Question

Industry standards suggest that 10 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:

a)  0.3874

b)  0.3874

c)  0.1722

d) Mean and Standard Deviation = 0.9

Step-by-step explanation:

This is binomial distribution problem that has formula:

$P(x=r)=nCr*p^{r}*q^{n-r}$

Here p is probability of success = 10% = 0.1

q is probability of failure, that is 90% = 0.9

n is total number, which is 9, so n = 9

a)

The probability that none requires warranty is r = 0, we substitute and find:

$P(x=r)=nCr*p^{r}*q^{n-r}\\P(x=0)=9C0*(0.1)^{0}*(0.9)^{9-0}\\P(x=0)=0.3874$

Probability that none of these vehicles requires warranty service is 0.3874

b)

The probabilty exactly 1 needs warranty would change the value of r to 1. Now we use the same formula and get our answer:

$P(x=r)=nCr*p^{r}*q^{n-r}\\P(x=1)=9C1*(0.1)^{1}*(0.9)^{9-1}\\P(x=1)=0.3874$

This probability is also the same.

Probability that exactly one of these vehicles requires warranty is 0.3874

c)

Here, we need to make r = 2 and put it into the formula and solve:

$P(x=r)=nCr*p^{r}*q^{n-r}\\P(x=2)=9C2*(0.1)^{2}*(0.9)^{9-2}\\P(x=2)=0.1722$

Probability that exactly two of these vehicles requires warranty is 0.1722

d)

The formula for mean is

Mean = n * p

The formula for standard deviation is:

Standard Deviation = $\sqrt{n*p*(1-p)}$

Hence,

Mean = 9 * 0.1 = 0.9

Standard Deviation = $\sqrt{9*0.1*(1-0.1)}=0.9$