SUMMARY 1
Probability Distributions
The probability distribution of a discrete random variable lists all the possible values that the random
variable can assume and their corresponding probabilities.
It is useful to have standard probability models for certain types of problems. Some of the most
important probability models are:
For discrete variables:
1. The binomial probability distribution
2. The Poisson distribution
3. The Geometric distribution
4. The multinomial distribution
5. The Hyper-geometric distribution
For continuous variables:
1. Normal distributions
2. The sampling distribution of the mean
3. The student's t distribution
4. The Chi-square distribution
5. The F distribution
Binomial distribution Model:
Characteristics
1. There is a fixed number of trials n.
2. Each trial will result in either a success or a failure.
3. The probability of success p for each trial remains the same from trial to trial.
(q is the probability of failure and q = 1- p)
4. The trials are independent.
5. The variable x measures the number of successes in n trials.
( x=0, 1, 2, 3, . . . , n)
Properties:
1. The variable x is a discrete variable.
2. The expected value or average of x is given by   E(x)  np
3. The variance of the variable x can be obtain from the formula
  npq
4. Binomial probabilities can be obtained by using the formula below or by using tables or
calculators.
2
binom (n, p; x) 
n!
p xqn  x , q  n  x
x! n  x !
5. Binomial probability distributions are said to have two parameters: n and p.
Example. A baseball player has a batting average of 0.25. What is the probability that he will have 0
hits in 10 trips to the plate, 3 hits in 10 trips to the plate?
10!
0.25 0 (0.75)10  0.056
0!10!
10!
binom ( 10, 0.25, 3) 
0.253 (0.75) 7  0.2502822876
3!7!
binom (10, 0.25; 0) 
Example. In the previous problem, what is the average and the standard deviation for the number of
hits in 10 trips to the plate.
  np  10(0.25)  2.5 and   npq  10(0.25)(0.75)  1.369
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Geometric probability distributions:
The variable x measures the number of trials needed to obtain the first success. Geometric
probabilities can be obtained from the formula
geometric( p, x)  q x 1 p,
  E ( x) 
1
1 p
, 2 
p
p2
Example. A medical diagnostic test could fail to detect a certain disease. Suppose that a given patient
has a certain disease and that the probability it will be detected by a certain test is 0.2. What is the
probability that a positive result will not be obtain until the fifth test.
4
x 1
p  0.8 (0.2)  0.08182
1
1
Note: it will take an average of  

 5 tests before a positive result is obtained from a
p 0.2
geometric( 0.2, 5)  q
patient with the disease. The standard deviation is

1 p
p2

1  0.2
(0.2) 2
 4.47
Poisson Distribution:
In a Poisson distribution we are given the expected number of successes  =E(x) and no limit on
the
number of trials.
 x e 
Poisson probabilities values are obtained from the formula P( x) 
, e  2.718
x!
   and   
2
Example: The A & P company receives 5.3 orders a day. Use the Poisson distribution to find the
probability that the company will receive exactly 7 orders.
7
x 
 5.3
Poison ( x  7) 
 e
x!

5.3
( 2.718)
7!
 0.1164
Poisson approximation to the binomial model:
When n is large and p is small a good approximation to binomial probabilities can be obtained
by using the Poisson distribution with  = u = np
Example. In a binomial problem n = 120 and p = 0.015. Use the Poisson approximation to the
binomial to find the probability of exactly 3 successes.
3
x 
1.8
1.8 (2.718)
 e

 0.1607005084
x!
3!
Note : binom (120, 0.015; 3)  0.1617229705 which is a good approximat ion .
  np  120(0.015)  1.8, Poisson( x  3) 
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Hyper-geometric distributions: to determine the number of successes x in n dependent trials when
sampling without replacement from a population of size N with contains M successes.
Population N
Sample
M
M
successes
ess
size n
x successes
successe
s
The probability of x successes in a sample of size n taken from a p0pulation of size N with M
successes is given by the formula
 M  N  M 
 

x  n  x   M C x    N  M C n  x 

P( x) 

