1. **Stating the problem:** We have a survey where 75% of Information Technology students have eWallets. From a random sample of 150 students, we want to find: i. The probability that at most 70% have eWallets. ii. The probability that at least 35% did not have eWallets. 2. **Understanding the problem:** Let $p=0.75$ be the true proportion of students with eWallets. Sample size $n=150$. 3. **Using the normal approximation to the binomial distribution:** Since $n$ is large, the sampling distribution of the sample proportion $\hat{p}$ is approximately normal with mean $\mu = p = 0.75$ and standard deviation $\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.75 \times 0.25}{150}}$. Calculate $\sigma$: $$\sigma = \sqrt{\frac{0.75 \times 0.25}{150}} = \sqrt{\frac{0.1875}{150}} = \sqrt{0.00125} \approx 0.03536$$ 4. **Part i: Probability at most 70% have eWallets** We want $P(\hat{p} \leq 0.70)$. Calculate the z-score: $$z = \frac{0.70 - 0.75}{0.03536} = \frac{-0.05}{0.03536} \approx -1.414$$ Using standard normal tables or calculator, $P(Z \leq -1.414) \approx 0.0785$. 5. **Part ii: Probability at least 35% did not have eWallets** Since 35% did not have eWallets, the proportion with eWallets is at most 65%. So, $P(\hat{p} \leq 0.65)$. Calculate the z-score: $$z = \frac{0.65 - 0.75}{0.03536} = \frac{-0.10}{0.03536} \approx -2.828$$ Using standard normal tables or calculator, $P(Z \leq -2.828) \approx 0.0023$. **Final answers:** - Probability at most 70% have eWallets is approximately 0.0785. - Probability at least 35% did not have eWallets is approximately 0.0023.