CONTINUOUS DISTRIBUTIONS
The random variable can assume literally any value in some range
Ø if x can be any value between, say, 0 and 12, the probability of any particular value is zero
12
Albert
Beauregard
6
Ø But a value between 2 and 6 is more likely than a value between 1 and 1.25
„ in the childhood game of “spin-the-bottle,” the probability of being kissed depends on
the size of the person
„ intuitively, Beauregard is more likely to be kissed than Slim Albert
Define a probability density function for each different kind of distribution
probabilities will be represented by areas under the graph of the probability density function
The Uniform Distribution: the spin-the-bottle distribution
simplest possible continuous distribution
graph:
vertical axis does NOT represent probability but rather probability density
normal probabilities, page 1
density measure along the vertical axis is chosen so that the area under the graph = 1: here,
(1/12) × 12 = 1
the region marked P(1 ≤ x ≤ 1.5) has area (1/12) × 0.5 = 1/24, while that marked P(6 ≤ x ≤
9) has area (1/12) × 3 = 1/4, and so on
the probability of x values occurring in a particular range is given by the area under the curve in
that range; that area is also the proportion of the total area in that range
THE NORMAL DISTRIBUTION
Reminder: the purpose of these various distributions is to represent mathematically some real-world
processes. The normal distribution can represent (or approximate) a startling variety of such
processes
General characteristics
• continuous
• symmetric about its mean
• values of x close to the mean are more likely than those further out
formula:
f ( x) =
1
2πσ

2
(
x
−
µ
)

−

2σ 2 

×e
where π = 3.14159… and e = 2.718281828…
Graph: a "bell curve"
height of the curve, as such, means little: probabilities are represented as areas under the
curve.
normal probabilities, page 2
on a distribution with µ = 200 and σ = 30, the probability of values between 225 and 250 is
given by the shaded area
the only things in the normal formula subject to change are µ and σ
π and e are natural constants
changing µ and σ will generate a whole family of distributions
changing µ shifts location:
increasing σ increases the dispersion in the distribution, as below
Examples of normal (or "Gaussian") processes
„ measures of humanity
weight
height
IQ scores
„ expectations of inflation held by individuals
„ length, weight, diameter, etc. of manufactured parts
weight of contents of cereal boxes, soap boxes, oil cans
normal probabilities, page 3
diameters of ball bearings, tires, etc.
„ deviations of prices from expected values
„ changes in prices of a corporation's stock shares
„ the deflection of an electron struck by gamma radiation
„ sample means of repeated samples drawn from the same population
Comment: Laplace could arrive at such a complex formula because he knew what he was looking
for
FINDING NORMAL PROBABILITIES
There are infinitely many normal distributions, one for each µ, σ pair
integration techniques?
in practice, tremendously useful fact about the normal distribution: all that really matters in
calculating probabilities on a normal distribution is the number of standard deviations from
the mean
for any two normal distributions, no matter how different their µ and σ, the same proportion of
the distributions lie in the range µ to µ + r × σ where r is any real number
Example: weights of adult human males are normally distributed with µ = 165 and σ =
15; a particular size of ball bearing has normally distributed diameters with µ = 10 mm
and σ = 0.01. Then the same proportion of males have weights between 165 and 180
(34.3% as it happens) as ball bearings have diameters between 10 and 10.01 mm;
likewise, the same proportion of males have weights between 135 and 150 as ball
bearings have diameters between 9.98 and 9.99 mm.
Allows us to lay out a table for only one distribution
w Distribution actually used to derive probabilities has µ = 0 and σ = 1, the "standard normal
distribution"
w This table may be arranged in several ways: Appendix E
Table E.2a: Standardized Normal Distribution Table gives areas between the mean and the number
of standard deviations given in the z column (extreme left hand)
the row across the top allows the extension to hundredths
„ P(0 ≤ z ≤ 1.5) = 0.4332
„ P(0 ≤ z ≤ 1.54) = 0.4382
normal probabilities, page 4
Table E.2b: Cumulative Standardized Normal Distribution gives cumulative less-than-or-equal
probabilities
in the sketch, the crosshatched area is P(z ≤ 1.5)
Examples:
„ P(z ≤ −2) = 0.0228
„ P(z ≤ 1.5) = 0.9332
notice that this is exactly 0.5 more than the probability found above
that is because P(z ≤ 0) = P(z ≥ 0) = 0.5.