,
N
N Cn
 
n 
  E ( x) 
nM
nM  N  n ( N  M )
and  2 

N  1
N
N2
Example. In a lot of 20 radios been inspected there are 4 defective items. If an inspector inspects 5
radios, what is the probability that 2 of them are defective. N=20, M=4, N-M=16, n-5, x=2, n-x=3
 4 16 
  
2 3
6  560
P( x  2)     
 0.2167182663
15504
 20 
 
5 
Normal Distributions:
1. There are infinitely many normal distributions, one for each pair of combinations
of the two parameters  and  .
2. The normal distribution with  = 0 and  = 1 is called the standard normal distribution.
3. The area below any normal distribution is 1.
4. Probability = Area under the curve = proportion.
5. The variable x is a continuous variable.
6. Every normal distribution can be standardized by converting x to the z-scale.
7. How to find probabilities under a normal curve using the tables (is the same as finding areas.)
Step 1. Convert x to z by using the formula z 
x

. Round z to two decimal places
Step 2. Using z and the standard normal distribution table, find the area.
Step 3. Carefully conclude, answer the question precisely.
8. To find the value of x that corresponds to a given area or proportion,
Step 1. Carefully read the problem and draw a diagram indicating the given area
Step 2. Using the appropriate table area, find the corresponding z-value.
Step 3. Use the formula x    z to obtain x.
9. When n is large we usually compute binomial probabilities by using the normal approximation with
 = np,  2 = npq and a +0.5 continuity correction factor on x. This also provides a fairly good
approximation even when n is small, provided p is reasonable close to 1/2.
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10. Solving problems using the TI-83
Problem- The variable X is normally distributed with   48 and   3.1
a) Find the ordinate (the y-value on the normal curve) when x=55
 2nd DISTR normalpdf (55, 48, 3.1)
answer: 0.0100541648
b) the ordinate (the y-value on the normal curve) at the mean value
 2nd DISTR normalpdf (48, 48, 3.1)
answer: 0.1286910582
c) Find the probability that a value is greater than 55?
 2nd DISTR normalcdf (55, 10^9, 48, 3.1)
answer: 0.0119707811
d) Find the probability that x is between 40 and 51?
 2nd DISTR normalcdf (40, 51, 48, 3.1)
answer: 0.8284825481
e) Find the value of x above which there is a 0.43 probability.
 2nd DISTR invNorm(0.57,48, 3.1)
answer: 48.54675989
f) Find the 90th percentile of the distribution.

2nd DISTR invNorm(0.90, 48, 3.1)
answer: 51.97280986
Problem. In a binomial problem n=45, p=0.6
a) Find the probability of exactly 30 successes
 2nd DISTR binompdf(45, 0.6, 30)
answer: 0.0818633582
b) Find the probability of at most 30 successes
 2nd DISTR binompcdf(45, 0.6, 30)
answer: 0.856960038
c) Find the probability of at least 30 successes
 1 - 2nd DISTR binomcdf(45, 0.6, 30)
answer: 0.143039962
Problem. In a Poisson problem with mean equal to 4.2 successes.
a) Find the probability of less than 5 successes
 2nd DISTR poissoncdf(4.2, 4)
answer: 0.5898270221
b) Find the probability of more than 5 successes
 2nd DISTR 1 - poissoncdf(4.2, 5)
answer: 0.2468571111
Problem. In a binomial problem n=1500 and p=0.12
a) Find the probability of at least 170 successes
 2nd DISTR 1 - binomcdf (1500, 0.12, 169)
answer: 0.7970876646
b) Use the Poisson approximation with   np  1500  0.12  180 to find the probability of at least
170 successes.
 2nd DISTR 1 - poissoncdf(180, 169)
answer: 0.7816766956
c) Use the Normal approximation with
  np  1500  0.12  180 ,   npq  1500(0.12)(0.88)  12.5857 and the 
correction factor to find probability of at least 170 successes.
 2nd DISTR normalcdf(169.5, 10^9, 180, 12.5857)
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continuity
answer: 0.7979384958