topics covered discrete probability distributions –the uniform distribution –the binomial...
TRANSCRIPT
Topics Covered
• Discrete probability distributions
– The Uniform Distribution
– The Binomial Distribution
– The Poisson Distribution
• Each is appropriately applied in certain situations and to particular phenomena
Continuous Random Variables
• Continuous random variable can assume all real
number values within an interval (e.g., rainfall, pH)
• Some random variables that are technically discrete
exhibit such a tremendous range of values, that is
it desirable to treat them as if they were continuous
variables, e.g. population
• Continuous random variables are described by
probability density functions
• Probability density functions are defined using the same rules required of probability mass functions, with some additional requirements:
• The function must have a non-negative value throughout the interval a to b, i.e.
f(x) >= 0 for a <= x <= b• The area under the curve defined by f(x), within
the interval a to b, must equal 1
Probability Density Functions
x
area=1f(x)
a b
The Normal Distribution
• The most common probability distribution is the normal distribution
• The normal distribution is a continuous distribution that is symmetric and bell-shaped
Source: http://mathworld.wolfram.com/NormalDistribution.html
The Normal Distribution
• Most naturally occurring variables are distributed normally (e.g. heights, weights, annual temperature variations, test scores, IQ scores, etc.)
• A normal distribution can also be produced by tracking the errors made in repeated measurements of the same thing; Karl Friedrich Gauss was a 19th century astronomer who found that the distribution of the repeated errors of determining the position of the same star formed a normal (or Gaussian) distribution
• The probability density function of the normal
distribution:
• You can see how the value of the distribution at x
is a f(x) of the mean and standard deviation
The Normal Distribution
2)(5.0
2
1)(
x
exf
Source: http://en.wikipedia.org/wiki/Normal_distribution
λ
The Poisson Distribution & The Normal Distribution
The probability density function of a normal distribution approximating the probability mass function of a binomial distribution
Source: http://en.wikipedia.org/wiki/Normal_Distribution
The Normal Distribution
• As with all frequency distributions, the area under the curve between any two x values corresponds to the probability of obtaining an x value in that range
• The total area under the curve is equal to one
Source: http://en.wikipedia.org/wiki/Normal_Distribution
The Normal Distribution
• The way we use the normal distribution is to find the probability of a continuous random variable falling within a range of values ([c, d])
The probability of the variable between c and d
is the area under the curve
• The areas under normal curves are given in tables such as that found in Table A.2 in Appendix A. Textbook (Rogerson)
x
f(x)
c d
Normal Tables
• Variables with normal distributions may have an infinite number of possible means and standard deviations
Source: http://en.wikipedia.org/wiki/Normal_distribution
Normal Tables
• Variables with normal distributions may have an infinite number of possible means and standard deviations
• Normal tables are standardized (a standard normal distribution with a mean of zero and a standard deviation of one)
• Before using a normal table, we must transform our data to a standardized normal distribution
Standardization of Normal Distributions
• The standardization is achieved by converting the data into z-scores
s
xxi z-score
• The z-score is the means that is used to transform our normal distribution into a standard normal distribution ( = 0 & = 1)
StandardizationMonth T (°F) Z-score
J 39.53 -1.56
F 46.36 -1.03
M 46.42 -1.02
A 60.32 0.05
M 66.34 0.51
J 75.49 1.22
J 75.39 1.21
A 77.29 1.36
S 68.64 0.69
O 57.57 -0.16
N 54.88 -0.37
D 48.2 -0.89
Mean = 59.70Standard deviation = 12.97
Original data: Z-score: Mean = 0Standard deviation = 1
• Example I – A data set with = 55 and = 20:
• Using z-scores in conjunction with standard normal tables (like Table A.2 on page 214 of Rogerson) we can look up areas under the curve associated with intervals, and thus probabilities (P(X>1.75) or P(X<1.75))
x - Z-score = = x - 5520
• If one of our data values x = 90 then:
x - Z-score = = 90 - 5520
= 3520
= 1.75
Standardization of Normal Distributions
Look Up Standard Normal Tables
