Question

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.06 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel. (a) What is the probability that there are no surface flaws in an auto's interior? (b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws? (c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws? Round your answers to four decimal places (e.g. 98.7654).

Answer 1

Answer:

a) 0.7403

b)0.0498

c)0.2240

Step-by-step explanation:

Given: The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution

We know that to calculate expected number of flaws use

expected number of flaws =10×0.03 =0.3

a) probability that there are no surface flaws in an auto's interior =P(X=0)=

e^-0.3 =0.7408

b) probability that none of the 10 cars has any surface flaws =(e^-0.3)^10 =0.0498

c) probability that at most one car has any surface flaws            =P(X<=1)=P(X=0)+P(X=1)

this means

=10C_0(1-0.7408)^0(0.7408)^10+10C_1(1-0.7408)^1(0.7408)^9=0.2240

Answer 2

Answer:

a) 0.4065

b) 0.000126

c) 0.001921

Step-by-step explanation:

Given data:

Poisson distribution mean = 0.06

Amount of plastic panel = 10 square panel

Probability mass function given by Poisson distribution is

$P(X =x) = \frac{e^{-\lambda} \lambda^x}{x!}$       x: 0,1,2,......

where $\lambda$ is parameter

here lambda is 0.06 per  square ft

$\lambda = 0.09\times 10 = 0.9 pe\ 10\ square\ foot$

a)  from Poisson distribution

$P(X =x) = \frac{e^{-\lambda} \lambda^x}{x!}$    

            $= \frac{e^{-0.9} \lambda^0}{0!}$

            $= e^{-0.9} = 0.4065$

B) PROBABILITY OF NO SURFACE FLAWS IN CAR

N = 10

p = 1 - 0.406 = 0.5934

q = 0.4065

$P(X =x) =\binom{n}{x} p^x q^{n-x}$

            $=\binom{10}{0} 0.59^0 0.4065^{10-0}$

             = 0.000126

c) probability of 1 car flaw

$P(X \leq 1) = P(X = 0) + P(X = 1)$

                 $= ^{10}C_0 (1 - 4065)^0 (0.4065)^{10} + ^{10}C_1 (1 - 4065)^1 (0.4065)^{9}$

                 $= 1 \times 1.231\times 10^{-4} + 10\times 0.5935\times 3.030\times 10^{-4}$

                  = 0.001921