Continuous Random Variables

1
Chapter 5

• Continuous Random Variables

2
Continuous Probability Distributions

• Continuous Probability Distribution areas under
  curve correspond to probabilities for x
• Area A corresponds to the probability that x lies
  between a and b
• Do you see the similarity in shape between the
  continuous and discrete probability distributions?

3
The Uniform Distribution
The Uniform Distribution

• Uniform Probability Distribution distribution
  resulting when a continuous random variable is
  evenly distributed over a particular interval

4
The Normal Distribution

• A normal random variable has a probability
  distribution called a normal distribution
• The Normal Distribution
• Bell-shaped curve
• Symmetrical about its mean µ
• Spread determined by the value
• of its standard deviation s

5
The Normal Distribution

• The mean and standard deviation affect the
  flatness and center of the curve, but not the
  basic shape

6
The Normal Distribution

• The function that generates a normal curve is of
  the form
• where
• ? Mean of the normal random variable x
• ? Standard deviation
• ? 3.1416
• e 2.71828
• P(xlta) is obtained from a table of normal
  probabilities

7
The Normal Distribution

• Probabilities associated with values or ranges of
  a random variable correspond to areas under the
  normal curve
• Calculating probabilities can be simplified by
  working with a Standard Normal Distribution
• A Standard Normal Distribution is a Normal
  distribution with ? 0 and ? 1
• The standard normalrandom variable is denoted
  by thesymbol z

8
The Normal Distribution

• Table for Standard Normal Distribution contains
  probability for the area between 0 and z
• Partial table shows
• components of table

9
The Normal Distribution

• What is P(-1.33 lt z lt 1.33)?
• Table gives us area A1
• Symmetry about the meantell us that A2 A1
• P(-1.33 lt z lt 1.33) P(-1.33 lt z lt 0) P(0 lt z lt
  1.33) A2 A1 .4082 .4082 .8164

10
The Normal Distribution

• What is P(z gt 1.64)?
• Table gives us area A2
• Symmetry about the meantell us that A2 A1 .5
• P(z gt 1.64) A1 .5 A2.5 - .4495 .0505

11
The Normal Distribution

• What is P(z lt .67)?
• Table gives us area A1
• Symmetry about the meantell us that A2 .5
• P(z lt .67) A1 A2 .2486 .5 .7486

12
The Normal Distribution

• What is P(z gt 1.96)?
• Table gives us area .5 - A2.4750, so A2 .0250
• Symmetry about the meantell us that A2 A1
• P(z gt 1.96) A1 A2 .0250 .0250 .05

13
The Normal Distribution

• What if values of interest were not normalized?
  We want to knowP (8ltxlt12), with µ10 and s1.5
• Convert to standard normal using
• P(8ltxlt12) P(-1.33ltzlt1.33) 2(.4082) .8164

14
The Normal Distribution

• Steps for Finding a Probability Corresponding to
  a Normal Random Variable
• Sketch the distribution, locate mean, shade area
  of interest
• Convert to standard z values using
• Add z values to the sketch
• Use tables to calculate probabilities, making use
  of symmetry property where necessary

15
The Normal Distribution

• Making an Inference
• How likely is an observationin area A, given an
  assumed normal distribution with mean of 27 and
  standard deviation of 3?
• Z value for x20 is -2.33
• P(xlt20) P(zlt-2.33) .5 - .4901 .0099
• You could reasonably conclude that this is a rare
  event

16
The Normal Distribution

• You can also use the table in reverse to find a
  z-value that corresponds to a particular
  probability
• What is the value of z that will be exceeded only
  10 of the time?
• Look in the body of the table for the value
  closest to .4, and read the corresponding z value
• Z 1.28

17
The Normal Distribution

• Which values of z enclose the middle 95 of the
  standard normal z values?
• Using the symmetry property,z0 must correspond
  with a probability of .475
• From the table, we find that z0 and z0 are 1.96
  and -1.96 respectively.

18
The Normal Distribution

• Given a normally distributed variable x with
  mean 550 and standard deviation of 100, what
  value of x identifies the top 10 of the
  distribution?
• The z value corresponding with .40 is 1.28.
  Solving for x0
• x0 550 1.28(100) 550 128 678

19
Descriptive Methods for Assessing Normality

• Evaluate the shape from a histogram or
  stem-and-leaf display
• Compute intervals about mean and corresponding
  percentages
• Compute IQR and divide by standard deviation.
  Result is roughly 1.3 if normal
• Use statistical package to evaluate a normal
  probability plot for the data

20
Approximating a Binomial Distribution with a
Normal Distribution

• You can use a Normal Distribution as an
  approximation of a Binomial Distribution for
  large values of n
• Often needed given limitation of binomial tables
• Need to add a correction for continuity, because
  of the discrete nature of the binomial
  distribution
• Correction is to add .5 to x when converting to
  standard z values
• Rule of thumb interval ?3? should be within
  range of binomial random variable (0-n) for
  normal distribution to be adequate approximation

21
Approximating a Binomial Distribution with a
Normal Distribution

• Steps
• Determine n and p for the binomial distribution
• Calculate the interval
• Express binomial probability in the form P(xlta)
  or P(xltb)P(xlta)
• Calculate z value for each a, applying continuity
  correction
• Sketch normal distribution, locate as and use
  table to solve

22
The Exponential Distribution

• Used to describe the amount of time between
  occurrences of random events
• Probability Distribution, for an Exponential
  Random Variable x Probability Density function
• Mean
• Standard Deviation

23
The Exponential Distribution

• Shape of the distributionis determined by the
  value of ?
• Mean is equal to
• Standard deviation

24
The Exponential Distribution

• To find the area A to the right of a,
• A can be calculatedusing a calculator orwith
  tables

25
The Exponential Distribution

• If ? .5, what is the p(agt5)?
• From tables, A .082085
• Probability that a gt 5 is.082085

26
The Exponential Distribution

• For a given distribution,
• If ? .16, what are the ? and ??
• What is the p(0ltalt5)?
• What is the p(?2?ltalt ?2?)?

27
The Exponential Distribution

• a) ? ? 1/ ? 1/.16 6.25
• b) P(xgta) e-?a
• P(xgt5) e-(.16)5 e-.8
• .449329
• P(xlt5) 1-P(xgt5)
• 1-.449329 .550671

28
The Exponential Distribution

• c) What is the p(?2?ltalt ?2?)?
• Find the complement of the area above ?2?
• P 1-P(xgt18.75) 1- e-?(18.75) 1-
  e-.16(18.75) 1- e-3 1- .049787
• .950213