Download Notes on Conditional Probability

Notes on Conditional Probability
1) Let A1, A2, …., An be n mutually exclusive events
where P(A1)+P(A2)+…+P(An)=1 (and collectively
exhaustive).
Then
P(B)=P(B|A1)P(A1)+P(B|A2)P(A2)+…+P(B|An)P(An)
2) The joint probability
P(A and B)=P(B|A)P(A)=P(A|B)P(B)
So the conditional probability becomes
P(A|B) = P(B|A)P(A)/P(B)
Example: (problem 35, chapter 5)
A team plays 70% of their games at night and 30% of their games
during the day.
A team wins 50% of their night games and 90% of their day games.
What is the probability that a randomly chosen game is a win?
Let W be the event the chosen game is a win.
Let N be the event that the game is a night game.
Let D be the event that the game is a day game.
P(W) = P(W|N)P(N) + P(W|D)P(D)
= (0.5)(0.7) + (0.9)(0.3) = 0.62
Example: (problem 35, chapter 5)
If yesterday’s game is a win, what is the probability that the game
was played at night.
Let W be the event the chosen game is a win.
Let N be the event that the game is a night game.
Let D be the event that the game is a day game.
P(N|W) = P(W and N)/P(W)
= P(W|N)P(N)/P(W)
= (0.5)(0.7)/0.62 = 0.5645
Baye’s Theorem: Notes on Conditional
Probability
P(B)=P(B|A1)P(A1)+P(B|A2)P(A2)+…+P(B|An)P(An)
and
P(Ai|B) = P(B|Ai)P(Ai)/P(B)
P( Ai ) P( B / Ai )
 P( Ai | B) 
P( A1 ) P( B | A1 )  P( A2 ) P( B | A2 )  ...  P( An ) P( B | An )
for i = 1,2,…,n.
Double check that A1,… An are collectively exhaustive by
checking for P(A1)+P(A2)+…+P(An)=1
Examples on Discrete Probability
Distributions.
Example: Toss a coin three times and let X be the number of
heads. What is the PDF and CDF of X?
Outcome
Prob.
X
HHH
1/8
3
HHT
1/8
2
HTH
1/8
2
HTT
1/8
1
THH
1/8
2
THT
1/8
1
TTH
1/8
1
TTT
1/8
0
x

P(X = x) F(x)=P(X ≤ x)
0
1/8
1/8
1
3/8
=1/8+3/8=1/2
2
3/8
=1/2+3/8=7/8
3
1/8
1
Examples on Discrete Probability
Distributions.
Example: Let X be the number of trials we roll a die until we get a 1. What is
the PDF of X?
P(X=1)=1/6
P(X=2)=(5/6)(1/6)
P(X=3)=(5/6)(5/6)(1/6)
P(X=4)=(5/6)3(1/6)
.
.
.
 P(X=i)=(5/6)(i-1) (1/6) for i=1,2,3,….
A Probability distribution function is
summarized by
Summarized
by its
MEAN –Long-run average value of the Random Variable
STANDARD DEVIATION.
Expected Value (Mean)
 Mathematically:
The expected value (or mean) of a RV X is
µ = E(X) =
 xp(x)
all x
 Sometimes
 Additivity:
write µX
E(X + Y) = E(X) + E(Y)
Expected Value (Mean)
 How
many heads “on average” will we get when tossing a coin
three times?
 If we repeat this many times, we expect that
 In
1/8 of the times, we’ll get 0
 In 3/8 of the times, we’ll get 1
 In 3/8 of the times, we’ll get 2
 In 1/8 of the times, we’ll get 3
 So
“on average” we’ll get (Average number of heads)
1
3
3
1
3
xp(x)
 0   1  2   3  

