Respuesta :
Answer:
Follows are the solution to this question:
Step-by-step explanation:
The Ford car has an issue rate of 1.27.
That means that [tex]\lambda = 1.27[/tex]
Here, [tex]X \sim poisson (\lambda = 1.27)[/tex]
In option (A) :
To make a random variable X a Poisson variable, Parameter:
[tex]\to \lambda \leq 0 .[/tex]
The likelihood of success decreases with increased test numbers.
Mathematics, as [tex]n \longrightarrow \infty, \ \ \ and \ \ \p \longrightarrow 0[/tex].
In option (B) :
You must find it: [tex]P(X =0)[/tex]
[tex]\to P(X=0) = \frac{e^{-\lambda}\cdot \lambda^{0} }{0!} \\\\\to \lambda= 1.27 \\\\=e^{-1.27} \\\\= 0.280832[/tex]
In Option (C):
You must find it: [tex]P( X \leq 2) \\[/tex]
[tex]\to P( X \leq 2) = P(X=0) +P(X=1)+P(X=2) \\\\= \sum_{x= 0}^{2} P(X=x) \\\\= \sum_{x= 0}^{2} \frac{e^{- \lambda} \lambda^{x} }{x!} \\\\= e^{-1.27}\sum_{x= 0}^{2} \frac{ 1.27^{x} }{x!} \\\\ = 0.863964[/tex]
In Option (d):
Its first rating is essential because failures in production could be detected but before the total product becomes defective, this can be reduced.
In Option (e) :
The Toyata car has an issue rate of 1.12
That means that [tex]\lambda = 1.12[/tex]
Here, [tex]Y \sim poisson ( \lambda= 1.12)[/tex]
In Option (f):
You must find it: [tex]P(Y=0)[/tex]
[tex]\to P(Y=0)= \frac{e^{- \lambda } \lambda^{0}}{0!} \\\\\to \lambda = 1.12 \\\\=e^{-1.12}\\\\= 0.32628 \\[/tex]
You must find it: [tex]P(X \leq 2)[/tex]
[tex]\to P( X \leq 2) = P(Y=0) +P(Y=1)+P(Y=2) \\\\= \sum_{x= 0}^{2} P(Y=y) \\\\= \sum_{x= 0}^{2} \frac{e^{- \lambda} \lambda^{y} }{y!} \\\\= e^{-1.12}\sum_{x= 0}^{2} \frac{ 1.27^{y} }{y!} \\\\= 0.896356[/tex]
Comparing [a], [b], for the ford car and [b] and [c] for Toyot effectiveness
Car, its probability that less faults with Toyota car are strong, that's why it's better
