(?) + 𝑷
?
(?) + 𝑷
?
(?)
? −
probability of acceptance a student to a collage
? = ?, ?
?
−
number of Bernoulli trails
? = ?
→
?
students apply
?
−
number of success in Bernoulli trails
? = ?, ?, ?
𝑷(?) = 𝑷(𝑿 ≤ ?) = 𝑷
?
(?) + 𝑷
?
(?) + 𝑷
?
(?) =
probabilities from the table
values in red
(below)
= ?, ???? + ?, ???? + ?, ???? = ?, ???
𝑿 −
a random variable has a binomial distribution
𝑿 ∶ ? ( ??,
√???
)
so:
µ = ?𝑿 = ?? = ? ∙ ?, ? = ?, ?
it means that the most expected value of acceptance
the students to a prestigious college is one and half student
𝑷
?
(?)
Binomial Probabilities
𝑷
?
(?) = (
?
?
) ∙ ?
?
∙ ?
?−?
<fragment>
n
k
0.01
0.05
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
5
0
0.95099
0.773781
0.59049
0.32768
0.16807
0.07776
0.03125
0.01024
0.00243
0.00032
1
0.04803
0.203627
0.32805
0.4096
0.36015
0.2592
0.15625
0.0768
0.02835
0.0064
2
0.00097
0.021434
0.0729
0.2048
0.3087
0.3456
0.3125
0.2304
0.1323
0.0512
3
9.8E-06
0.001128
0.0081
0.0512
0.1323
0.2304
0.3125
0.3456
0.3087
0.2048
4
4.95E-08
2.97E-05
0.00045
0.0064
0.02835
0.0768
0.15625
0.2592
0.36015
0.4096
5
1E-10
3.13E-07
0.00001
0.00032
0.00243
0.01024
0.03125
0.07776
0.16807
0.32768

Application of Mathematical Statistics (4)
Iwona Nowakowska
5
Case Problem:
In a certain company there were noticed that the average number of
accidents in a month is 2 and the distribution of the number of accidents in
this company can be described by
Poisson
distribution.
Let
𝑿 −
be a random variable describing the number of accidents in a month.
1)
Find the probability that there will be no accidents in a randomly
chosen month.
2)
Find the probability that there will be 4 accidents in a randomly chosen
month.
3)
Find the probability that the number of accidents in a month will be at
most 2.
Solution:
𝑿 −
the number of accidents in a month
Let us notice that theoretically the random variable
𝑿
can assume
any value
0,1,2,3, ….
𝑿 −
has Poisson distribution, so we know that:
𝑿 ∶ 𝑷 ( ?,
√?
)
and:
𝑷(𝑿 = ?) = ?
−?
∙
?
?
?!
where
? −
the average number of events that occur during the given
interval
? = ?, ?, ?, ?, …
,
? −
the number of successes
1)
In our case:
? = ?
,
? −
no accidents
? = ?
⇒
the average number of accidents in a month is 2

Application of Mathematical Statistics (4)
Iwona Nowakowska
6
𝑷(𝑿 = ?) = 𝑷(𝑿 = ?) = ?
−?
∙
?
?
?!
= ?
−?
∙
?
?
?!
= ?
−?
∙ ? =
?
?
?
=
=
?
(?, ??)
?
=
?
?, ????
= ?, ???
2)
In our case:
? = ?
,
? −
4 accidents in a month
? = ?
⇒
the average number of accidents in a month is 2
𝑷(𝑿 = ?) = 𝑷(𝑿 = ?) = ?
−?
∙
?
?
?!
= ?
−?
∙
?
?
?!
= ?
−?
∙
??
??
=
?
?
∙
?
?
?
=
=
?
?
∙
?
(?, ??)
?
=
?
?
∙
?
?, ????
=
?
??, ????
= ?, ??
3)
In our case:
? −
number of accidents
? = ?
or
? = ?
or
? = ?
? = ?
⇒
the average number of accidents in a month is 2
𝑷(𝑿 = ?) = ?
−?
∙
?
?
?!
We
look for the probability:
𝑷(𝑿 ≤ ?) = ? ? ?
𝑷(𝑿 ≤ ?) = 𝑷(𝑿 = ?) + 𝑷(𝑿 = ?) + 𝑷(𝑿 = ?) = ? ? ?
𝑷(𝑿 = ?) = ?
−?
∙
?
?
?!
𝑷(𝑿 ≤ ?) = 𝑷(𝑿 = ?) + 𝑷(𝑿 = ?) + 𝑷(𝑿 = ?) =
= ?
−?
∙
?
?
?!
+ ?
−?
∙
?
?
?!
+ ?
−?
∙
?
?
?!
=

Application of Mathematical Statistics (4)
Iwona Nowakowska
7
=
?
?
?
+ ? ∙
?
?
?
+
?
?
∙
?
?
?
=
?
?
?
+ ? ∙
?
?
?
+ ? ∙
?
?
?
=
=
?
?
?
=
?
(?, ??)
?
=
?
?, ????
= ?, ???? ≈ ?, ???
All above probabilities we can find in the table of
Cumulative Poisson
Probability Distribution.
1)
𝑷(𝑿 = ?)
The probability
𝑷(𝑿 = ?)
must be replaced by the probability
𝑷(𝑿 ≤ ?)
→
the variable
𝑿
does not assume negative values.
In the table we must find the value of the parameter
? = ?
(in a column)
and
the value of
? = ?
(in a row)
We have immediately:
𝑷(𝑿 = ?) = ?, ????
value in red in the table
(below)
2)
𝑷(𝑿 = ?)
The probability
𝑷(𝑿 = ?)
must be replaced by the probability
𝑷(𝑿 ≤ ?) − 𝑷(𝑿 ≤ ?)
→
only such probabilities we have in the table
We have immediately:
𝑷(𝑿 ≤ ?) − 𝑷(𝑿 ≤ ?) = ?, ???? − ?, ???? = ?, ???? ≈ ?, ??

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- Fall '13