Question statement: A tobacco company claims that the amount of nicotine in one of its cigarettes is a random variable with mean 2.2 mg and standard deviation 0.3 mg. However, the average nicotine content of 100 randomly chosen cigarettes was 3.1mg. Approximate the probability that the average would have been as high as or higher than 3.1 if the company’s claims were true.
My attempt: Let $X$ be the average amount of nicotine in the cigarettes, measured in mg. If the claim of the company is true, then X has mean $2.2$ and variance $\dfrac{0.3^2}{100}$ or a standard deviation of $\dfrac{0.3}{10}=0.03.$ Thus, the probability requested is $$\mathbb{P}(X>3.1)=\mathbb{P}\left(\dfrac{X-2.2}{0.03}>\dfrac{3.1-2.2}{0.03}\right).$$ Using the Central Limit Theorem, the quantity can be approximated by $$1-\Phi(30) \approx 0.$$
Now, the solution gives the same answer, but the method is different. In particular, they divide by $\sqrt{0.03}$ rather than $0.03$ as I do. But why is this?
Thank you.