• Using our example z-score of 1.75, we find the position of 1.75 in the table and use the value found there
Look Up Standard Normal Tables
μ = 0
f(x)
+1.75
.0401 P(Z > 1.75) = 0.0401
P(Z <= 1.75) = 0.9599
μ = 55
f(x).0401
+90
P(X > 90) = 0.0401
P(X <= 90) = 0.9599
• Example II – If we have a data set with = 55 and = 20, we calculate z-scores using:
• Using z-scores in conjunction with standard normal tables (like Table A.2 on page 214 of Rogerson) we can look up areas under the curve associated with intervals, and thus probabilities (P(X>20) or P(X<=20))
x - Z-score = = x - 5520
• If one of our data values x = 20 then:
x - Z-score = = 20 - 5520
= -3520
= -1.75
Standardization of Normal Distributions
μ = 0
f(x)
+1.75-1.75
Look Up Standard Normal Tables• Using our example z-score of -1.75, we find the position of 1.75 in the table and use the value found there; because the normal distribution is symmetric the table does not need to repeat positive and negative values
Look Up Standard Normal Tables
μ = 0
f(x)
+1.75-1.75
4.01% (.0401)4.01% (.0401)
Look Up Standard Normal Tables
μ = 0
f(x)
μ = 55
f(x)
c d
-1.75
4.01% (.0401)
4.01% (.0401)
P(Z <= -1.75) = 0.0401
P(Z > -1.75) = 0.9599
P(X <= 20) = 0.0401
P(X > 20) = 0.9599
Finding the P(x) for Various Intervals
a
3.
a
2.
a
1.
P(0 Z a) = [0.5 – (table value)]•Total Area under the curve = 1, thus the area above x is equal to 0.5, and we subtract the area of the tail
P(Z a) = [1 – (table value)]•Total Area under the curve = 1, and we subtract the area of the tail
P(Z a) = (table value)• Table gives the value of P(x) in the tail above a
Finding the P(x) for Various Intervals
a
6.
a
5.
a
4.
P(Z a) = [1 – (table value)]• This is equivalent to P(Z a) when a is positive
P(Z a) = (table value)• Table gives the value of P(x) in the tail below a, equivalent to P(Z a) when a is positive
P(a Z 0) = [0.5 – (table value)]• This is equivalent to P(0 Z a) when a is positive
Finding the P(x) for Various IntervalsP(a Z b) if a < 0 and b > 0
a
7. b
• With this set of building blocks, you should be able to calculate the probability for any interval using a standard normal table
or= [0.5 – (table value for a)] + [0.5 – (table value for b)]= [1 – {(table value for a) + (table value for b)}]
= (0.5 – P(Z<a)) + (0.5 – P(Z>b))= 1 – P(Z<a) – P(Z>b)
Finding the P(x) – Example
• Suppose we are in charge of buying stock for a shoe store. We will assume that the distribution of shoe sizes is normally distributed for a gender. Let’s say the mean of women’s shoe sizes is 20 cm and the standard deviation is 5 cm
• If our store sells 300 pairs of a popular style each week, we can make some projections about how many pairs we need to stock of a given size, assuming shoes fit feet that are +/- 0.5 cm of the length of the shoe (because of course we must use intervals here!)
Finding the P(x) – Example• 1. How many pairs should we stock of the 25
cm size?• We first need to convert this to a range
according to the assumption specified above, since we can only evaluate P(x) over an interval P(24.5 cm ≤ x ≤ 25.5 cm) is what we need to find
• Now we need to convert the bounds of our interval into z-scores:
x - Zlower = = 24.5 - 205
= 4.55
= 0.9
x - Zupper = = 25.5 - 205
= 5.55
= 1.1
Finding the P(x) – Example
• 1. How many pairs should we stock of the 25 cm size?
• We now have our interval expressed in terms of z-scores as P(0.9 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables
1.1
0.9
• We can calculate this area by finding P(0.9Z 1.1) = P(Z0.9) - P(Z1.1)
Finding the P(x) – Example• 1. How many pairs should we stock of the 25
cm size?• We now have our interval expressed in terms of
z-scores as P(0.9 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables
1.1
0.9 • We can calculate this area by finding P(0.9Z 1.1) = P(Z0.9) - P(Z1.1) = 0.1841 – 0.1357 = 0.0484
• Now multiply our P(0.9 ≤ Z ≤ 1.1) by total sales per week to get the number of shoes we should stock in that size: 300 х 0.0484 = 14.52 ≈ 15
Finding the P(x) – Example• 2. How many pairs should we stock of the 18 cm
size? (assuming shoes fit feet that are +/- 0.5 cm of the length of the
shoe) • Again, we convert this to a range according to the
assumption specified above, since we can only evaluate P(x) over an interval P(17.5 cm ≤ x ≤ 18.5 cm) is what we need to find
• Now we need to convert the bounds of our interval into z-scores:
x - Zlower = = 17.5 - 205
= -2.55
= -0.5
x - Zupper = = 18.5 - 205
= -1.55
= -0.3
Finding the P(x) – Example
• 2. How many pairs should we stock of the 18 cm size?
