CHAPTER 7
Continuous Probability Distributions
.
Chapter 7 - Learning Objectives
•
Differentiate between the normal and the exponential distributions.
•
Use the standard normal distribution and z-scores to determine
probabilities associated with the normal distribution.
•
Use the normal distribution to
approximate the binomial distribution.
•
Use the exponential distribution to
determine related probabilities.
Chapter 7 - Key Terms
•
Probability density function
•
Probability distributions
– Standard normal distribution
» Mean, variance, applications
– Exponential distribution
» Mean, variance, applications
•
Normal approximation to the
binomial distribution
Chapter 7 - Key Concept
•
The area under a probability density
function between two bounds, a and b, is the probability that a value will occur within the
bounded interval
between a and b.
The Normal Distribution
• An important family of continuous distributions
• Bell-shaped, symmetric, and asymptotic
• The mean, median and mode are at the same position on the horizontal axis
• The curve is asymptotic
• The total area beneath the curve is 1.0
• To specify a particular distribution in this family, two parameters must be given:
– Mean
– Standard deviation
Areas under the Normal Curve
Use the standard normal table to find:
• The z-score such that the area from the midpoint to z is 0.20.
In the interior of the standard normal table, look up a value close to 0.20.
The closest value is 0.1985, which occurs at
z = 0.52
.20%
of the area
z
Areas under the Normal Curve
Use the standard normal table to find:
• The probability associated with z: P(0 z 1.32).
Locate the row whose header is 1.3. Proceed along that row to the column whose
header is .02. There you find the value .4066, which is the amount of area capture between the mean
and a z of 1.32.
Answer: 0.4066
z = 1.32
Areas under the Normal Curve
Use the standard normal table to find:
• The probability associated with z: P(–1.10 z 1.32).
Find the amount of area between the mean and z = 1.32 and add it to the amount of area between the mean and z = 1.10*.
0.3643 + 0.4066 = 0.7709
z = 1.32 z = –1.10
Area 1 Area 2
Areas under the Normal Curve -
Dealing with Negative Z’s
•
Note: Because the normal curve is symmetric, the amount of
area between the mean and z = –1.10 is the same as the
amount of area between the
mean and z = +1.10.
Areas under the Normal Curve
Use the standard normal table to find:
• The probability associated with z: P(1.00 z 1.32).
Find the amount of area between
the mean and z = 1.00 and subtract it from the amount of area
between the mean and z = 1.32.
0.4066 – 0.3413 = 0.0653
z = 1.32 z = 1.00
Standardizing Individual Data Values on a Normal Curve
•
The standardized z-score is how far above or below the individual value is compared to the population mean in units of standard deviation.
– “How far above or below”= data value – mean
– “In units of standard deviation”= divide by
•
Standardized individual value z data value mean
standard deviation x –
Standard Normal Distribution:
An Example
• It has been reported that the average hotel check-in time, from curbside to delivery of bags into the room, is 12.1 minutes. Mary has just left the cab that
brought her to her hotel. Assuming a normal
distribution with a standard deviation of 2.0 minutes, what is the probability that the time required for Mary and her bags to get to the room will be:
a) greater than 14.1 minutes?
b) less than 8.1 minutes?
c) between 10.1 and 14.1 minutes?
d) between 10.1 and 16.1 minutes?
An Example, cont.
Given in the problem:
µ = 12.1 minutes, = 2.0 minutes
•
a) Greater than 14.1 minutes
P(x > 14.1) = P(z > 1.00)
= .5 – .3413 = 0.1587 z x– 14.1–12.1
2.0 1.00
z = 1.00
An Example, cont.
Given in the problem:
µ = 12.1 minutes, = 2.0 minutes
•
b) Less than 8.1 minutes
P(x < 8.1) = P(z < –2.00)
= .5 – .4772 = 0.0228 z x– 8.1–12.1
2.0 –2.00
z = –2.00
An Example, cont.
Given in the problem:
µ = 12.1 minutes, = 2.0 minutes
•
c) Between 10.1 and 14.1 minutes
P(10.1 < x < 14.1)
= P(–1.00 < z < 1.00)
= 0.3413 + 0.3413 = 0.6826 z lower x – 10.1–12.1
2.0 –1.00 zupper x–
14.1–12.1
2.0 1.00
z = 1.00 z = –1.00
Area 1 Area 2
An Example, cont.
Given in the problem:
µ = 12.1 minutes, = 2.0 minutes
•
d) Between 10.1 and 16.1 minutes
P(10.1 < x < 16.1)
= P(–1.00 < z < 2.00)
= 0.3413 + 0.4772 = 0.8185
zlower x – 10.1–12.1
2.0 –1.00
zupper x–
16.1–12.1
2.0 2.00
z = 2.00 z = –1.00
Area 1 Area 2
Example: Using Microsoft Excel
• Problem: What is the probability that the time required for Mary and her bags to get to the room will be:
a) greater than 14.1 minutes?
In a cell on an Excel worksheet, type
=1-NORMDIST(14.1,12.1,2,true) and you will see the answer: 0.1587
b) less than 8.1 minutes?
In a cell on an Excel worksheet, type
=NORMDIST(8.1,12.1,2,true) and you will see the answer: 0.0228
Example: Using Microsoft Excel
• Problem: What is the probability that the time required for Mary and her bags to get to the room will be:
c) between 10.1 and 14.1 minutes?
In a cell on an Excel worksheet, type all on one line
=NORMDIST(14.1,12.1,2,true)- NORMDIST(10.1,12.1,2,true)
and you will see the answer: 0.6826 d) between 10.1 and 16.1 minutes?
In a cell on an Excel worksheet, type all on one line
=NORMDIST(16.1,12.1,2,true)- NORMDIST(10.1,12.1,2,true)
and you will see the answer: 0.8185
The Exponential Distribution
where = mean and variance of a Poisson distribution.
1/ = mean and standard deviation of the corresponding exponential distribution
x – the length of interval between occurrences
e = 2.71828, a constant
Probability:
e
xx
f ( )
–P(x k) e
–kExponential Distribution:
Problem 7.56
A taxi dispatcher has found that successive calls for a taxi are exponentially distributed, with a
mean time between calls of 5.30 minutes. The
dispatcher must disconnect the telephone system for 3 minutes in order to have the push-button
mechanism repaired. What is the probability that a call will be received while the system is out of service?
Answer:
Given x is exponentially distributed with a mean of 1/ = 5.3 minutes:
and
k
k 5.3
P(x k) e e
3
0.566
P(x 3) 1 P(x 3) 1 e5.3 1 e 1 0.5678 0.4322
Normal Approximation to the Binomial Distribution
Can use whenever np and n(1-p) ≥ 5.
Useful when values of n and p are not listed in the binomial probability tables.
Mean = np
Standard deviation = [np(1-p)]^0.5
Because we are transforming from a discrete into a
binomial distribution, this method involves a correction for continuity. So, the probability for an individual
value of x is expressed as an interval from [x-0.5] to [x+0.5].
Normal Approximation to the Binomial Distribution: Problem 7.45
Of all individual tax returns field in the U.S.A. during 1999 tax filing season, 13.7% were prepared by H&R Block.
For a randomly selected sample of 1000 tax returns field during this
period, determine the probability that between 110 and 140 of these
returns were prepared by H&R Block.
Answer:
Let x be the number of tax returns in this group; x ~ bin (1000,0.137).
Mean=np=1000*0.137=137
S.d.=[1000*0.137*(1- 0.137)]^0.5=10.873
P(110 ≤ x ≤ 140)= P(109.5 ≤ x ≤ 140.5) this is equivalent to
