Population Distributions

1
Chapter 7

• Population Distributions

2
Chapter 7 Population Distributions

• 7.1 Describing the Distribution of Values in a
  Population
• 7.2 Population Models for Continuous Numerical
  Variables
• 7.3 Normal Distributions

3
7.1 Describing the Distribution of Values in a
Population

• An Example How is the performance of an online
  registration system for college courses?
• Population All the students who use the system
• Variable of Interest Time to complete
  registration
• A variable associates a value with each
  individual or object in a population.
• The distribution of all the values of a numerical
  variable or all the categories of a categorical
  variable is called a population distribution.

4
Categorical Variables Example

• In a study of factors related to air quality,
  monitors were posted at every entrance to a
  California university campus on October 10, 2003.
  From 6am to 10 pm, monitors recorded the mode of
  transportation for each person entering the
  campus. Based on the information collected, the
  population distribution of the variable
  
  x mode of transportation was
  constructed.

5
Numerical Variables

• Numerical variables can be either discrete or
  continuous.
• A discrete numerical variable is one whose
  possible values are isolated points along the
  number line.
• A continuous numerical variable is one whose
  possible values form an interval along the number
  line.
• A discrete numerical variable can be summarized
  by a relative frequency histogram.
• A continuous numerical variable can be summarized
  by a density histogram.

Therefore, relative frequency
(density)(interval width).
6
Example Pet Ownership

• The Department of Animal Regulation released
  information on pet ownership for the population
  of all households in a particular county. The
  variable considered was
• x number of licensed dogs or cats for a
  household.
• Summarize the population distribution in a
  relative frequency histogram.
• Is x discrete or continuous?
• What is the probability of observing a household
  with 3 or more licensed dogs or cat?

Answer on the next slide
7
Answer to the Example Pet Ownership

• x number of licensed dogs or cats for a
  household
• Possible x values are 0, 1, 2, 3, 4 and 5. These
  are isolated points along the number line, x is a
  discrete variable.
• The probability of observing a household with 3
  or more licensed dogs or cats is
• P ( x 3 ) P ( x 3 ) P ( x 4 ) P ( x
  5 )
• 0.9 .03 .01 0.13
• Exercises
• What is the probability of observing a household
  with at most 2 licensed dogs or cat?
• What is the probability of observing a household
  with at least 2 licensed dogs or cat?

8
Example Birth Weights

• Birth weight was recorded for all full-term
  babies born during 2002 in a semirural county.
  The variable x birth weight for a full-term
  baby in this county is an example of a continuous
  numerical variable.
• We can construct a density histogram of describe
  the population distribution of x values.

• Density the height of each rectangle
• Relative frequency Probability
• (density)(interval width)
• (height)(interval width) area

9
The Mean (µ) and Standard Deviation (s) of A
Numerical Variable

• The mean value of a numerical variable x, denoted
  by µ, describes where the population distribution
  of x is centered.
• The standard deviation of a numerical variable x,
  denoted by s, describes variability in the
  population distribution.
• When s near 0, the values of x tend to be close
  to µ (little variability) when s is large, there
  is more variability in the population of x
  values.

10
(No Transcript)
11
Which distribution has the largest standard
deviation?Which distribution has the largest
mean?Which distribution has a mean of about
5?Which distribution has the smallest standard
deviation?
12

• The area of any rectangle in a density
  histogram can be interpreted as the probability
  of observing a variable value in that interval.
• In (a), P ( 4.5 lt x lt 5.5 ) height (density) x
  width (.05)(1) .05.

13
A smooth curve specifies a continuous probability
distribution

• An observation of the four density histograms on
  the preceding slide shows
• A density histogram based on a small number of
  intervals can be quite jagged.
• As the number of interval increases, the
  resulting histograms become much smoother in
  appearance.
• A smooth curve superimposed over a density
  histogram such as the one shown on the right, is
  called a continuous probability distribution.

A smooth superimposed over the density histogram
(d) of the preceding slide
14
7.2 Population Models for Continuous Numerical
Variables

• A continuous probability distribution is a smooth
  curve, called a density curve, that serves as a
  model for the population distribution of a
  continuous variable.
• Properties of continuous probability
  distributions are
• The total area under the curve is equal to 1
• The area under the curve and above any particular
  interval is interpreted as the (approximate)
  probability of observing a value in the
  corresponding interval when an individual or
  object is selected at random from the population.

15

• Let x be a continuous numerical variable.
• For any particular number a, P( x a ) 0.
• (Because there is 0 area under the density curve
  above a single x value.)
• For any particular numbers a and b,
• P( x b ) P( x lt b )
• P( x a ) P( x gt a )
• P( a lt x lt b ) P( a x b )
• The above are NOT true for discrete numerical
  variables!