all x
8
8
8
8
2
Variance and Standard Deviation
A
measure of the variability of a RV is its Variance
 To compute the variance of a discrete RV X
 Compute
µ
 For each possible x, compute (x – µ)2 p(x)
 Add up these values
 It helps to construct a table
 In
a formula:
σ 2  Var(X)  (x  μ)2 p(x)
all x
OR
σ 2  Var(X)   x 2 p(x)   2
all x
 Standard
Deviation (SD): σ  Var(X)
Variance and Standard Deviation
 In
the three-tosses example:
x
0
1
2
3
p(x)
1/8
3/8
3/8
1/8
 Had
µ = 3/2
 Using the first formula,
Var(X) = (0 – 3/2)2(1/8) + (1 – 3/2)2 (3/8) + (2 – 3/2)2 (3/8)
+ (3 – 3/2)2 (1/8) = 3/4
 Using
the second formula,
Var(X) = 02(1/8) + 12 (3/8) + 22 (3/8) + 32 (1/8) – (3/2)2 = 3/4
Discrete Probability Distributions.
Example: Let X the numbers of cars sold. The PDF of X is given in the table
below. Find the mean and variance of X.
Find the Mean of X.
Find the Variance of X.
Variance and Standard Deviation
 When
X and Y are independent,
Var(X + Y) = Var(X) + Var(Y)
 Note: it is the variances that add up, not the SDs:
X+Y  X + Y
 Both
variances and SDs are always positive
The Binomial Distribution
Let X be the number of “successes” in n independent “trials,” each with
success probability p,
 Such an X is a Binomial R.V. with parameters n and p

 n x
p( x)  P( X  x)    p (1  p) n  x
 x
where
n
n!
  
 x  x!(n  x)!
n is the number of trials
x is the number of observed successes, x=0…n
p is the probability of success on each trial

In the book: the probability p is denoted by π,
p( x)  n Cx (1   )
x
n x
where
n!
n Cx 
x!(n  x)!
The Binomial Distribution
Example: Let X = number of 1’s after rolling a die 5 times. What
is P(X = 2) = ?
X is a Binomial RV with 5 independent trials (n =5) and a
probability of success of 1/6 (p =1/6). The PDF of X is:
 n x
5 1 x
1 5 x
n x
P( X  x)    p (1  p)   ( ) (1  )
6
 x
 x 6
for x = 0,1,2,3,4,5
2
3
5
  1   5 
 P(X  2)        0.161
 2  6   6 
The Binomial Distribution
Example: A study concluded that 76.2 percent of drivers used
seat belts. A sample of 12 vehicles is selected. What is the
probability the drivers in at least 7 of the 12 vehicles are
wearing seat belts?
If X is the number of drivers wearing seatbelts then X is a
binomial RV with parameters n=12 and success probability
p=0.762.
P(X=7) = 12C7(0.762)7(1-0.762)(12-7) = 0.0902
P(X≥7) = P(X=7) + P(X=8) + P(X=9) + P(X=10)+ P(X=11)+ P(X=12)
=0.0902 + 0.1805 + 0.2569 +0.2467+ 0.1436 + 0.0383
= 0.9563
The Binomial Distribution
 An
important part of understanding probability/statistics is
recognizing a “binomial situation”
 Binomial
example
 Number

n = number of items, p = probability of a product being defective
 Number

of students in this class who are in senior year
n = number of students in this class, p = probability of a student being a senior.
 Number

of defective products in a sample of items.
of no-shows for a flight
n = number of passengers, p = probability of a no show flight
 Number
of times next week I’ll get stuck in traffic on my way to
school

