DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION
SYSTEMS
Introduction to Business Statistics
QM 120
Chapter 5
Discrete random variables and their probability
distribution
Spring 2008
Dr. Mohammad Zainal
Chapter 5: Random Variables
2
Random variables
# of PC’s owned
Frequency
Relative Frequency
0
120
.12
1
180
.18
2
470
.47
3
230
.23
23
N = 1000
Sum = 1.000
Let x denote the number of PCs owned by a family. Then x
can take any of the four possible values (0, 1, 2, and 3).
¾
A random variable (RV) is a variable whose value is
determined by
y the outcome of a random experiment.
p
¾
Chapter 5: Random Variables
3
Discrete random variable
A random variable that assumes countable values is called a
discrete random variable.
¾
¾
Examples of discrete RVs:
¾
Number of cars sold at a dealership during a week
¾
Number of houses in a certain block
¾
Number of fish caught
g on a fishing
g trip
p
¾
Number of costumers in a bank at any given day
Continuous random variable
Continuous random variable
A random variable that can assume any value contained in
one or more intervals is called a continuous random variable.
¾
Chapter 5: Random Variables
4
¾
Examples of continuous RVs:
¾
Height of a person
¾
Time taken to complete a test
¾
Weight
g of a fish
¾
Price of a car
Example: Classify each of the following RVs as discrete or E
l Cl if
h f h f ll i RV
di
continuous.
¾
Th number
The
b off new accounts
t opened
d att a bank
b k during
d i a week
k
¾
The time taken to run a marathon
¾
The price of a meal in fast food restaurant
¾
The score of a football game
¾
The weight of a parcel
Chapter 5: Probability Distribution of a Discrete RV
5
The probability distribution of a discrete RV lists all the
possible
bl
value
l
that
h
the
h
RV can assume and
d their
h
corresponding probabilities.
¾
Example: Write the probability distribution of the number of
PCs owned by a family.
# of PC’s owned
Frequency
Relative Frequency
0
120
.12
1
180
.18
2
470
.47
3
230
.23
N 1000
N = 1000
S
Sum =
1 000
1.000
Chapter 5: Probability Distribution of a Discrete RV
6
The following two characteristics must hold for any discrete probability distribution:
b bl d
b
¾
¾
The probability assigned to each value of a RV x lies in the range 0
to 1; that is 0 ≤ P(x) ≤ 1 for each x.
¾
The sum of the probabilities assigned to all possible values of x is
equal to 1.0; that is ΣP(x) = 1.
Example: Each of the following tables lists certain values of x
and their probabilities. Determine whether or not each table
represents a valid probability distribution.
x
P(x)
x
P(x)
x
P(x)
0
.08
0
.25
4
.2
1
.11
1
.34
5
.3
2
.39
39
2
.28
28
6
.6
6
3
.27
3
.13
8
‐.1
Chapter 5: Probability Distribution of a Discrete RV
7
Example: The following table lists the probability distribution of a discrete RV x.
x
0
P(x)
1
2
3
4
5
6
.11 .19 .28 .15 .12 .09 .06
a) P(x = 3)
a)
P(
3)
b) P( ≤ 2)
b) P(x ≤
c) P(x ≥
) P( ≥ 4)
d) P(1 ≤ x ≤
d) P(1 ≤
≤ 4)
e) Probability that x assumes a value less than 4
f) Probability that x assumes a value greater than 2
f) Probability that x assumes a value greater than 2
g) Probability that x assumes a value in the interval 2 to 5
Chapter 5: Probability Distribution of a Discrete RV
8
Example: For the following table
x
1
2
3
4
5
P(x)
8
20
24
16
12
a) Construct a probability distribution table. Draw a graph of a)
Construct a probability distribution table. Draw a graph of
the probability distribution.
b) Find the following probabilities
i. P(x = 3)
ii. P(x < 4)
iii. P(x ≥ 3)
iv. P(2 ≤ x ≤ 4)
Chapter 5: Mean of a discrete RV
9
Mean of a discrete RV
The mean μ ‐or expected value E(x)‐ of a discrete RV is the
value that you would expect to observe on average if the
experiment
i
t is
i repeated
t d again
i and
d again
i
¾
¾
It is denoted by E (x ) = ∑ xp (x )
Illustration: Let us toss two fair coins, and let x denote the
number of heads observed.
observed We should have the following
probability distribution table
¾
x 0
P(x) 1/4
1
2
1/2
1/4
Suppose we repeat the experiment a large number of times,
say n =4,000.
