Question

A consumer organization inspecting new cars found that many had appearance defects (dents, scratches, paint chips, etc.). While none had more than three of these defects, 7% had exactly three, 11% exactly two, and 21% only one defect. Find the expected number of appearance defects in a new car and the standard deviation.

Answer 1

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) = $\nu$ = $= \sum X P(X) =3\times 0.07 + 2\times0.11 + 3\times0.21 + 0\times0.61$ = 0.64

$\sigma^{2}$  = variation of E(X) = $\sum (X - \nu) ^{2}P(X)$ = $(3 - 0.64) ^{2}\times0.07 + (2 - 0.64) ^{2}\times0.11 + (1 - 0.64) ^{2}\times0.21 + (0 - 0.64) ^{2}\times0.61$

 = $(2.36) ^{2}\times0.07 + (1.36) ^{2}\times0.11 + (.36) ^{2}\times0.21 + ( 0.64) ^{2}\times0.61$ = 0.8704

standard deviation  =  $\sqrt{\sigma^{2}}$ = $\sqrt{0.8704^{2}}$ = 0.93(approximately)