Answer:
(1) 0.058
(2) 601
Step-by-step explanation:
(1)
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:

The standard deviation of this sampling distribution of sample proportion is:

As the sample size is large, i.e. <em>n</em> = 350 > 30, the Central limit theorem can be used to approximate the sampling distribution of sample proportion of adults in a certain country work from home.
Compute the probability that fewer than 42 out of a random sample of 350 adults will work from home:
Sample proportion: 

Thus, the probability that fewer than 42 out of a random sample of 350 adults will work from home is 0.058.
(2)
The (1 - <em>α</em>)% confidence interval for population proportion is:

The margin of error for this interval is:

Given:
MOE = 0.04
Confidence level = 95%
Assume that the sample proportion is 50%.
The critical <em>z</em>-value for 95% confidence level is 1.96.
Compute the required sample size as follows:

![n=[\frac{z_{\alpha/2}\cdot\sqrt{\hat p(1-\hat p)}}{MOE}]^{2}\\\\=[\frac{1.96\times \sqrt{0.50(1-0.50)}}{0.04}]^{2}\\\\=600.25\\\\\approx 601](https://tex.z-dn.net/?f=n%3D%5B%5Cfrac%7Bz_%7B%5Calpha%2F2%7D%5Ccdot%5Csqrt%7B%5Chat%20p%281-%5Chat%20p%29%7D%7D%7BMOE%7D%5D%5E%7B2%7D%5C%5C%5C%5C%3D%5B%5Cfrac%7B1.96%5Ctimes%20%5Csqrt%7B0.50%281-0.50%29%7D%7D%7B0.04%7D%5D%5E%7B2%7D%5C%5C%5C%5C%3D600.25%5C%5C%5C%5C%5Capprox%20601)
Thus, the required sample size is 601.