=4 000 We should expect to have approximately
Chapter 5: Mean of a discrete RV
10
1 thousand zeros, 2 thousand ones, and 1 thousand twos. Then th
the average value of x would equal l
f
ld
l
Sum of measurements 1, 000 (0) + 2, 000 (1) + 1000 (2)
=
4 000
4,
n
⎛1⎞
⎛1⎞
⎛1⎞
= ⎜ ⎟ (0) + ⎜ ⎟ (1) + ⎜ ⎟ (2)
⎝4⎠
⎝2⎠
⎝4⎠
Similarly, if we use the E (x ) = ∑ xp (x ), we would have
E (x ) = 0P (0) + 1P (1) + 2P (2)
= 0(1/ 4) + 1(1/ 2) + 2(1/ 4)
=1
Chapter 5: Mean of a discrete RV
11
Example: Recall “ number of PC’s owned by a family” example. Find the mean number of PCs owned by a family.
l F d h
b
f PC
db
f
l
Chapter 5: Mean of a discrete RV
12
Example: In a lottery conducted to benefit the local fire
company 8000 tickets are to be sold at $5 each.
company,
each The prize is a
$12,000. If you purchase two tickets, what is your expected
gain?
Chapter 5: Mean of a discrete RV
13
Example: Determine the annual premium for a $1000
insurance policy covering an event that over a long period of
time, has occurred at the rate of 2 times in 100. Let
x : the yearly financial gain to the insurance company
resulting from the sale of the policy
C : unknown premium
Calculate the value of C such that the expected gain E(x) will
equal to zero so that the company can add the administrative
costs and profit.
Chapter 5: Mean of a discrete RV
14
Example: You can insure a $50,000 diamond for its total value
by
yp
paying
y gap
premium of D dollars. If the p
probability
y of theft in
a given year is estimated to be .01, what premium should the
insurance company charge if it wants the expected gain to
equal $1000.
Chapter 5: Standard Deviation of a Discrete RV
15
¾
¾
¾
The standard deviation of a discrete RV x, denoted by σ,
measures the spread of its probability distribution.
A higher value of σ indicates that x can assume values over
a larger range about the mean. While, a smaller value
indicates that most of the values that can x assume are
clustered closely around the mean.
The standard
Th
d d deviation
d i i σ can be
b found
f
d using
i the
h following
f ll i
formula:
σ=
¾
∑x
2
p (x ) − μ 2
Hence, the variance σ2 can be obtained by squaring its
standard deviation σ.
σ
Chapter 5: Standard Deviation of a Discrete RV
16
Example: Recall “ number of PC’s owned by a family”
example. Find the standard deviation of PCs owned by a
family.
Chapter 5: Standard Deviation of a Discrete RV
17
Example: An electronic store sells a particular model of a
computer notebook. There are only four notebooks in stocks,
and the manager wonders what today’s demand for this
particular model will be. She learns from marketing
department that the probability distribution for x,
x the daily
demand for the laptop, is as shown in the table.
x
P(x)
1
2
3
4
5
.40 .20 .15 .10 .05
Find the mean, variance, and the standard deviation of x. Is it
lik l that
likely
th t five
fi or more costumers
t
will
ill wantt to
t buy
b a laptop?
l t ?
Chapter 5: Standard Deviation of a Discrete RV
18
Example: A farmer will earn a profit of $30,000 in case of
heavy rain next year, $60,000 in case of moderate rain, and
$15,000 in case of little rain. A meteorologist forecasts that the
probability is .35 for heavy rain, .40 for moderate rain, and .25
for little rain next year.
year Let x be the RV that represents next
year’s profits in thousands of dollars for this farmer. Write the
probability distribution of x. Find the mean, variance, and the
standard deviation of x.
Chapter 5: Standard Deviation of a Discrete RV
19
Example: An instant lottery ticket costs $2. Out of a total of
10,000 tickets printed for this lottery, 1000 tickets contain a
prize of $5 each. 100 tickets contain a prize of $10 each, 5
tickets contain a prize of $1000 each, and 1 ticket has the prize
of $5000.