16
Example Departure Delays of A Commuter Train

• The length of time that elapses between the
  scheduled departure time and the actual departure
  time is recorded on 200 occasions, and the
  resulting observations are summarized in the
  density histogram.

The histogram in (a) is fairly flat, a reasonable
model for the population distribution is uniform
distribution in (b). The height of the density
curve is uniformly chosen to be 0.1, so that the
total area under the curve is equal to 1.
17
Example 7.6 Priority Mail Package Weights

• Develop a reasonable model for the population
  distribution.
• The shape of the sample density histogram
  suggests that a reasonable model for the
  population is a triangular distribution. (See
  next slide.)
• 2. How to choose the height of the triangle so
  that the total area under the probability
  distribution curve 1?
• Total area of triangle
• ½ ( base )( height ) 1.
• With base 2.0, the height must be equal to
  1.

• 200 packages shipped using the
  Priority Mail rate for packages under 2 lb were
  weighed, resulting the following density
  histogram.
• Let x package weight (in pounds).

18
Example Priority Mail Package Weights
Figure (a) histogram of package weight values
(b) continuous probability distribution for
package weight.

• Example Find the probability that a package
  selected at random weighs less than 1.5 lb.
• First using similar triangles to find the height
  h at x 1.50 is 0.75.
• Then P( x lt1.5 ) ½ (1.5)(0.75) .5625.
• Exercise Find P( x 1.5 ) and P( x 1.5 ).

19
Example Service Times of An Airlines Reservation

• An airlines toll-free reservation number
  recorded the length of time required to provide
  service to each of 500 callers. Let x service
  time.

• What is the population?
• Develop a model for the population distribution.

20
Example Service Times of An Airlines Reservation
Figure (a) histogram of service times (b)
continuous distribution of service times.

• What is the probability that the service time for
  a randomly selected caller lasts less than 3
  minutes?
• P ( x lt 3 ) ( height )( width ) (1/24)(3)
  1/8
• 2. What is the probability that the service time
  for a randomly selected caller lasts greater than
  8 minutes?
• 3. What is the probability that the service time
  for a randomly selected caller lasts between 2
  and 4 minutes?

21
Example Telephone Registration Times

• Students at a university use a telephone
  registration system to register for courses. The
  variable
• x length of time required for a student to
  register.
• The general form of the density histogram can be
  described as bell shaped and symmetric.
• The probability model of this problem is an
  example of a type of symmetric bell-shaped
  distribution known as a normal probability
  distribution.

The superimposed smooth curve is a
reasonable model for the population distribution.
22
7.3 Normal Distributions

• Normal distributions are widely used for two
  reasons
• They provide a reasonable approximation to the
  distribution of many different variables.
• They play a central role in many of the
  inferential procedures.
• Normal distributions are distinguished by two
  important parameters
• The mean µ where the normal curve is centered.
• The standard deviation s how much the curve
  spreads out around the center.

23
Normal Distributions

• Normal distributions are continuous probability
  distributions that are (1) bell shaped, (2)
  symmetric and (3) the two tails die out quickly.

24
The Standard Normal Distribution

• The normal distribution with µ 0 and s 1.
• The term z curve is used for the standard normal
  curve.
• P ( z lt z ) the cumulative area of z.

25
Using Appendix Table 2 on page 706 and page 707
(also inside the back cover) to find standard
normal curve areas

• For any number z between -3.89 and 3.89 and
  rounded to two decimal places, Appendix Table 2
  gives
• (area under z curve to the left of z) P ( z lt
  z ) P ( z z )
• where z represents a variable whose distribution
  is standard normal distribution .
• To find this probability, locate the following
• The row labeled with the sign z and the digit to
  either side of the decimal point (e. g., -1.7,
  0.5, 3.6, etc.)
• The column identified with the second digit to
  the right of the decimal point in z (e. g., .06
  if z -1.76)
• The number at the intersection of this row and
  column is the desired probability, P ( z lt z ) .
• Example Find P ( z lt -1.76).
• Because -1.76 is a negative number, we
  use the table on page. 706. First locate the row
  labeled with -1.7 in the first column (z
  column). Then we find P ( z lt -1.76) .0392 at
  the intersection of this row and the column
  labeled .06.
• Note You may use an online z table instead of
  Table 2, but you have to understand how to use it
  because it may have a different design.

