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i have a probability problem and i dont know if im using the right way to find a solution to it.

Here the problem :

A person drink in average 490ml of water from a bottle of 500ml and i have a standard deviation of 4 ml. It is specified that this data is normally distributed

So :

$$\mu = 490$$ $$\sigma = 4$$

I want to calculate the probability that he will drink 500 ml of water (the entire bottle) using the mean and the standard deviation as information.

I tried to use the z-score, so i did $$z = \frac{500 - \mu}{\sigma} = 2,5$$ then i can know that there are 0,62% of data over 500 ml but this isn't the probability to drink 500 ml right ? What i'am not understanding ?

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  • $\begingroup$ Your understanding looks superior to that of the originator of the problem: you have discovered that for this question (about the chance of emptying the bottle) it just isn't plausible that the distribution is Normal. $\endgroup$ – whuber Feb 23 at 21:06
  • $\begingroup$ Perhaps they have an extra source of water if they finish the first bottle. $0.62\%$ looks like the answer the question wants you to give $\endgroup$ – Henry Feb 23 at 22:55
  • $\begingroup$ Thanks for your responses, I understand @whuber, it does not make sense to specify that the data is normally distributed if they ask for the probability of only one value, but still my variable is continuous, I think that the problem I received was not correctly formulated and as Henry said they certainly wanted to ask about the probability to drink 500 ml or more. $\endgroup$ – Wolt Feb 24 at 23:20
  • $\begingroup$ If someone's desired amount of water is $X \sim \mathsf{Norm}(\mu=490, \sigma=4),$ and you seek $P(X >500),$ then the answer obtained using R without standardization is computed with code 1 - pnorm(500, 490, 4) which returns 0.006209665, where pnorm is a normal CDF. So it is reasonable that standardizing and using normal tables you will get about $0.0062 = 0.62\%.$ $\endgroup$ – BruceET Feb 25 at 0:02

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