8) = ?\)

\(\frac{x}{b-a}\bigg\rvert_2^8 = \frac{8}{24} - \frac{2}{24}= \frac{1}{4}\)

\(P(X<8) - P(X<2)\)

punif(q = 8, min = 0, max = 24) -
  punif(q = 2, min = 0, max = 24)
[1] 0.25

Expected Value

What is the expected value of the time that the phone alarm will go on?

\(E(X) = \frac{b+a}{2}\)

\(E(X) = \frac{24+0}{2} = 12\)

We found that the expected value is 12. Does that mean that every day we would expect the alarm to go on at noon (12 pm)?

No! Some days the alarm might go on at 9 am, some days at 3 pm (15). Some days it might go on at 9:15 pm (21.25). We would expect the average to be 12 in the long run.

Variance

We already pointed out that there is variance to the time the alarm goes on. It is not at noon every day. What is the variance of the time that the phone alarm will go on?

\(Var(X) = \frac{(b-a)^2}{12}\)

\(Var(X) = \frac{(24-0)^2}{12}\)

\(Var(X) = 48\)

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