• We now have our interval expressed in terms of z-scores as P(-0.5 ≤ Z ≤ -0.3) for use in a standard normal distribution, which we can evaluate using our standard normal tables
-0.5-0.3
• We can calculate this area by finding P(-0.5Z-0.3) = P(Z-0.3) - P(Z-0.5)
Finding the P(x) – Example• 2. How many pairs should we stock of the 18 cm
size?• We now have our interval expressed in terms of z-
scores as P(-0.5 ≤ Z ≤ -0.3) for use in a standard normal distribution, which we can evaluate using our standard normal tables
-0.5-0.3 • We can calculate this area by finding
P(-0.5Z-0.3) = P(Z-0.3) - P(Z-0.5) = 0.3821 – 0.3085 = 0.0736• Now multiply our P(-0.5 ≤ Z ≤ -0.3) by total sales
per week to get the number of shoes we should stock in that size: 300 х 0.0736 = 22.08 ≈ 22
Finding the P(x) – Example
• 3. How many pairs should we stock in the 18 to 25 cm size range?
• As always, we convert this to a range according to the assumption specified above, since we can only evaluate P(x) over an interval P(17.5 cm ≤ x ≤ 25.5 cm) is what we need to find
• We have already found the appropriate Z-scores
x - Zlower = = 17.5 - 205
= -2.55
= -0.5
x - Zupper = = 25.5 - 205
= 5.55
= 1.1
Finding the P(x) – Example• 3. How many pairs should we stock in the 18 to
25 cm size range?• We now have our interval expressed in terms of z-
scores as P(-0.5 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables
-0.51.1
• We can calculate this area by finding P(-0.5Z1.1) = 1 – [P(Z-0.5) + P(Z1.1)] = 1 – [0.3085 + 0.1357] = 1 – 0.4442 = 0.5558• Now multiply our P(-0.5 ≤ Z ≤ 1.1) by total sales
per week to get the # of shoes we should stock in that range: 300 х 0.5558 = 166.74 ≈ 167
Commonly Used Probabilities
-3σ -2σ -1σ μ +1σ +2σ +3σ
f(x)
68%
95%
99.7%
-σ μ +σ
f(x)
Commonly Used Probabilities
1)(
xz
• Standard normal distribution:
P(z > 1) = 0.1587 p(-1 ≤ z ≤ 1) = 1 – 2*0.1587 = 0.6826
• Normal distribution as illustrated in the Figure:
P(μ-σ ≤ x ≤ μ-σ) = 0.6826 ≈ 0.68
• For a normal distribution, 68% of the observations lie within about one standard deviations of the mean
-2σ μ +2σ
f(x)
Commonly Used Probabilities
2)2(
xz
• Standard normal distribution:
P(z > 2) = 0.0228 p(-2 ≤ z ≤ 2) = 1 – 2*0.0228 = 0.9543
• Normal distribution as illustrated in the Figure:
P(μ-2σ ≤ x ≤ μ-2σ) = 0.9543 ≈ 0.95
• For a normal distribution, 95% of the observations lie within about two standard deviations of the mean
-3σ μ +2σ
f(x)
Commonly Used Probabilities
3)3(
xz
• Standard normal distribution:
P(z > 3) = 0.0013 p(-3 ≤ z ≤ 3) = 1 – 2*0.0013 = 0.9974
• Normal distribution as illustrated in the Figure:
P(μ-3σ ≤ x ≤ μ-3σ) = 0.9974 ≈ 0.997
• For a normal distribution, 99.7% of the observations lie within about two standard deviations of the mean
Commonly Used Probabilities
-3σ -2σ -1σ μ +1σ +2σ +3σ
f(x)
68%
95%
99.7%