$5000 Let x be the RV that denotes the net amount player
wins by playing this lottery. Write the probability distribution
of x. Determine the mean and the standard deviation of x.
Chapter 5: Standard Deviation of a Discrete RV
20
Example: Based in its analysis of future demand for its
products, the financial department a company has determined
that there is a .17 probability that the company will lose $1.2
million during the next year, a .21 probability that it will lose
$ 7 million,
$.7
million a .37
37 probability that it will make a profit of $0.9
$0 9
million, and a .25 probability that it will make a profit of $2.3
million.
a) Let x be a RV that denotes the profit earned by this
company during the next year. Write the probability
distribution of x.
x
Chapter 5: Standard Deviation of a Discrete RV
21
b) Find the mean and standard deviation of the probability of
part a.
Chapter 5: Factorials & Combinations
22
Factorials
¾ The symbol
y
n!, reads as “n factorial,” represents
p
the
product of all integers from n to 1. In other words,
n! = n(n ‐ 1)(n ‐ 2)(n ‐ 3)…3.2.1
Example: Evaluate 7!
7! = 7.6.5.4.3.2.1 = 5040
Example: Evaluate (12‐4)!
(12‐4)! = (8)! = 8.7.6.5.4.3.2.1 =40,320
Chapter 5: Factorials & Combinations
23
Example: Evaluate (8‐8)!
(8‐8)!
(8
8)! = (0)! = 1
Example:
p Find the value of 8!
By using factorial Table or your calculator.
Locate 8 in the column labeled n.
Then read the value for n! next
to 8.
Chapter 5: Factorials & Combinations
24
Combinations
¾ Combinations g
give the number of ways
y x elements can be
selected from n elements. The notation used to denote the
total number of combinations is
n
¾
C x = C xn = ( nx )
It can be found using the following formula
C xn = n C x =
¾
n!
x!(n − x)!
Remember: n is always greater than or at least equal to x.
Chapter 5: Factorials & Combinations
25
Example: Three members of a jury will be randomly selected
from five people. How many different combinations are
possible?
Chapter 5: Factorials & Combinations
26
Using the table of combinations
¾ Values of Table III in Appendix C lists the number of combinations of n elements selected x at a time.
Example: Evaluate
the followings
a) 5C3
b) 8C4
c) ) 9C5
d) 5C5
e) 5C0
e) Chapter 5: The Binomial Probability Distribution
27
The most widely used discrete probability distribution.
¾
It is applied to find the probability that an event will occur
x times in n repetitions of an experiment (under certain
conditions).
¾
Suppose the probability that a VCR is defective at a factory
is .05. We can apply the binomial probability distribution to
fi d the
find
th odds
dd off getting
tti exactly
tl one defective
d f ti VCR outt three
th
VCRs.
Chapter 5: The Binomial Probability Distribution
28
Binomial experiment
¾ An experiment
p
that satisfies the following
g four conditions is
called a binomial experiment.
¾
¾
There are n identical trials
¾
Each trail has only two possible outcomes.
¾
The probabilities of the two outcomes remain constant.
¾
The trials are independent.
independent
A success does not mean that the corresponding outcome is
considered favorable. Similarly, a failure doesnʹt necessarily
refer to unfavorable outcome.
Chapter 5: The Binomial Probability Distribution
29
Example: Consider the experiment consisting of 10 tosses of a
coin. Determine whether or not it is a binomial experiment.
Solution:
Chapter 5: The Binomial Probability Distribution
30
The binomial probability distribution and its formula.
¾ The RV x that represents
p
the number of successes in n trials
for a binomial experiment is called a binomial RV.
¾
The probability distribution of x is called the binomial
distribution.
¾
Consider the VCRs example.
p
Let x be the number of
defective VCRs in a sample of 3, x can assume any of the
values 0, 1, 2, 3.
Chapter 5: The Binomial Probability Distribution
31
¾
For a binomial experiment, the probability of exactly x
successes in n trials is given by the binomial formula
P ( x ) = n C x p x q n− x
=
where
n!
p x q n− x
x !( n − x )!
n = totall number
b off trials,
i l
p = probability of success,
q = probability of failure 1 – p,
p
x = number of successes in n trials,
n – x = number of failures in n trials
Chapter 5: The Binomial Probability Distribution
32
Example: Find P(2) for a binomial RV with n = 5 and p =0.1
Solution:
Chapter 5: The Binomial Probability Distribution
33
Example: Over a long period of time it has been observed that
a given sniper can hit a target on a single trial with a
probability = .8. Suppose he fires four shots at the target.
a) What is the probability that he will hit the target exactly two
times?
b) What is the probability that he will hit the target at least
once?
