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The toughness of a metal has a gaussian distribution with avarage 70 and standard deviation of 3. Suppose that we denote by m the average of a poblation of this metal.We want to make sure that our average of metals does not differ from m more than 0.5 with a probability of 98%. How many metals do we have to study?

Thanks for your help I am struggling a lot with this problem.

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    Would you mind adding your work so far on the problem? – 2017-01-26
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    The standard error $E=\frac{\sigma}{\sqrt{n} $ and I think that I have that E=0.5, but I don't want the error to be standard, I want a probability of 98% and the standard error is 68% I believe. – 2017-01-26
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    I think your are confusing _standard error_ and _standard deviation._ In a normal distribution about 68% of the probability is within one _standard deviation_ on either side of the mean. (Sometimes this is called the Empirical Rule.) See my Answer for what I believe you are asking. Please leave a Comment if that is not it. (_Poblation_ of metal is not exactly a statistical term.) – 2017-01-27

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I interpret your question as follows: You have observations $X_1, X_2, \dots, X_n$ randomly sampled from a population distributed as $Norm(\mu = 70, \sigma=3).$ Denote the sample mean by $\bar X,$ which has distribution $Norm(\mu=70,\sigma=3/\sqrt{n}).$ You seek $P(|\bar X - 70| \le 0.5) = .98.$

This amounts to $$P(|\bar X - 70| \le 0.05) = P(69.5 \le \bar X \le 70.5) \\ = P\left(\frac{69.5 - 70}{3/\sqrt{n}} \le Z \le \frac{70.5 - 70}{3/\sqrt{n}} \right) = 0.98,$$

where $Z$ has a standard normal distribution. From printed tables of the standard normal distribution we find that $\frac{70.5 - 70}{3/\sqrt{n}} =2.236.$ Solve for $n$ and round up to the nearest integer.

Note: In your Comment, I believe you may misinterpret the term standard error. The standard error of in this situation is $SD(\bar X) = \sigma/\sqrt{n}.$ You can view this problem in terms of a 98% confidence interval, in which you want the margin of error $2.236\sigma/\sqrt{n} = 0.5.$ (It is common to use 95% confidence intervals, but here you are asked for a 98% CI.)