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}](https://tex.z-dn.net/?f=P%28x%3Dr%29%3DnCr%2Ap%5E%7Br%7D%2Aq%5E%7Bn-r%7D)
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](https://tex.z-dn.net/?f=P%28x%3Dr%29%3DnCr%2Ap%5E%7Br%7D%2Aq%5E%7Bn-r%7D%5C%5CP%28x%3D0%29%3D9C0%2A%280.1%29%5E%7B0%7D%2A%280.9%29%5E%7B9-0%7D%5C%5CP%28x%3D0%29%3D0.3874)
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](https://tex.z-dn.net/?f=P%28x%3Dr%29%3DnCr%2Ap%5E%7Br%7D%2Aq%5E%7Bn-r%7D%5C%5CP%28x%3D1%29%3D9C1%2A%280.1%29%5E%7B1%7D%2A%280.9%29%5E%7B9-1%7D%5C%5CP%28x%3D1%29%3D0.3874)
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](https://tex.z-dn.net/?f=P%28x%3Dr%29%3DnCr%2Ap%5E%7Br%7D%2Aq%5E%7Bn-r%7D%5C%5CP%28x%3D2%29%3D9C2%2A%280.1%29%5E%7B2%7D%2A%280.9%29%5E%7B9-2%7D%5C%5CP%28x%3D2%29%3D0.1722)
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)}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%2Ap%2A%281-p%29%7D)
Hence,
Mean = 9 * 0.1 = 0.9
Standard Deviation = ![\sqrt{9*0.1*(1-0.1)}=0.9](https://tex.z-dn.net/?f=%5Csqrt%7B9%2A0.1%2A%281-0.1%29%7D%3D0.9)