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Answer:
The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:
![CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:
![MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The information provided is:
<em>σ</em> = $60
<em>MOE</em> = $2
The critical value of <em>z</em> for 95% confidence level is:
![z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3Dz_%7B0.05%2F2%7D%3Dz_%7B0.025%7D%3D1.96)
Compute the sample size as follows:
![MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}](https://tex.z-dn.net/?f=n%3D%5B%5Cfrac%7Bz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Csigma%20%7D%7BMOE%7D%5D%5E%7B2%7D)
![=[\frac{1.96\times 60}{2}]^{2}](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B1.96%5Ctimes%2060%7D%7B2%7D%5D%5E%7B2%7D)
![=3457.44\\\approx 3458](https://tex.z-dn.net/?f=%3D3457.44%5C%5C%5Capprox%203458)
Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
Yes because only 1/3 of the days are rainy days, which is about 4.67 days of rain.
Answer:
-3
Step-by-step explanation:
simplify the expression