n = number of work days per week, p = probability I get stuck in traffic
The Binomial Distribution
 Often
define and use q = 1 – p
 When
X is a Binomial RV with parameters n and p,
E(X) = np
Var(X) = npq,
 In
the “rolling a die 5 times” example
E(X) = 5(1/6) = 5/6
The Binomial Distribution
Example: A salesperson is about to visit 25 potential customers;
the probability of a successful visit (a “sale”) is 0.3,
independent of other visits. Let X be the number of sales.
X is a binomial RV with parameters n=25 and p=0.3.
1.
2.
3.
4.
p(8) = P(X = 8) = ?
F(8) = P(X ≤ 8) = ?
E(X) = µ = ?
SD(X) =  = ?
If Y is the number of unsuccessful visits, then what is the
distribution of Y?
The Poisson Distribution
If X is a Poisson RV then the PDF of X is:
p ( x)  P( X  x) 
 xe
x!
Approximation a Binomial distribution
with a Poisson distribution
o Consider a Binomial RV X with parameters n and p, X~Bin(n,p).
If n is large enough and for a small value of p, we can approximate
X with a Poisson RV with parameter µ=np, X~ Pois(µ).
o Recall that the mean of a Binomial RV is np and the mean of the
Poisson RV is µ.
o If n is large and p is small, then use a Poisson Random variable by
setting µ=np.
o The mean of the Poisson process becomes µ=np and the variance
is also equal to µ=np.
Example: Assume baggage is rarely lost by the Airlines. Suppose a
random sample of 1,000 flights shows a total of 300 bags were lost.
Assume that the number of lost bags per flight follows a Poisson
distribution. What is the PDF? What is the probability that 0 bags
are lost per flight? 3 bags are lost per flight?
Let X be the number of bags lost per flight. Assume X is Poisson.
The mean of X is 300/1000=0.3 bags. We know that µ=the mean of the Poisson
process, so µ=0.3.
o The number of lost bags per flight follows a Poisson distribution with mean = 0.3.
o If X is the number of lost bags then the PDF of X is:
0.3x e 0.3
P( X  x) 
x!
0.30 e 0.3
P( X  0) 
 0.741
0!
0.33 e 0.3
P( X  3) 
 0.003
3!
Example: An Emergency Room (ER) is located in a town with a
population of n=50,000. The probability that a town resident will
need to enter the ER during any chosen day is p=0.0001.
What is the distribution of the number of ER patients per day?
What is the probability that on a certain day the ER is empty?
What is the probability that on a certain day the ER has more
than three patients?
Let X be the of patients entering the ER during a day.
The mean of X is µ=np=(50000)(0.0001)=5 patients per day.
The PDF of X is:
5x e5
P( X  x) 
x!
50 e5
P( X  0) 
 0.006738
0!
P( X  3)  1  ( P( X  0)  P( X  1)  P( X  2)  P( X  3))
 1  (.006738  .03369  .084224  .140374)  0.735
The Poisson Distribution
The Poisson probability distribution describes
the number of times some event occurs during
a specified interval. The interval may be time,
distance, area, or volume.
Examples:
o
o
The number of patients that enter the ER from 1pm to 2pm.
The number of defects on a cable within 1 meter.
Assumptions of the Poisson Distribution:
o
o
The probability is proportional to the length of the interval.
The intervals are independent.
Continuous Probability Distributions

For Discrete RV X, the pdf is given by p(x)=P(X=x) for all possible values
of x.

For a Continuous RV X, P(X=x)=0 for all values of x.
 Example: If X is the amount of time you wait in line at Starbucks then
P(X=30.567… seconds)=0.

The pdf of a continuous RV is represented by a function p(x) for all values
of x where the area under p(x) is 1.

The Uniform and Normal Distributions are commonly used Continuous
Distributions.
Uniform Distribution
 The
simplest distribution for a continuous random variable.
 Rectangular in shape, constant (uniform) height
 Defined by minimum and maximum values a and b.
 Areas within the distribution represent probabilities
 Example:

Time to fly on MEA from Beirut to Paris ranges from 4 hrs to 5hrs. Random
variable is flight time; it is continuous.
P(x)
A continuous
Uniform Distribution
1/(b-a)
a
b
x
Uniform Distribution
 Mean:
ab

2
 SD:
(b  a ) 2

12
 Height:
1
if a≤ x ≤b,
P( x ) 
ba
0
elsewhere.
Uniform Distribution
Area=
(height)(base)=
1
(b  a )
(b  a )
Example 1: If a uniform distribution ranges from 10 to 15, the height is .20,
found by 1/(15-10). The base is 5, found by 15-10. The total area is [1/(1510)](15-10)=1.00
Example 2: The Logan Transit Department (LTD) provides free bus service to
Logan residents. A bus arrives at the Transit Center every 30 minutes
between 6 A.M. and 9:30 P.M. during weekdays. People arrive at the Transit
Center at random times. The time that a person waits is uniformly distributed
from 0 to 30 minutes.
Uniform Distribution
 QUESTIONS:
 Draw
a graph for this distribution.
 Show that the area is 1.00
 How long will a person “typically” have to wait for a bus? In other
words what is the mean waiting time? What is the SD of the
waiting times?
 What is the probability a person will wait more than 25 minutes?
 What is the probability a person will wait between 10 and 20
minutes?
Uniform Distribution
The random variable is time a person must wait. Time is measured on a
continuous scale, and wait time ranges from 0 minutes to 30 minutes.
 To draw the uniform distribution, we start by finding the height P(x).
 P(x)=1/(30-0)= .033
Area calculation: the time people must
wait for the bus is uniform over the
interval from 0 to 30 minutes so, a=0
and b=30.
Area= (height)(base)=[1/(30-0)](30-0)
=1.00


