1. **Problem statement:** The weekly sales amount of a skin care product follows a normal distribution with mean $\mu=5000$ and standard deviation $\sigma=400$. We need to find probabilities for sales amounts: (i) More than 6000 (ii) More than 4000 (iii) Between 4800 and 5800 2. **Formula and rules:** For a normal distribution, the probability for a value $X$ is found using the standard normal variable $Z$: $$Z=\frac{X-\mu}{\sigma}$$ We use standard normal tables or a calculator to find probabilities from $Z$ values. 3. **(i) Probability sales > 6000:** Calculate $Z$: $$Z=\frac{6000-5000}{400}=\frac{1000}{400}=2.5$$ Probability $P(X>6000)=P(Z>2.5)$. From standard normal tables, $P(Z>2.5)=1-P(Z\leq2.5)=1-0.9938=0.0062$. 4. **(ii) Probability sales > 4000:** Calculate $Z$: $$Z=\frac{4000-5000}{400}=\frac{-1000}{400}=-2.5$$ Probability $P(X>4000)=P(Z>-2.5)$. Since $P(Z>-2.5)=1-P(Z\leq-2.5)$ and $P(Z\leq-2.5)=0.0062$, then $$P(Z>-2.5)=1-0.0062=0.9938$$ 5. **(iii) Probability sales between 4800 and 5800:** Calculate $Z$ for both limits: $$Z_1=\frac{4800-5000}{400}=\frac{-200}{400}=-0.5$$ $$Z_2=\frac{5800-5000}{400}=\frac{800}{400}=2$$ Probability $P(4800