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```Continuous probability distributions
Uniform probability distribution (ASW, section 6.1)
Normal probability distribution (ASW, section 6.2)
Table – Appendix B and inside front cover
Excel – Appendix 6.2
Bring the text to class on Wednesday, October 1. We
will be using Table 1 of Appendix B of ASW.
Notes for October 1, 2008
Probability density function
• For a continuous random variable x, the probability
that x takes on a specific value is zero. As a result,
probabilities for values of x are assigned across an
interval of values of x. The probability function f(x)
that is used to assign these probabilities is termed
the probability density function.
• If a variable x has probability density function f(x),
the probability that x takes on values between a and
b is the area under the graph of f(x) that lies between
a and b (or the integral of f(x) over the range a to b).
Uniform probability distribution (ASW, 226)
• This is a distribution where the probability density
function f(x) has the same value across all the
values of x for which it is non-zero.
 1
 b  a for a  x  b
f ( x)  

 0 elsewhere
Diagram of a uniform probability distribution
If x represents the variable being considered, the distribution
has density f(x) = 1/(b – a) over the range from a to b and
density of 0 elsewhere.
f(x)
1/(b-a)
x
a
b
Example – cost of travel
Suppose a firm reimburses employees at the rate of 40 cents per
kilometre when an employee uses his or her own automobile
for company travel. Over the past years, the number of
kilometres reimbursed in this manner has been between
100,000 and 150,000. The probability distribution of
anticipated annual travel costs for the firm is considered to be
a uniform distribution over this range.
At the lower bound of 100,000 km., the total cost would be
\$40,000 and at the upper bound of 150,000 km., the total cost
would be \$60,000. Since travel is not anticipated to be less
than 100,000 km. nor greater than 150,000 km., cost is zero
outside the lower and upper bound.
Uniform probability distribution for travel example
Let x be the anticipated total cost of annual travel
for the firm in thousands of dollars. Then the
uniform probability distribution is:
 1
 60  40 for 40  x  60
f ( x)  


0 elsewhere
Uniform probability distribution of total
expected travel cost for the firm
Let x represents the total anticipated annual travel costs in
thousands of dollars, the distribution has density f(x) =
1/(60 - 40) = 1/20 over the range from 40 to 60 and
density of 0 elsewhere.
f(x)
1/20
x
40
60
Travel cost in thousands
of dollars
Area of the distribution
f(x)
1/20
x
40
60
Travel costs
Note that the total area under the density function f(x)
between 40 and 60 equals 1.
Area = height x length = (1/20) x (60–40) = (1/20) x 20 = 1 and
this equals the probability that some value within the
range from 40 to 60 occurs.
Area and probability over an interval
f(x)
1/20
x
40
50
60
Travel costs
What is the probability that travel costs are between 40 and
50?
Area under the curve between 40 and 50
= height x length = (1/20) x (50–40) = (1/20) x 10 = 0.5
and this is the required probability.
Expected value and variance for a uniform
probability distribution
For the travel cost example:
ab
E ( x) 
2
(b  a )
Var ( x)   
12
ba

12
2
E(x) = (40 + 60)/2 = 50
2
Var(x) = (60–40)2/12 = 33.333
Standard deviation of x is the
square root of 33.333 or 5.774
Firm’s anticipated travel costs
If the uniform distribution applies, then the firm might
predict that travel costs will be \$50,000 annually
(the mean or expected value). However, there is
variability in anticipated costs, with the standard
deviation being \$5,774.
The probability that travel costs are within one
standard deviation of the mean of \$50,000 turns out
to be 0.5774. This is the area under the line
between 50,000 - 5,774 = 44,226 and 50,000 + 5,774
= 55,774. This is a distance of 11,548. The area
under the line is (11,548/20,000) = 0. 5774.
Note that the area under the curve within two standard
deviations of the mean is the whole distribution.
Features of a continuous probability distribution
ASW (228)
• For a continuous probability distribution, the probability for
the random variable must be defined over an interval, not at a
single value of the variable.
• The probability that the random variable takes on values
within an interval is the area under the curve of the density
function f(x) across that interval. (Or it equals the integral of
f(x) from the lower to upper bounds of the interval).
• Also note that the total area under the curve of the density
function equals one.
• Most continuous distributions do not have the linear or
straight line characteristic of the uniform distribution, but will
be nonlinear or curved. Tables of these distributions are
often available. These tables give the required areas or
probabilities.
Normal probability distribution
The normal probability distribution is the most
common and important of the continuous
probability distributions used in statistical and
econometric work.
Other names for the normal distribution are the
bell curve, since it has a sort of bell shape, and
the Gaussian distribution, after Gauss, who is
considered to be the first to have described
and used the distribution.
Formula and parameters for the normal
distribution
There are many normal distributions, but any normal
distribution can be described and graphed with two
parameters (µ and σ) and the following formula.