Probability

.033
0
10
20
30
Length of wait time (minutes)
Mean calculation:
µ = (a+b)/2= (0+30)/2 = 15
the typical wait time for an LTD bus is 15 minutes.
Uniform Distribution
(30  0) 2
 8.66
12
(b  a)2


12
Probability
Probability a person waits more than 25 minutes:
 This probability is graphically represented by the area within the
distribution for the interval 25 to 30.
From the area formula:
P(25< wait time <30)= (height)(base) = [1/(30-0)](5)= .1667
The probability a person waits more than 25 minutes is .1667
Area= .1667
.033
0
10
μ=15
20
25
30
Uniform Distribution
 Probability
a person waits between 10 and 20 minutes:
P(10< wait time <20) = (height)(base)
=[1/(30-0)](10) = .3333
Probability
Area= .333
.033
0
10
20
25
μ=15
Length of wait time (minutes)
3
0
The Normal Distribution
A distribution defined by its mean and its standard deviation
1
P( x ) 
e
 2
( x  2 )
[
]
2 2
o Is bell-shaped and has a single peak at the center of the distribution.
o Is symmetrical about the mean.
o Is asymptotic. That is the curve gets closer and closer to the X-axis but
never actually touches it.
o Has its mean µ to determine its location and its standard deviation, σ, to
determine its dispersion.
l
i
:
,
The Normal Distribution
Theoretically,
curve extends to
infinity
Normal curve
is
symmetrical
.
- 5
Mean, median, and
mode are equal
x
The Normal Distribution
 Number
of normal probability distributions is unlimited.
 Each has a different mean µ and different SD σ.
 Providing tables for each is impossible.
 Standard normal distribution
 Used
to determine the probabilities for all the normal families
 With
one table for a standard normal distribution, values can be
located easily.
 Any normal distribution can be converted into a standard
normal distribution: next
 Obtain
z-values (z-scores, standard normal values, etc.)
The Standard Normal Distribution
The standard normal distribution is a normal distribution with
a mean of 0 and a standard deviation of 1.
It is also called the z distribution.
A z-value is the signed distance between a selected value,
designated X, and the mean µ, divided by the standard
deviation, σ. The formula is:
z
X 

The Standard Normal Distribution
A
standard normal distribution has a:
 Mean
of 0
 SD of 1
 z values on horizontal scale
 Standard
normal distribution table (Appendix B.1 pages 784)
lists the probabilities for the distribution.
Example: If Z~Normal(0,1)
 P(0<Z<1.5)
z
0.00
0.01
0.02
…
=P(1.5<Z<0) …
1.5
0.4332
0.4345
0.4352
The Standard Normal Distribution
Example: The bi-monthly starting salaries of recent MBA
graduates follow the normal distribution with a mean of $2,000
and a standard deviation of $200.
What is the z-value for a salary of $2,200?
z=
$2,200 - $2000
$200
=1
What is the z-value for a salary of $1,700?
z=
$1,700 - $2000 =-1.5
$200
A z-value of 1 indicates
that the value of $2,200 is
one standard deviation
above the mean of $2,000.
A z-value of –1.50
indicates that $1,700 is 1.5
standard deviation below
the mean of $2000.
The Normal Distribution
Example: The m&m’s factory supervisor wants to know the
probability that m&m’s bags weigh between 283 and 285.4
grams. The bag weight of m&m’s follow a normal probability
distribution with a mean of 283 grams and a SD of 1.6 grams.
 We
want to know the area under the curve between the mean,
283, and 285.4 grams. Or P(283< weight <285.4)
 We convert the x values into z values
The Normal Distribution
z value for 283:
 z= (x-μ)/σ= (283-283)/1.6 = 0
 z value for 285.4:
 z= (285.4-283)/1.6 = 1.5


Using Appendix B.1
The value in the standardized normal probability table corresponding to
our z values is 0.4332. This means that the area under the curve, between
0.00 and 1.5 is 0.4332. It also means that the probability of a bag,
selected at random, will weigh between 283 and 285.4 is 0.4332.
Empirical Rule
About 68% of the area under the normal curve is within 1 SD of the
mean; i.e.,μ ± σ.
 About 95% of the area under the normal curve is within 2 SD’s of
the mean; i.e., μ ± 2σ.
 Practically all of the area under the normal curve is within 3 SD’s of
the mean; i.e., μ ± 3σ.

