Answer:
1.an=45+(n-1)-5
2.Gn=-3r^n-1
Step-by-step explanation:
Check for answer 1
Take n=2 a2=45+(2-1)-5=40
Take n=3 n3=45+(3-1)-5=35
Check for answer 2
Take n=2 G2=-3×-2^1=6
Take n=3 G3=-3×-2^2=12
Answer:
![\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997](https://tex.z-dn.net/?f=%5Cmu%20_%7B%5Chat%7Bp%7D%7D%3D%200.275%5C%5C%5C%5C%20%5Csigma_%7B%5Chat%7Bp%7D%7D%3D0.1997)
Step-by-step explanation:
We know that the mean and standard deviation of the sampling distribution of the sample proportion(
) is given by :-
![\mu _{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}](https://tex.z-dn.net/?f=%5Cmu%20_%7B%5Chat%7Bp%7D%7D%3Dp%5C%5C%5C%5C%20%5Csigma_%7B%5Chat%7Bp%7D%7D%3D%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
, where p= Population proportion and n = sample size.
Let p be the proportion of blue chips.
As per given , we have
p= 0.275
n= 5
Now , the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be :
![\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=\sqrt{\dfrac{ 0.275(1- 0.275)}{5}}\\\\=0.19968725547\approx0.1997](https://tex.z-dn.net/?f=%5Cmu%20_%7B%5Chat%7Bp%7D%7D%3D%200.275%5C%5C%5C%5C%20%5Csigma_%7B%5Chat%7Bp%7D%7D%3D%5Csqrt%7B%5Cdfrac%7B%200.275%281-%200.275%29%7D%7B5%7D%7D%5C%5C%5C%5C%3D0.19968725547%5Capprox0.1997)
Hence, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 are :
![\mu _{\hat{p}}= 0.275\\\\ \sigma_{\hat{p}}=0.1997](https://tex.z-dn.net/?f=%5Cmu%20_%7B%5Chat%7Bp%7D%7D%3D%200.275%5C%5C%5C%5C%20%5Csigma_%7B%5Chat%7Bp%7D%7D%3D0.1997)
Answer:
8
Step-by-step explanation: