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.

Answer to b) enter image description here

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    $\begingroup$ It is a basic courtesy in this site to accept and upvote answers. It is the only way you reward the user who given the time and effort to answer your question. You have not done so in many of your posts. $\endgroup$ Apr 13 at 12:06
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    $\begingroup$ @Mr.GandalfSauron I always try to upvote as much as I can. It is not that I didn't find it helpful, it is that I usually forget but you are right, the least I can do is to upvote but I wish there was a reminder or something like that. I will try to be better at it, thanks for your input and sorry to everyone that might have felt unappreciated because I really do appreciate yout help. $\endgroup$ Apr 13 at 12:11

1 Answer 1


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 $$\Pr\left\{ 0.66<X\leq0.70\right\} =0.$$ 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.


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