Empirical Rule
 To
translate the empirical rule into standard normal deviates
terms (z values):
between z=-1 (or X= µ - 1σ) and z=+1 (or X= µ +1 σ)
 95% between z=-2 (or X= µ - 2σ) and z=+2 (or X= µ +2 σ)
 ≈all between -3 and +3
 68%
The Normal Distribution
Example: Suppose the distribution of the annual incomes of a group
of middle-management employees at HSBC approximates a
normal distribution with a mean of $47,200 and a SD of $800.
About 68% of the incomes lie between what two amounts?
 About 95% of the incomes lie between what two amounts?
 Virtually all the incomes lie between what two amounts?
 What are the median and the modal values?
 Is the distribution of incomes symmetrical?

The Normal Distribution
About 68% of the incomes lie between
$47,200 –1($800) = $46,400
and $47,200 +1($800) = $48,000
About 95% of the incomes lie between
$47,400 – 2($800) = $45,600
and $47,400 + 2($800) = $48,800
Almost all of the incomes lie between
$47,400 – 3($800) = $44,800
and $47,400 + 3($800) = $49,600
The mean, the median, and the mode are equal for
normal distributions; in this ex. $47,200. The distribution
is symmetrical.
The Normal Distribution
 Example:
The distribution of weekly income of workers in a
computer assembly company is normal with a mean of $1,000
and a SD of $100.
 What
is the probability of selecting a worker whose income is less
than $790? Or P(weekly income < $790)?
z value for $790= (790-1000)/100=-2.1
From Appendix B.1, the area between 0 and -2.1=.4821
Area between 0 and -∞ = .5
Area beyond -2.1 = .5-.4821 = .0179.
The Normal Distribution
Example (continued): What is the probability of selecting, at random, a
worker whose income is between $840 and $1,200? Or P($840<
weekly income < $1,200)?

z value of $840 = (840-1000)/100=-1.6.

Using Appendix B.1, area between 0 and -1.6 is .4452.

z value of $1,200 = 2.00.

Using Appendix B.1, area between 0 and 2.00 is .4772.

P($840 < weekly income < $1,200)= .4452+.4772 = .9224;
i.e., 92.24% of workers have weekly incomes between $840
and $1,200
The Normal Distribution
Example: A tire company wishes to set a minimum mileage
guarantee on its new line of tires. Tests reveal that the mean
mileage is 67,900 and SD is 2,050 miles. The distribution of
miles is normal. They want to set the minimum guaranteed
mileage so that no more than 4% of tires are replaced. What
minimum guaranteed mileage should they announce?
 The
minimum guaranteed number of mileage is x.
 P( mileage < x), or the area under the normal curve is 4% or
.0400
The Normal Distribution
 Area
between 0 and x =.5000-.0400 =.4600
We look for .4600 in Appendix B.1, the closest value is .4599
and its corresponding z value is 1.75. Because the value is to
the left of the mean z is -1.75 .

z
= (x-µ)/σ
-1.75 = (x-67,900)/2,050
x= 64,312
 The
tire company should announce that it will replace for free any
tire that wears out before it reaches 64,312 miles. Under this plan
only 4% of the tires will be replaced.
EXAMPLES
Doug scored a 57 on the Miller Analogies Test which
has a mean of 50 and a SD of 5. Jennifer scored 120
on the WISC (intelligence test) which has a mean of
100 and a SD of 15. Compare their scores. Who had a
better score?

To compare the 2 scores, first convert each to a standard z score.
x
In Doug’s distribution:
z
z = (57 – 50)/5 = 1.4
x 
z
 In Jennifer’s distribution:

z = (120 – 100)/15 = 1.33
 Doug’s score is higher, therefore better than Jennifer’s.

If x is a normally distributed variable with a mean of 30
and a SD of 6, what is the probability of x falling
above 30; i.e. P(x>30)?
 The
z score that corresponds to x=30 is z=0.
 Before starting with your calculations notice that any
value of x falling above 30 will have a positive z
score.
 P (x>30) = P (z>0) = .5000
 The
assembly times required for the manufacturing of a certain
product are normally distributed with a mean of 400 seconds
and a SD of 50 seconds. An item is selected at random. Find the
probability that its assembly time is between 360 and 440
seconds.

