# Expected value and average value problem

One has a very large batch of tablets. The weight (unit: gram) of a randomly selected tablet can be accurately considered as a normally distributed random variable X with mean μ and standard deviation $$0.02$$. (Here, μ is the mean weight in the batch.) For weight control, a number of tablets are now taken and weighed. Assume $$μ = 0.65$$.

a) Calculate the probability that the weight of a randomly selected tablet lies outside the interval $$(0.60, 0.70)$$.

b) Calculate the probability that the arithmetic mean X of the weights of 30 randomly selected tablets lies outside $$(0.64, 0.66)$$.

My question: I am a little bit lost at b) look at the answer below. It doesn't feel like they are correct. Why are they using the value of 0.7 from from a). I think it is an error but maybe I didn't understood something so I would appreciate it if someone could explain this to me.

If $$1-\Pr\left\{ 0.64\leq X\leq0.66\right\}$$ $$=1-\Pr\left\{ \dfrac{0.64-\mu}{\sigma/\sqrt{n}}\leq\dfrac{X-\mu}{\sigma/\sqrt{n}}\leq\dfrac{0.70-\mu}{\sigma/\sqrt{n}}\right\},$$ then $$\Pr\left\{ 0.66 However, this cannot be true for $$X\sim\textrm{Normal}\left(\mu,\sigma^{2}\right).$$ Therefore, the value of 0.70 is most likely a typo.