Answer:
B
Step-by-step explanation:
Answer:I’m guessing the 3rd one
Step-by-step explanation:
Answer: D
Step-by-step explanation:
Answer:
So whats the question?
Step-by-step explanation:
Answer:
![Mean=p=0.75](https://tex.z-dn.net/?f=Mean%3Dp%3D0.75)
![sd=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204](https://tex.z-dn.net/?f=sd%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B450%7D%7D%3D0.0204)
Step-by-step explanation:
1) Data given
represent the population proportion of complaints settled for new car dealers
represent the sample of complaints involving new car dealers.
2) Find the distribution of ![\hat p](https://tex.z-dn.net/?f=%5Chat%20p)
First we can begin with the expected value
and that represent the mean
Now we can find the variance for ![\hat p](https://tex.z-dn.net/?f=%5Chat%20p)
When we use a proportion p, when we draw n items each from a Bernoulli distribution. The variance of each Xi distribution is p(1−p) and hence the standard error is p(1−p)/n. for this reason the variance for
is given by:
![Var(\hat p)= \frac{p(1-p)}{n}](https://tex.z-dn.net/?f=Var%28%5Chat%20p%29%3D%20%5Cfrac%7Bp%281-p%29%7D%7Bn%7D)
So then the deviation would be given by:
![Sd(\hat p)=\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=Sd%28%5Chat%20p%29%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
The sample distribution of the sample proportion
is normal, so then we have this:
![\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})](https://tex.z-dn.net/?f=%5Chat%20p%20%5Csim%20N%28p%2C%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%29)
3) Calculating the mean and standard deviation
We can replace the values given in order to find the mean and deviation:
![Mean=p=0.75](https://tex.z-dn.net/?f=Mean%3Dp%3D0.75)
![sd=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204](https://tex.z-dn.net/?f=sd%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B450%7D%7D%3D0.0204)