For x = 360:
z

x
For x = 440:
z


360  400

 .80
50
x

440  400

 .80
50
P(360 < x < 440) = P( -.8 < x < .8)
= .2881 + .2881
= .5762
In a large section of a Western Civilization course, test grades are
normally distributed with a mean of 70 and a SD of 7. Grades are to be
assigned according to the following rule. Find the numerical limits for
each letter grade.
A
top 10%
B
between the top 10% and 30%
C
scores between the top 30% and
the bottom 30%
D
between the bottom 10% and 30%
F
bottom 10%

 Construct
a standard normal curve.
 The top 10% refers to the area above .4000 or above a z
score of 1.28.
 The top 30% refers to the area above .2000 or a z value of
.52.
 The bottom 30% refers to the area below .2000 or a z
value of -.52.
 The bottom 10% refers to the area below .4000 or a z
value of -1.28.
Using the formula x = µ+ z.σ, we convert each z score into an x score.
x = 70 + 7(1.28) = 78.96 or 79
x = 70 + 7(.52) = 73.64 or 74
x = 70 + 7(-.52) = 66.36 or 66
x = 70 + 7(-1.28) = 61.04 or 61
Therefore,
A
79 or above
B
74 – 78
C
66 – 73
D
61 – 65
F
60 and below


a)
b)
Quick Start Company makes 12-volt car batteries. After many years
of product testing, the co. knows that the average life of a Quick Start
battery is normally distributed, with a mean of 45 months and a SD of
8 months.
If Quick Start guarantees a full refund on any battery that fails within
the 36-month period after purchase, what percentage of its batteries
will the co. expects to replace?
If Quick Start does not want to replace more than 10% of its batteries
under the full-refund guarantee policy, for how long should the co.
guarantee the batteries (to the nearest month)?
a)
Find the z value that corresponds to x=36.
z= (36-45)/8 = -1.125
 We need to find P (z<-1.125). Since -1.125 is midway
between -1.12 and -1.13 we need to take the average of the 2
areas as our estimate. The areas are .3686 and .3708; their
average is .3679.
 P (z<-1.125)= .5000 - .3679 = .1303
 They will replace about 13% of their batteries.

b)
x is the life span of the battery.
 Find value of x so that 10% will no longer work and 90% still
work.
 Find the z score with 10% of the area under the curve falling to
its left.
 This means that we need to find the z score for the left area of
.4000.
 z = -1.28
 Convert z to x = µ + z.σ
= 45 + (-1.28)8 = 34.76
or 35 months

 The
resting heart rate for an adult horse should average
about µ=46 beats per minute with (95% of data) range
from 22 to 70 beats per minute, based on information
from the Merck Veterinary Manual (a classic vet.
reference). Let x be a random variable that represents the
resting heart rate for an adult horse. Assume that x has a
distribution that is approximately normal.
a)
Estimate the SD of the x distribution.
A good estimate is:
95% fall between -2σ and +2σ
we can get 1σ by dividing the range by 4.
(70-22)/4 = 12
σ= 12 beats per minute
b)
What is the probability that the heart rate is less
than 25 beats per minute?
Find the z score that corresponds to x=25.
z = (25-46) / 12 = -1.75
Use the table to find the area under the curve for z
= -1.75. It is .4599.
P (z< -1.175) = .5000 - .4599
= .0401
c)
What is the probability that the heart rate is greater
than 60 beats per minute?
Find the z score that corresponds to x=60.
z = (60-46) / 12 = 1.17
Use the table to find the area under the curve for z
= 1.17. It is .3790.
P (z > 1.17) = .5000 - .3790
= .1210
d)
What is the probability that the heart rate is
between 25 and 60 beats per minute?
Add .4599 and .3790
The probability is .8389
e)
A horse whose resting heart rate is in the upper 10% of the
probability distribution of heart rates may have a secondary
infection or illness that needs to be treated. What is the heart
rate corresponding to the upper 10% cutoff point of the
probability distribution?
First, find the z score that separates the lower 90% from the
upper 10% of the data.
from the table, find the z score that corresponds to .4000
(closest is .3997). z=1.28. So x= 46+(1.28)12 = 61.36
A horse with a heart rate of 61 or more should be further
examined for illness.