Question

Industry standards suggest that 8 percent of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 9 Nissans yesterday. (Round your mean answer to 2 decimal places and the other answers to 4 decimal places.)
a. What is the probability that none of these vehicles requires warranty service?
b. What is the probability exactly one of these vehicles requires warranty service?
c. Determine the probability that exactly two of these vehicles require warranty service.
d. Compute the mean and standard deviation of this probability distribution.

Answer 1

Answer:

Step-by-step explanation:

Let X be the no of vehicles that require warranty service within I year out of vehicles sold (9) yesterday.

Each vehicle is independent of the other to get warranty service

Hence X is binomial with p=8% = 0.08

So P(X=x) = $9Cx (0.08)^x (1-0.08)^{9-x} ,x=0,1,2...9$

$a) P(X=0) = 0.472161\\b) P(x=1) = 0.3695$

c) $P(X=2) = 0.1285$

d) Mean = Mean of binomial = np =

$9(0.08) = 0.72$

Var (x) = npq = $0.6624$

Std dev = square root of variance = 0.8139

Answer 2

Answer:

p(warranty service) = 0.08  q(no warranty service) = 0.92  n = 9

Step-by-step explanation:

using the binomial distribution

x=0

a) 9C0 x (0.08)^0 (0.92)^9

= 1 x 0.4716

= 08464

x=1

b) 9C1 (0.08)^1 (0.92)^8

= 9 x 0.08 x 0.5132

= 0.3695

x=2

c) 9C2 (0.08)^2 x (0.92)^7

= 36 x 0.0064 X 0.5578

= 0.1285

d) mean = np

= 9 x 0.08

= 0.72

Standard deviation = Sqrt(npq)

= (9 x 0.08 x 0.92)^0.5

= 0.8139