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probability on how do you do a poisson distributuion

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    The edit seems to have dramatically changed the question; as it is not, it is very unclear what you actually mean. You should take a little time to carefully formulate the question you have and edit this to reflect it. Include as much detail as you can to provide context. – 2017-01-30

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Let's call $X$ the number of items that are purchased in 7 days. With the salesman's assumption, $X\sim\text{Poisson}(14)$. Then what we are looking for is the least number of items $I$ such that $$P(X\leq I)\geq 0.95.$$

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    You should use a table of Poisson probabilities or a computer statistics package to find this, using the formulation above. – 2017-01-29
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You can use a normal approximation in the method of @RideTheWavelet's response (+1) to get close to the answer. Unless you are expected to use software or have extensive Poisson tables in your text, my guess is you are supposed to make such an approximation.

With $X \sim Pois(\lambda = 14 \text{ per week}),$ you have $E(X) = Var(X) = \lambda = 14.$ Then you seek

$$P(X \le x) = P\left(\frac{X - \lambda}{\sqrt{\lambda}} \le \frac{x-14}{\sqrt{14}}\right) \approx P\left(Z \le \frac{x-14}{\sqrt{14}} = 1.645\right) = 0.95,$$ where $Z$ is standard normal and 1.645 is from a printed normal table. Solving for $x,$ one gets about 20, maybe rounding up to 21 to be safe.

Exact computation in R statistical software shows that 20 would suffice:

ppois(20, 14)  # in R 'ppois' is the Poisson CDF ...
## 0.9520916
qpois(.95,14)  # ... and 'qpois' in the inverse CDF or 'quantile' function
## 20

The sketch below shows the Poisson distribution and the density function of the 'best fitting' normal distribution.

enter image description here

Note: As a very rough approximation using the Empirical Rule, about 16% of the probability lies above $\mu + \sigma = 14 + 3.74 = 17.74$ and about 2.5% of the probability lies above $\mu + 2\sigma = 21.5,$ so it seems reasonable that the answer (for 95%) is between those two values.