probability and statistics: poisson distribution

Question

probability on how do you do a poisson distributuion

Answer 1

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 $$P(X\leq I)\geq 0.95.$$

Answer 2

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.

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.