1
f ( x) 
e
 2
( x )2
2 2
where µ is the mean of the normal distribution
σ is the standard deviation of the normal distribution
π is 3.14159
e is 2.71828, the base of the natural logarithms
Some characteristics of the normal distribution
• The curve is entirely described by µ, the mean, and σ, the
standard deviation, using the formula above.
• The curve peaks at the mean, µ, so the mode also equals µ.
• The distribution is symmetric about the centre, µ, so the
median is also µ. The distribution is not skewed.
• The tails of the distribution never quite reach the horizontal
axis, but get closer and closer to this axis the further away
from the centre x is. This characteristic means that the
distribution is said to be asymptotic to the horizontal axis.
• The probability that a normally distributed variable x takes on
values in the range from a to b is the area under f(x) between
a and b.
• The total area under the curve is 1; the area under the curve
to the left of centre is 0.5 and the area right of centre is 0.5.
Reasons for using the normal distribution
• Describes some characteristics of populations. Eg. Height,
weight, and perhaps weight of packaged foods and travel time
to work. Some consider intelligence and ability to be normally
distributed. Grades for a large number of students across
classes are often normally distributed.
• Characteristics such as incomes, wealth, assets and debts,
farm size, and stock prices are usually not normal. But it is
sometimes possible to transform these to the normal.
• The normal provides an approximation to probabilities such
as the binomial when n is large, is the limiting distribution of
the t distribution, and forms the basis for other distributions.
• Many statistics obtained from random samples have a normal
distribution. In particular, when n is large, the sample means
from randomly selected samples haves a normal distribution
(ASW, 271).
Standard normal distribution (z)
• Each μ and σ define a different normal distribution for a
variable x.
• But any normally distributed variable can be transformed into
the standard normal variable (and vice-versa).
x    z
• The standard normal variable
has a mean of zero and a
standard deviation of 1 and is usually referred to as z.
• Any normally distributed variable x can be transformed into
the standard normal variable z by using the transformation
z
x

• The inverse transformation is x    z
Some probabilities for z
P(z < -1) = 0.1587
P(z > 1) = 1 – 0.8413 = 0.1587
P(z < -1.57) = 0.0582
P(z > 0.43) = 1 – 0.6664 = 0.3336
P (-1.37 < z < 1.75) = 0.9599 – 0.0853 = 0.8746
P (1.32 < z < 2.36) = 0.9909 – 0.9066 = 0.0843
P (-1 < z <1) = 0.8413 – 0.1587 = 0.6826
P (-2 < z < 2) = 0.9772 – 0.0228 = 0.9544
z values for areas
Area of 0.05 in the right tail of the distribution is obtained by
finding the z where the cumulative probability reaches 1 0.05 = 0.95, that is, at z = 1.64 or z = 1.65. For this area, z =
1.645 is often used.
Area of 0.025 in each tail of the distribution, or a total of 0.05 in
the two tails. The cumulative probability first reaches 0.025 at
z = -1.96. By symmetry, the z value in the right tail is a 1.96.
The interval (-1.96, 1.96) contains 95% of the distribution
leaving a total of 5% in the two tails of the distribution.
Total area of 0.01 in the two tails is given by the area to the left
of z = -2.575 and to the right of z = 2.575.
The above z values will be used extensively later in the semester.
Per cent of
<50
7.5
50-60
16.3
60-70
26.6
70-80
30.0
80-90
16.6
90+
3.0
Total
100.0
For this distribution, μ = 69 and σ = 14
Calculations for two intervals of grade
distribution
1. Grade less than x = 50?
z = (x-μ)/σ = (50 – 69)/14 = -1.36 and the cumulative probability
is 0.0869. If exactly normal, 8.7% of grades would be less than
50, whereas 7.5% actually were less than 50.
2. Grade of 80 to 90?
For x = 90, z = (x-μ)/σ = (90 – 69)/14 = 1.50. Cum P = 0.9332
For x = 80, z = (x-μ)/σ = (80 – 69)/14 = 0.79. Cum P = 0.7852
Area between these values is 0.9332 – 0.7852 = 0.1480 or
14.8%, which is a little less than the 16.6% who received
Comparing actual and normal distributions
Actual per
cent of
Per cent if
normally
distributed
<50
7.5
8.7
50-60
16.3
17.4
60-70
26.6
26.7
70-80
30.0
25.7
80-90
16.6
14.8
90+
3.0
6.7
Total
100.0
100.0
And the actual distribution is close to the normal distribution,
especially for grades up to 70. Note that fewer grades of 90 or
more were awarded than if the distribution was exactly normal.
If grades are normally distributed with μ = 69 and
σ = 14, what grade is required to:
1. Be in the upper 5% of all grades?
Upper 5% or 0.05 begins where the cumulative probability
reaches 1 - 0.05 = 0.95 and this is at z = 1.645. Rearranging
the formula z = (x-μ)/σ to solve for x gives
x = μ + (zσ) = 69 + (1.645 x 14) = 69 + 23.03 or x = 92.
2. Not be in the lower 20% for all grades?
The cumulative probabilities first reach 0.20 at z = -0.84. Using
the same formula as above to transform this z into an x gives
x = μ + (zσ) = 69 + (-0.84 x 14) = 69 – 11.76 or x = 57.24
and a grade of 58 would ensure that one is not in the lower
20% or one-fifth of the distribution.
Note that the z value is equivalent to the number of standard
deviations the value of the normal variable is from the mean
(ASW, 238).
Most of the distribution is within 3 standard deviations or 3 z
values of the mean. That is, the probability of any normal
variable being more than 3 z values from the mean is 0.003.
Excel can be used to obtain normal probabilities. See ASW, 255.
We will study section 6.3 of the text, normal approximation of
binomial probabilities, when we study sections 7.6 and 8.4 of
the text.
Skip section 6.4.
Next day
• Sampling and sampling distributions – ASW,
chapter 7.
• Begin interval estimation – ASW, chapter 8.
```
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