„ P(z ≥ x) = 1 − P(z ≤ x)
„ P(z ≥ 1.5) = 1 − P(z ≤ 1.5) = 1 − 0.9332 = 0.0668.
„ P(z0 ≤ z ≤ z1) = P(z ≤ z1) − P(z ≤ z0)
P(−1.58 ≤ z ≤ 2.1) = P(z ≤ 2.1) − P(z ≤ −1.58) = 0.9821 − 0.0571 = 0.925
Reverse of this process is sometimes useful, that is,
Find z0 such that P(z ≤ z0) has some given value
Examples:
„ find z0 such P(z ≤ z0) = 0.4
find 0.4 in the body of the table (well, 0.3974)
and read backwards to find z0 = − 0.26
„ find z* such that 80% of all z's are less than z*
find 0.8 in the body of the table (0.7995)
and read backwards to find z = 0.84
note that 0.84 is the 80-th percentile of the z distribution
„ find zH such that there is only a 4% probability that z > zH
P(z ≤ zH) = 1 − 0.04 = 0.96, so zH = 1.75
„ find zL and zH such that the middle 90% of the distribution is between these values
by symmetry zL = − zH; since 90% is between, there must be 5% in either tail, so that
the required values imply P(z ≤ zL) = 0.05 and P(z ≤ zH) = 0.95. Accordingly, zL = −
1.64 and zH = +1.64
normal probabilities, page 5
MORE GENERAL NORMAL DISTRIBUTIONS
z values are numbers of standard deviations on a particular normal distribution; the probabilities will be
the same for the same number of standard deviations on any normal distribution.
To find P(x ≤ x0) we need only find out how many standard deviations x 0 is from the mean of
its distribution and look up the result in the z table.
We’ve seen this before: z = (x − µ)/σ expresses the difference between x and µ in terms of σ
FUNDAMENTAL EQUATIONS FOR USE WITH THE NORMAL DISTRIBUTION
To convert values of any normal distribution into values on the standard normal:
z=
x −µ
σ
Expresses deviations from the mean in numbers of standard deviations
Examples:
„ adult males have mean weight µ = 165 lb., with σ = 15 lb. Charley weighs 172.5
lb. His z score is z = (x − µ)/σ = (172.5 − 165)/15 = 0.5
Put another way, Charley's weight is greater than the mean by 0.5 standard deviation
„ Sam weighs 135 lb. His z score is z = (135 − 165)/15 = −2
Sam's weight is 2 standard deviations below the mean
z values may be positive or negative, depending on whether a given x value is greater or less
than the mean
To find an x value which is a given number of standard deviations from the mean:
x = µ+ z × σ
used when we want to find an x range that has a given probability
Examples:
„ adult females have a mean weight of 132 lb. with a standard deviation of 14 lb. Find
the weights that are one standard deviation above and below the mean weight.
xL = µ + z × σ = 132 + (−1) × 14 = 118
xH = µ + z × σ = 132 + 1 × 14 = 146
comment: by principles we’ve seen before 68.26% of all adult females have weights between
132 and 146 lb.
normal probabilities, page 6
FINDING PROBABILITIES ON NORMAL DISTRIBUTIONS
Procedure
convert x values on a given distribution to z scores
use z table to find appropriate probabilities
Examples:
Ø the random variable x is normally distributed with µ = 320 and σ = 12.
„ Find P(x ≤ 340)
z = (x − µ)/σ = (340 − 320)/12 = 20/12 = 1.67
P(z ≤ 1.67) = 0.9525
„ find P(x ≥ 330)
z = (330 − 320)/12 = 0.83
P(z ≥ 0.83) = 1 − P(x ≤ 0.83) = 1 − 0.7967 = 0.2033
Ø with µ = 132 and σ = 14, find the proportion of females whose weight exceeds 154
lb.