Chapter 5: The Binomial Probability Distribution
34
Example: Five percent of all VCRs manufactured by a large
factory are defective. A quality control inspector selects three
VCRs from the production line. What is the probability that
exactly one of these three VCRs is defective
Solution:
Chapter 5: The Binomial Probability Distribution
35
Example: At the Express Delivery Services (EDS), providing
high‐quality service to its customers is the top priority of the
management. The company guarantees a refund of all charges
if a package is not delivered at its destination by the specified
time It is known from the past that despite all efforts,
time.
efforts 2% of
the packages mailed through this company does not arrive on
time. A corporation mailed 10 packages through EDS.
a) Find the probability that exactly 1 of these 10 packages will
not arrive on time
b) Find the probability that at most 1 of these 10 packages will
not arrive on time.
time
Chapter 5: The Binomial Probability Distribution
36
Example: According to the U.S. Bureau of Labor Statistics, 56%
of mothers with children under 6 years of age work outside
their homes. A random sample of 3 mothers with children
under 6 years of age is selected. Let x denote the number of
mothers who work outside their homes.
homes Write the probability
distribution of x and draw a graph of the probability
distribution.
Solution:
Chapter 5: The Binomial Probability Distribution
37
Using the table of binomial probabilities
¾ We can use the tables of binomial p
probabilities (Table IV)
given in Appendix C to calculate the required probabilities.
P
n
x
.05
.1
.2
…
.95
1
0
.9500
.9000
.8000
…
.0500
1
.0500
.1000
.2000
…
.9500
0
.9025
.8100
.6400
…
.0025
1
.0950
.1800
.3200
…
.0950
2
.0025
.0100
.0400
…
.9025
0
.8574
.7290
.5120
…
.0001
1
.1354
.2430
.3840
…
.0071
2
.0071
.0270
.0960
…
.1354
3
.0001
.0010
.0080
…
.8574
2
3
Chapter 5: The Binomial Probability Distribution
38
Example: Based on data from A peter D. Hart Research
Associates’ poll on consumer buying habits and attitudes, It was
estimated that 5% of American shoppers are status shoppers,
that is, shoppers who love to buy designer labels. A random
sample of eight American shoppers is selected.
selected Using Binomial
Table, answer the following:
a)) Find the p
probability
y that exactly
y 3 shoppers
pp
are status
shoppers.
b) Find the probability that at most 2 shoppers are status
shoppers.
pp
Chapter 5: The Binomial Probability Distribution
39
c) Find the probability that at least 3 shoppers are status
shoppers.
d) Find the probability that 1 to 3 shoppers are status shoppers.
e) Let x be the number of status shoppers, Write the probability
distribution of x and draw a graph of this probability.
Chapter 5: The Binomial Probability Distribution
40
Probability of success and the shape of the binomial
distribution
For any number of trials n
¾
The binomial probability distribution is symmetric if p = .5
0.4
p
0.35
x
.05
.1
…
.5
…
.9
.95
0.3
4
0
.8145
.6561
…
.0625
…
.0001
.0000
0.25
1
.1715
.2916
…
.2500
…
.0036
.0005
2
.0135
.0486
…
.3750
…
.0486
.0135
0.1
0.05
3
.0005
.0036
…
.2500
…
.2916
.1715
4
.0000
.0001
…
.0625
…
.6561
.8145
P(x)
n
0.2
0.15
0
0
1
2
x
3
4
Chapter 5: The Binomial Probability Distribution
41
The binomial probability distribution is skewed to the right if p is less than .5
0.7
p
0.6
x
.05
.1
…
.5
…
.9
.95
0.5
4
0
.8145
.6561
…
.0625
…
.0001
.0000
0.4
1
.1715
.2916
…
.2500
…
.0036
.0005
2
.0135
.0486
…
.3750
…
.0486
.0135
3
.0005
.0036
…
.2500
…
.2916
.1715
4
.0000
.0001
…
.0625
…
.6561
.8145
P(x)
n
0.3
0.2
0.1
0
0
1
2
3
4
x
Chapter 5: The Binomial Probability Distribution
42
The binomial probability distribution is skewed to the left if p is greater than .5
0.7
p
0.6
x
.05
.1
…
.5
…
.9
.95
0.5
4
0
.8145
.6561
…
.0625
…
.0001
.0000
0.4
1
.1715
.2916
…
.2500
…
.0036
.0005
2
.0135
.0486
…
.3750
…
.0486
.0135
3
.0005
.0036
…
.2500
…
.2916
.1715
4
.0000
.0001
…
.0625
…
.6561
.8145
P(x)
n
0.3
0.2
0.1
0
0
1
2
3
x
4
Chapter 5: The Binomial Probability Distribution
43
Mean and standard deviation of the binomial distribution
¾ Although
g we can still use the mean and standard deviations
formulas learned in 5.3 and 5.4, it is more convenient and
simpler to use the following formulas once the RV x is
known to be a binomial RV
μ = np and σ = npq
Chapter 5: The Binomial Probability Distribution
44
Example: Refer to the mothers with children under 6 years of
age example, find the mean and the standard deviation of the
probability distribution.