26
Using Table 2 (p. 706 p. 707) to Find Standard
Normal Probabilities ( Cumulative z Curve Areas)

• Examples Find the following probability
• 1. P( z lt -1.76) .0392
• Exercise P( z 0.58)
• P( z lt -4.12)
• P( z lt 4.18)
• 2. P( zgt1.96) 1 - P( zlt 1.96) 1 - .9750
  .0250
• Another method P( zgt1.96) P( zlt -1.96)
  .0250
• Exercise P(zgt -1.28)
• P(zgt 3.9)
• 3. P(-1.76lt z lt0.58) P( z 0.58) - P( z lt
  -1.76)
• .7190 - .0392 .6798
• Exercise P(-2.00lt z lt2.00)

Answer to the exercises P( z 0.58) .7190
P( z lt -4.12) 0 P( z lt 4.18) 1 P( z gt
-1.28) .8997 P( z gt 3.9) 0 P(-2.00 lt z lt
2.00) .9544.
27
Example Identifying Extreme Values

• Find z such that P( zlt z ).02.
• Figure (a) shows that the cumulative area
  for z is .02. (The area of .02 lt 0.5, so z must
  be a negative number.) Therefore, we look for an
  area of .02 in the body of Appendix Table 2. The
  closest area in the table is .0202 in the -2.0
  row and .05 column. So z -2.05.

• Find z such that P( z gt z ).05.
• P( z gt z ) .05 indicates that the area to
  the right of z is .05 in Figure (b). Area to the
  left of z is 1 -.05 .95. In Table 2 on page
  707 .95 falls exactly between .9495
  (corresponding to a z value of 1.64) and .9505
  (corresponding to a z value of 1.65). So z
  ½(1.64 1.65)1.645.

28

• Exercise Find the values that make up the most
  extreme 5 of the standard normal distribution.
• We need to separate the middle 95 from the
  extreme 5. Because the standard normal
  distribution is symmetric, the most extreme 5 is
  equally divided between the high side and the low
  side of the distribution, resulting in an area of
  .025 for each of the tails of the z curve.
  Symmetry about 0 implies that if z denotes the
  value that separate the largest 2.5, the value
  that separate the smallest 2.5 is simply -z.

Complete the problem by finding z.
Answer z 1.96
29
Other Normal Distributions

• Let x be a variable whose behavior is described
  by a normal distribution with mean µ and standard
  deviation s. To calculate probabilities for x, we
  standardize the relevant values and then use the
  table for z curve areas.
• P( xltb ) P( zltb )
• P( altx ) P( altz ) (equivalently, P( xgta )
  P( zgta) )
• P( altxltb ) P( altzltb ),
• where

30
Example Childrens Heights

• The height of a randomly selected 5-year-old
  child is a normal distribution with a mean of µ
  100 cm and standard deviation s 6cm. What
  proportion of the heights is between 94 and 112
  cm?
• Let x the height of a randomly selected
  5-year-old child.

About 82 of 5-year-old children have heights
between 94 and 112 cm. Exercise What is the
probability that a randomly selected 5-year-old
child will be taller than 110 cm?
Answer to Exercise 4.75
31
Example IQ Scores

• A commonly used IQ scale has a mean of 100 and a
  standard deviation of 15, and scores are
  approximately normally distributed. (IQ score is
  actually a discrete variable, but its population
  distribution closely resembles a normal curve.)
• What proportion of the population would qualify
  for Mensa membership, which requires an IQ score
  above 130?
• What proportion of the population with IQ score
  below 80?
• What proportion of the population with IQ score
  between 75 and 125?

32
Solution to Example Registration Times
Let x IQ score of a randomly selected
individual. Given µ 100 and s 15.

• 1.
• 2.
• 3. For you exercise.

Answer of 3. P ( 75 lt x lt 125 ) .9050
33
Example Registration Times

• The length of time (in minutes) required for
  students to complete telephone registration in a
  particular university can be well approximated by
  a normal distribution with mean µ 12 min and
  standard deviation s 2 min. The university
  would like to disconnect students automatically
  after some amount of time has elapsed. Determine
  the amount of time that should be allowed before
  disconnecting a student if the university wants
  only the largest 1 to be disconnected.

34
Solution of Registration Time Example

• Let x be the length of registration time for a
  randomly selected student.
• Given µ 12 minutes and s 2 minutes. Let x
  be the time (in minutes) the phone registration
  should be disconnected.
• P( x gt x ) 1, P( x lt x ) 99.
• The z value corresponding to the
• cumulative area .99 is 2.33.
• Therefore,

35
Example Motor Vehicle Emissions

• The EPA has determined that the emissions of
  nitrogen oxides, which are major constituents of
  smog, can be modeled using a normal distribution
  with µ 1.6 and s 0.4. Suppose that the EPA
  wants to offer some sort of incentive to get the
  worst polluters off the road. What emission
  levels constitute the worst 10 of the vehicles?

36
Solution of Vehicle Emission Example

• Let x be the emission level of pollutant for a
  randomly selected vehicle.
• Given µ 1.6 and s 0.4. Let x be the
  emission level that constitutes the worst 10
  (the highest 10 emission level).
• P( x gt x ) 10, P( x lt x ) 90.
• The z value corresponding to area .9 is 1.28.
• Therefore,