I believe it's b. Hope this helps
Answer:
I would say that the Answer is 35 I hope this isnt wrong..
Answer: (141.1, 156.48)
Step-by-step explanation:
Given sample statistics : ![n=45](https://tex.z-dn.net/?f=n%3D45)
![\overline{x}=148.79\text{ lb}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D148.79%5Ctext%7B%20lb%7D)
![\sigma=31.37\text{ lb}](https://tex.z-dn.net/?f=%5Csigma%3D31.37%5Ctext%7B%20lb%7D)
a) We know that the best point estimate of the population mean is the sample mean.
Therefore, the best point estimate of the mean weight of all women = ![\mu=148.79\text{ lb}](https://tex.z-dn.net/?f=%5Cmu%3D148.79%5Ctext%7B%20lb%7D)
b) The confidence interval for the population mean is given by :-
, where E is the margin of error.
Formula for Margin of error :-
![z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%5Ctimes%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Given : Significance level : ![\alpha=1-0.90=0.1](https://tex.z-dn.net/?f=%5Calpha%3D1-0.90%3D0.1)
Critical value : ![z_{\alpha/2}=z_{0.05}=\pm1.645](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3Dz_%7B0.05%7D%3D%5Cpm1.645)
Margin of error : ![E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69](https://tex.z-dn.net/?f=E%3D1.645%5Ctimes%5Cdfrac%7B31.37%7D%7B%5Csqrt%7B45%7D%7D%5Capprox%207.69)
Now, the 90% confidence interval for the population mean will be :-
![148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)](https://tex.z-dn.net/?f=148.79%5C%20%5Cpm%5C%207.69%20%3D%28148.79-7.69%5C%20%2C%5C%20148.79%2B7.69%29%3D%28141.1%2C%5C%20156.48%29)
Hence, the 90% confidence interval estimate of the mean weight of all women= (141.1, 156.48)
4/5 / 2/5
= 4/5 * 5/2
= 4/2
= 2
Answer: 2 pages per hour