Respuesta :
Answer:
The expected number of appearance defects in a new car = 0.64
standard deviation = 0.93(approximately)
Step-by-step explanation:
Given -
7% had exactly three, 11% exactly two, and 21% only one defect in cars.
Let X be no of defects in a car
P(X=3) = .07 , P(X=2) = .11 , P(X=3) = .21
P(X=0 ) = 1 - .07 - .11 - .21 = 0.61
The expected number of appearance defects in a new car =
E(X) = [tex]\nu[/tex] = [tex]= \sum X P(X) =3\times 0.07 + 2\times0.11 + 3\times0.21 + 0\times0.61[/tex] = 0.64
[tex]\sigma^{2}[/tex] = variation of E(X) = [tex]\sum (X - \nu) ^{2}P(X)[/tex] = [tex](3 - 0.64) ^{2}\times0.07 + (2 - 0.64) ^{2}\times0.11 + (1 - 0.64) ^{2}\times0.21 + (0 - 0.64) ^{2}\times0.61[/tex]
= [tex](2.36) ^{2}\times0.07 + (1.36) ^{2}\times0.11 + (.36) ^{2}\times0.21 + ( 0.64) ^{2}\times0.61[/tex] = 0.8704
standard deviation = [tex]\sqrt{\sigma^{2}}[/tex] = [tex]\sqrt{0.8704^{2}}[/tex] = 0.93(approximately)
