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:
[tex]P(x=r)=nCr*p^{r}*q^{n-r}[/tex]
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:
[tex]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[/tex]
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:
[tex]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[/tex]
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:
[tex]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[/tex]
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 = [tex]\sqrt{n*p*(1-p)}[/tex]
Hence,
Mean = 9 * 0.1 = 0.9
Standard Deviation = [tex]\sqrt{9*0.1*(1-0.1)}=0.9[/tex]