Solution:
Chapter 5: The Binomial Probability Distribution
45
Example: Let x be a discrete RV that posses a binomial
distribution. Using the binomial formula, find the following
probabilities.
a) P(x = 5) for
f n = 8 and
d p = .6
b) P(x = 3) for n = 4 and p = .3
c) P(x = 2) for n = 6 and p = .2
Verify your answer by using Table of Binomial Probabilities.
Chapter 5: The Binomial Probability Distribution
46
Example: Let x be a discrete RV that posses a binomial
distribution.
a) Table of Binomial Probabilities, write the probability
distribution for x for n = 7 and p = .3 and graph it.
b) What are the mean and the standard deviation of the
probability distribution developed in part a?
Chapter 5: The Binomial Probability Distribution
47
¾
An experiment that satisfies the following four conditions is
called a binomial experiment.
¾
There are n identical trials
¾
Each trail has only two possible outcomes.
¾
The probabilities of the two outcomes remain constant.
¾
The trials are independent. Chapter 5: The Hypergeometric Probability Distribution
48
¾
We learned that one of the conditions required to apply the
binomial distribution is that the trials are independent so
the probability of the two outcomes remain constant.
¾
What if the probability of the outcomes is not constant?
¾
In such cases we replace the binomial distribution by the
hypergeometric
yp g
probability
p
y distribution.
¾
Such a case occurs when a sample is drawn without
replacement from a finite population.
Chapter 5: The Hypergeometric Probability Distribution
49
Hypergeometric probability distribution
population
p
Let
N = total number of elements in the p
r = number of successes in the population
N – r = number of failures in the population
n = number of trials (sample size)
x = number
b off successes in
i n trials
ti l
n – x = number of failures in n trials
¾
The probability of x successes Th
b bili
f
in n trials is given by
P (x ) =
rC x
N −r C n −x
N Cn
The mean and the variance Th
d th
i
are given by
⎛ r
⎝N
⎛ r
σ2 =n⎜
⎝N
μ =n⎜
⎞
⎟
⎠
⎞⎛ N − r ⎞⎛ N − n ⎞
⎟⎜
⎟⎜
⎟
⎠⎝ N ⎠⎝ N − 1 ⎠
Chapter 5: The Hypergeometric Probability Distribution
50
Example: Dawn corporation has 12 employees who hold
managerial positions. Of them, 7 are female and 5 are male.
The company is planning to send 3 of these 12 managers to a
conference. If 3 mangers are randomly selected out of 12,
a) find the probability that all 3 of them are female
Chapter 5: The Hypergeometric Probability Distribution
51
b) find the probability that at most 1 of them is a female
Chapter 5: The Hypergeometric Probability Distribution
52
Example: A case of soda has 12 bottles, 3 of which contain diet
soda. A sample of 4 bottles is randomly selected from the case
a) find the probability distribution of x, the number of diet
sodas in the sample
Chapter 5: The Hypergeometric Probability Distribution
53
b) what are the mean and variance of x?
