Respuesta :
Answer:
a)0.5611
b)0.1828
c)0.6217
d)0.4389
e)6.25
Step-by-step explanation:
Let's define the following event :
X : ''Calls involving a fax message''
The random variable X can be modeled as a Binomial random variable.
N ~ Bi(n,p)
N ~ Bi(25,0.25)
Where n is the sample and p is the success probability
Let's also define nCr as the combinatorial number :
[tex]nCr=\frac{n!}{r!(n-r)!}[/tex]
The probability function for X is :
[tex]P(X=k)=nCk.p^{k}.(1-p)^{n-k}[/tex]
Where k is the number of success
a) [tex]P(X\leq 6)[/tex]
[tex]P(X\leq 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)[/tex]
[tex]P(X\leq 6)=0.5611[/tex]
b)[tex]P(X=6)=25C6.(0.25)^{6}.(1-0.25)^{19}=0.1828[/tex]
c) [tex]P(X\geq 6)=1-P(X<6)=1-[P(X\leq 6)-P(X=6)]=1-(0.5611-0.1828)\\P(X\geq 6)=1-0.3783=0.6217[/tex]
d)[tex]P(X>6)=1-P(X\leq 6)=1-0.5611=0.4389[/tex]
e) They are asking us about the expected value for random variable X
In a Binomial random variable :
E(X) = μ = n.p
[tex]E(X)=25(0.25)=6.25[/tex]
Using the binomial distribution, it is found that:
a) 0.3783 = 37.83% probability that at most 6 of the calls involve a fax message.
b) 0.1645 = 16.45% probability that exactly 6 of the calls involve a fax message.
c) 0.7862 = 78.62% probability that at least 6 of the calls involve a fax message.
d) 0.6217 = 62.17% probability that more than 6 of the calls involve a fax message.
e)The expected number of calls involving a fax message is 6.25.
-------------------------------
For each call, there are only two possible outcomes. Either it involves a fax, or it does not. The probability of a call involving a fax is independent of any other call, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 25% of the calls involve fax messages, thus [tex]p = 0.25[/tex]
- Sample of 25 calls, thus [tex]n = 25[/tex].
Item a:
This probability is:
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.0008[/tex]
[tex]P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.0063[/tex]
[tex]P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.0251[/tex]
[tex]P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.0641[/tex]
[tex]P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.1175[/tex]
[tex]P(X = 5) = C_{25,5}.(0.25)^{5}.(0.75)^{20} = 0.1645[/tex]
Then
[tex]P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0008 + 0.0063 + 0.0251 + 0.0641 + 0.1175 + 0.1645 = 0.3783[/tex]
0.3783 = 37.83% probability that at most 6 of the calls involve a fax message.
Item b:
From item a, P(X = 6) = 0.1645, thus:
0.1645 = 16.45% probability that exactly 6 of the calls involve a fax message.
Item c:
This is:
[tex]P(X \geq 6) = 1 - P(X < 6)[/tex]
In which:
[tex]P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0008 + 0.0063 + 0.0251 + 0.0641 + 0.1175 = 0.2138[/tex]
[tex]P(X < 6) = 1 - P(X < 6) = 1 - 0.2138 = 0.7862[/tex]
0.7862 = 78.62% probability that at least 6 of the calls involve a fax message.
Item d:
This is:
[tex]P(X > 6) = 1 - P(X \leq 6) = 1 - 0.3783 = 0.6217[/tex]
0.6217 = 62.17% probability that more than 6 of the calls involve a fax message.
Item e:
The expected number of calls involving a fax is:
[tex]E(X) = np = 25(0.25) = 6.25[/tex].
A/ similar problem is given at https://brainly.com/question/13628006