„ z = (x − µ)/σ = (154 − 132)/14 = 22/14 = 1.57
„ P(z ≥ 1.57) = 1 − P(z < 1.57) = 1 − 0.9418 = 0.0582
„ a woman is selected at random. What is the probability that she weighs less than
123 lb.?
z = (123 − 132)/14 = −9/14 = −0.64
P(z < −0.64) = 0.2611
Ø x is normally distributed with µ = 0.6 and σ = 0.03. What is the probability that a
randomly selected x value will be between 0.53 and 0.62?
zL = (0.53 − 0.6)/0.03 = −2.33
zH = (0.62 − 0.6)/0.03 = 0.67
P(−2.33 ≤ z ≤ 0.67) = 0.7486 − 0.0099 = 0.7387
Ø with mean 165 and standard deviation 15, what proportion of adult males have
weights between 180 and 190 lb.?
zH = (190 − 165)/15 = 1.67, zL = (180 − 165)/15 = 1
P(1 ≤ z ≤ 1.67) = 0.9525 − 0.8413 = 0.1112
FINDING x VALUES FOR GIVEN PROBABILITIES
Procedure:
• find z value corresponding to given probability
z will be positive or negative as probability is > or < 0.5
• use second fundamental equation to find x value
or substitute appropriate values into first equation and solve for x
normal probabilities, page 7
Examples:
Ø x is normally distributed with µ = 23 and σ = 2.
„ Find x0 such that 60% of the distribution is less than x0
in the body of the z table find 0.6 (0.5987).
Read backwards to z = 0.25.
x = µ + z × σ = 23 + 0.25 × 2 = 23 + 0.50 = 23.50
„ Find x0 such that 32% are less than x0
find 0.32 probability and read backwards to z = − 0.47
x = µ + z × σ = 23 + (−0.47) × 2 = 23 − 0.94 = 22.06
Ø W is normally distributed with µ = 4000 and σ = 162. Find a symmetric interval that
contains 90% of the W values
Find a WH such that 95% of the W values are less than WH and a WL such that 5%
of the W's are less than W0
in the z table, find probability 0.05 (0.0495), then zL = −1.64
in the z table, find probability 0.95 (0.9495), then zH = +1.64. Since the z distribution
is symmetric, one or the other of these efforts is redundant.
WH = 4000 + 1.64 × 162 = 4265.68
WL = 4000 − 1.64 × 162 = 3734.32
ADDITIONAL EXAMPLES:
Ø Fruit Loops boxes are filled to a mean weight of 10 oz. with a standard deviation of 0.2 oz.
„ what proportion of these boxes have weights less than 9.8 oz? (0.1587)
„ Fruit Loopers want to guarantee that their boxes contain at least some minimum weight; it
would be prohibitively expensive to assure that every single box does contain that weight, but
they wish to have no more than 2% of boxes that have less than the guaranteed weight. With
machinery operating as above, what weight can Fruit Loopers safely guarantee?
put another way, find W0 such that only 2% of all boxes contain less than W0
answer: 9.588.
Ø Courageous Couriers Company has recorded data on delivery times of packages within the
city. For all times recorded, µ = 25 min, and σ = 4 min.
„ find the probability that a delivery time will be between 21 and 29 minutes (68.26%)
„ what proportion of delivery times exceed 31 min? (0.0668)
„ what time can CCC guarantee if they want to have to pay off on non-delivery no more than
1% of the time? (34.32 min.)
Ø granules of industrial diamond have a mean size of 1/10 carat with σ = 0.01 carat. Find C0
such that 95% of all granules are at least as big as C0.
C0 = µ + z.05 × σ = .1 + (−1.64) × 0.01 = 0.0836
normal probabilities, page 8
Ø Amalgamated Rat Trap has mean daily orders of 1920 cases of rat traps with µ = 300. ART
would like to guarantee same-day shipment of orders.
„ how much inventory must they hold to have no more than a 6% probability of being unable
to ship the day an order is received? (1920 + 1.56 × 300 = 2388)
„ how much inventory must ART hold to have no more than a 40% probability of being
unable to honor their guarantee? (1920 + 0.26 × 300 = 1998)
„ is the larger inventory worthwhile?
If cost of one case of rat traps = $100. holding an additional 390 cases means
additional $39,000 worth of inventory. If the relevant interest rate is 10%, ART must
incur $3900 a year of additional inventory cost to achieve the lower probability
normal probabilities, page 9