Chapter 5: The Hypergeometric Probability Distribution
54
Example: GESCO Insurance company has prepared a final list
of 8 candidates for 2 positions. Of the 8 candidates, 5 are
business majors and 3 are engineers. If the company manager
decides to select randomly two candidates from this list, find
the probability that
a) both candidates are business majors
b)) neither of the two candidates is a business major
j
c) at most one of the candidates is a business major
Chapter 5: The Poisson Probability Distribution
55
¾
The Poisson distribution is another discrete probability
distribution that has numerous practical applications.
¾
It provides a good model for data that represent the number
of occurrence of a specified event in a given unit of time or
space.
¾
Here are some examples of experiments for which the RV x
can modeled
d l d by
b the
h Poisson
P i
RV
RV:
¾
The number of phone calls received by an operator during a day
¾
Th
The number of customer arrivals at checkout counter during an hour
b
f
t
i l t h k t
t d i
h
¾
The number of bacteria per a cm3 of a fluid
¾
The number of machine breakdowns during a given day
The number of machine breakdowns during a given day
¾
The number of traffic accidents at a given time period
Chapter 5: The Poisson Probability Distribution
56
¾
The Poisson probability distribution is applied to
experiments with random and independent occurrences
¾
The occurrences are random in the sense they do not follow any pattern and, hence, the are unpredictable.
¾
IIndependence of occurrences means that the occurrence or d
d
f
th t th
nonoccurrence of an event does not influence the successive occurrence of that next event.
¾
The occurrences are always considered with respect to an
interval.
¾
The
h intervall may be
b a time interval,
l a space interval,
l or a
volume interval.
¾
If the
th average number
b off occurrences (λ) for
f a given
i
i t
interval
l
is known, then by using the Poisson probability we can
compute
p
the p
probability
y of a certain number of occurrences
x in that interval.
Chapter 5: The Poisson Probability Distribution
57
¾
¾
The following three conditions must be satisfied to apply
the Poisson probability distribution
¾
x is a discrete RV.
¾
The occurrences are random.
¾
The occurrences are independent.
The Poisson probability formula is given by
P (x ) =
¾
λ x e −λ
x!
where λ (pronounced lambda) is the mean number of
occurrences in that interval and the value of e is
approximately
i
l 2.71828.
2 71828
Chapter 5: The Poisson Probability Distribution
58
¾
¾
The mean number of occurrences, denoted by λ, is called
the parameter of the Poisson probability distribution or the
Poisson parameter.
Remember: The interval of λ and x must be of the same
length. If they are not, the mean λ should be redefined to
make
k them
h
equal.
l For
F instance,
i
if λ was given
i
i hours
in
h
and
d
we were asked to find the event in minutes, λ needs to be
divided by 60 to have both x and λ in the same interval
length
Chapter 5: The Poisson Probability Distribution
59
Example: The automatic teller machine (ATM) installed
outside Mansfield Savings and Loan is used on average by
five costumers per hour. The bank closed this ATM for one
hour for repairs. What is the probability that during that hour
eight customers came to use this ATM?
Solution:
Chapter 5: The Poisson Probability Distribution
60
Example: The average number of traffic accidents on a certain
section of highway is two per week. assume that the number of
accidents follows a Poisson distribution with λ = 2.
1. Find the probability of no accidents on this section of highway
during a 1‐week period.
2. Find the probability of at most three accidents on this section
of highway during a 2‐week period.
Chapter 5: The Poisson Probability Distribution
61
Using the table of Poisson probabilities
¾ The p
probabilities for Poisson distribution can also be found
using Poisson Probability Table
Chapter 5: The Poisson Probability Distribution
62
Example: On average, two new accounts are opened per day
at an Imperial Savings Bank branches. Using Table VI of
Appendix C, find the probability that on a given day the
number of new accounts opened at this bank will be
a) Exactly 6
b) At most 3
c) At least 7
Chapter 5: The Poisson Probability Distribution
63
Mean and standard deviation of the Poisson probability
distribution
¾ For
F the
th Poisson
P i
di t ib ti
distribution
μ = λ
σ2 = λ
=λ
Example: An auto salesperson sells an average of .9 cars per
day. Let x be the number of cars sold by this salesperson on
any given day. Using the Poisson probability table, write the
probability distribution of x, draw the probability distribution,
and
d find
fi d the
th mean and
d standard
t d d deviation
d i ti off x